Transcript
Contents
MBA
Test Prep
1.
Explanations: Fundamentals of Logical Reasoning and Data Interpretation
1
2.
Explanations: Fundamentals of Ratio & Proportion
71
3.
Explanations: Fundamentals of Time, Speed & Distance
93
4.
Explanations: Fundamentals of Grammar
119
Explanations: Fundamentals of Logical Reasoning & Data
Interpretation
Data Interpretation
150 − 45
× 100 = 233.3%
45
3. b
Growth rate of HLL =
4. e
Sale of P & G in 2002 = 25% of 150.
Sale of P & G in 2003 = 20% of 375.
Practice exercise – A1
From the pie charts, the sales figures in rupees (in crores)
are:
Ratio of sales in 2002 : 2003 =
Company
2002
2003
HLL
45
150
P&G
37.5
75
Henkel
22.5
75
Nirma
30
56.25
Others
15
18.75
Above table has been made to explain the things. You need
not make it. Also, in pie-charts of this sorts, it is very
advantageous to identify the ratio and percentage increase
of total size of pie in the year after year. In this case the ratio
of the total size of market is 6 : 15 i.e. 2 : 5
1. b
375 − 150
× 100 = 150%
150
Alternately, since we know the ratio of the pies
is 2 : 5, one just needs to do the following 25% of
2 : 20% of 5 i.e. 50 : 100
5. a
Detergent market grows by 10% of 150
= Rs. 15 crore.
6. a
It is equivalent to that of the whole market = 10%.
7. a
10% annual growth for 2 years means 21% growth
ab
).
100
Sales in 2005 = 1.21 × 375 = Rs. 453.75 crore.
(using the formula, a + b +
8. d
Since one would already have found the ratio of
the pies, before starting, as 2 : 5, one could also
Growth in sales of P & G = (0.2 × 375 – 0.25 × 150)
= Rs. 37.5 crore.
HLL = (0.4 × 375 – 0.25 × 150) = Rs.112.5 crore.
Henkel = (0.2 × 375 – 0.15 × 150) = Rs.52.5 crore.
Others = (0.05 × 375 – 0.1 × 150) = Rs.3.75 crore.
Costs of HLL in 2002 = 30 × 1.1 = Rs. 33 crore.
Profit = 45 – 33 = Rs. 12 crore.
Profit percentage =
5−2 3
= = 150%
have mentally just calculated
2
2
2. e
9. a
Explanations: Fundamentals of
Logical Reasoning & Data Interpretation
12
× 100 = 36.4%
33
Margin =
Pr ofit
12
× 100 =
× 100 = 26.67%
Sales
45
Henkel =
75 − 22.5
× 100 = 233.3%
22.5
P&G=
Alternative method:
Since growth in total sales has been 150% from
2002 to 2003.
So we can change the pie chart of 2003 as
HLL = 40 × 2.5 = 100%
P & G = 20 × 2.5 = 50%
Henkel = 20 × 2.5 = 50%
Nirma = 15 × 2.5 = 37.5%
Other = 5 × 2.5 = 12.5%
These percentages are of market size in 2002, i.e.
Rs. 150 crore.
Now, visually, one can figure out the minimum growth
in sales is in others.
0.25 × 150 1
= .
0.2 × 375 2
75 − 37.5
× 100 = 100%
37.5
⇒ Difference = 133.33%
Alternative method:
If we keep base for 2002 and 2003, then (As shown
in the solution of question 2)
Henkel =
50 − 15 7
= × 100 = 233%
15
3
P&G=
50 − 25
= 100%
25
MBA
Test Prep
Page: 1
10. c
The ratio will not change because if the total sales
are doubled, the sales of Nirma will also double for
both the years.
56.25
15
=
.
30
8
Again one could have used the ratio of pie 2 : 5
effectively and answer would have been 15% of
5 : 20% of 2 i.e. 75 : 40 i.e. 15 : 8
Hence, the ratio will be
11. c
Surf Excel = 30% of HLL.
HLL = 30% of detergent market.
Surf Excel = 30% of (30% of detergent market)
= 9% of detergent market.
12. c
Profit = 25%. Hence, margin = 20%.
Profit amount = 20% of sales = 0.2 × 22.5
= Rs. 4.5 crore.
13. a
Since all the companies have the same expenditure,
the company with the maximum sales will have the
highest profit as well as profit percentage.
14. d
Sale
Cost =
× 100
100 + Pr ofit percentage
150 × 15% 22.5
= 18 crore.
=
1.25
1.25
Henkel’s cost in 2003 = 19.8.
Henkel’s sale in 2003 = 75.
Cost of Henkel in 2002 =
Profit percentage =
15. c
change in any year with respect to the base year. An index
cannot give the actual value for any year unless the actual
value and the index of any one year is given to us.
e.g. if, for the given graph, actual sales in 1994 (sales index
120) is Rs. 144 crore, then we can say that:
Sales index of 120 = Sales of 144 crore.
144
Sales index of 1 = Sales of
= Rs. 1.2 crore.
120
Sales index of 100 = Sales of Rs. 120 crore.
Thus, we now have the conversion factors, i.e. to convert
sales index to sales : Multiply index with 1.2 and to convert
sales into sales index : Divide sales by 1.2.
Thus, now we can determine the index of sales for any
year if the sales (rupees in crores) is known and also the
sales (rupees in crores) of any year if the sales index is
known.
In the solutions below the abbreviations being used are:
SV = Sales value
CV = Cost value
PV = Profit value
1. e
Index does not give the actual value. The profit
index is 110.
2. a
Sales in 1993 = Rs. 500 crore = Index of 100.
In 1998, SI = 154, which is 54% above the index of
1993. Thus, sales of 1998 should be 54% more
than the sales in 1993.
Thus, sales in 1998 = 500 × 1.54 = Rs. 770 crore
OR
Index of 100 = Sales of Rs. 500 crore.
Index of 1 = Sales of Rs. 5 crore.
Index of 154 = Sales of 154 × 5 = Rs. 770 crore.
3. c
In 1996, CI = 110 and CV = Rs. 550.
55
× 100 = 278%.
19.8
S − 0.8S
× 100 = 25% .
0.8S
If both sales and costs increase by 10%, then sales
will be 1.1S and cost will be 1.1(0.8S) and hence
the profit percentage will remain at 25%.
P & G’s profit in 2003 =
In 2002, CI = 190. Thus, CV =
Practice exercise – A2
Index is a representation of actual value. A year is designated
as the base year and the index for the base year is taken as
100, e.g. in the given graph for sales, cost and profit, the
base year is 1993 (index = 100). For 1994, the sales index
is 120, which means that sales in 1994 is 20% more than
the sales in 1993 and sales in 1995 (index 131) is 31% more
than in the base year (1993) sales. Similarly, the cost index
of 98 in 1994 indicates that the cost has gone down by 2%
over the base year of 1993 and in 1996 the cost has gone
up by 10% over cost in 1993. (Index = 110)
SI = Sales index
CI = Cost index
PI = Profit index
550
× 190 = Rs. 950 .
110
4. e
It is not possible to find the value across indices.
With one value of cost we can find out all the values
of cost but none of the values of sales or profits.
5. e
Gross profit = (Sales – Cost), since we do not have
SV and CV for various years, the gross profit cannot
be calculated.
6. b
In 2003, PI = 130 and PV = Rs. 780.
780
× 110 = Rs. 660 .
In 2001, PI = 110. Thus, PV =
130
Index is used to compare data over a large number of years
and to determine trends. Because the base year value is
always taken as 100, it is easy to determine the percentage
Page: 2
MBA
Test Prep
Solution Book-2
7. a
In 1993, SI = CI = 100,SV = Rs. 300 and
CV = Rs.120. Thus, SV in 1998 = 154 × 3 = Rs. 462
and CV in 1998 = 153 × 1.2 = Rs. 183.6.
Gross profit = 462 – 183.6 = Rs. 278.4.
In the given question, we cannot determine the profit
because it is defined as (Sales – Cost – Tax) and
we do not have the tax for 1998.
8. b
In 1993, SI = CI = 100,SV = Rs. 300 and
CV = Rs.120. Thus, SV in 2003 = 188 × 3 = Rs. 564
and CV in 2003 = 166 × 1.2 = Rs. 199.2.
Gross profit = 564 – 199.2 = Rs. 364.8.
9. c
In 1993, SI = 100 and SV = Rs. 700.
Thus, SV in 1998 = 154 × 7 = Rs. 1078.
In 1996, CI = 110 and CV = 600.
600
= 834.54.
110
Gross profit = 1078 – 834.54 = Rs. 243.45.
Practice exercise – A3
1. e
We need the total number of students opting for
finance in three institutes to arrive at the average.
2. c
Average salary at IIML
(60 × 550000) + (40 × 420000) + (50 × 725000)
150
= 574000.
Rather than multiplying by 60,40 and 50 you can
also multiply by 6,4,5 and then divide by 6 + 4 + 5
= 15.
=
3. a
Calculating as above, average salary at IIMK
= 75900.
Average salary of IIMK – IIML = 759000 – 574000
= 185000.
4. b
Average salary at IIMB
Thus, CV in 1998 = 153 ×
10. b
100
Thus, new SI in 1993 =
× 100 = 83.33.
120
11. a
12. b
Only 3 years — 2000, 2001 and 2002 — have their
CI more than the CI for 2003. So for these years,
CI will be more than 100.
13. b
Total Sales Value for the period 1993-98
= 1200 + (1.2 × 1200) + (1.31 × 1200) +
(1.52 × 1200) + (1.62 × 1200) + (1.54 × 1200)
= Rs. 9828
1200
or (Sum of SI from 1993 to 1998) ×
.
100
Sales Value for the period 1993-98 = 9828.
(Refer question 13)
Total Cost Value for the period 1993-98
800
= (100 + 98 + 95 + 110 + 121 + 153) ×
100
= 5416.
Difference = 9828 – 5416 = Rs. 4412.
15. c
5. d
The average salary for finance in IIMB is significantly
lower than that for systems. Since the number in
finance has gone up, the average salary will be
lower than what we calculated for question 4.
Thus, the answer is 626. (Check by calculation)
6. c
Now since the number in systems has increased,
the average salary will be more than the average
salary in question 5 by a large margin.
So the answer is 667.
7. c
Average salary for finance in IIMA, IIMB and IIMC
Current PI in 1996 = 125 and PI in 2003 = 130.
New PI in 1996 = 100.
100
Thus, new PI in 2003 =
× 130 = 104.
125
14. a
(50 × 540000) + (45 × 453000) + (110 × 775000)
205
= 647000.
=
Current SI in 1993 = 100 and SI in 1994 = 120.
New SI in 1994 = 120.
1537
= 139.72
11
Average of Sales Index – Average of Cost Index
= 147.27 – 139.72 = 7.54
Average of Cost Index =
Explanations: Fundamentals of
Logical Reasoning & Data Interpretation
(540000 × 50) + (520000 × 90) + (725000 × 45)
185
= 624000
=
8. b
Average salary of the remaining students of
systems =
(500 × 850000 – 1000000 × 200)
300
= 750000.
Instead of the above calculation, one could also
have done the following mentally : Of 5 students
with average salary 850000, 2 are in dotcom
companies with average 1000000. Thus they
contribute 300000 more than overall average and
thus on an average each of the 3 non-dotcom
students would have a salary of (850000 – 100000)
= 750000.
MBA
Test Prep
Page: 3
9. b
Average salary of IIMA finance students placed in
(1 × 620 – 0.6 × 800)
= 350.
0.4
Alternately, as explained in the oral calculations
for solutions to 17, in this question 60% could have
been taken as the fraction 3/5 and thus 5 students
have average of 620. Of these 3 have average of
800 and thus contribute a total of 540 more to the
sum than the average. Thus the remaining 2 students
on an average has 620 – 270 = 350
6. a
Rephrase the question as, “Which crop has the
highest yield over the years?” Answer is barley.
7. b
Percentage increase in production of maize in
2000-01 = (9.99 – 8.06) × 100/8.06 = 24%.
Production in 2002-03 = 9.99 × (1.24)2 = 15.36 Mn T.
Yield = Production/Area.
Thus, yield = 15.36 × 1000/6.1 = 2518 kg/Ha.
India =
10. d
Alternative method:
We can approximate, though some risk is involved.
But whenever you approximate, you need to
understand how much error you are introducing and
whether your actual answer would be more than or
less than the calculated answer.
Percentage increase in production of maize in
Rather than taking number as 150 and 50, to reduce
calculations, we can take their ratio 3 : 1.
Hence, average =
(3 × 453 + 1 × 575)
= 483.
4
10 − 8
× 100 = 25%.
8
Production in 2002-03 = 10 × (1.25)2 = 10 × 1.5625
= 15.62 Mn T.
But actual answer would be less than this.
Let us say → 15.30 Mn T
2000-01 =
Practice exercise –A4
1. a
Land used for paddy in 2001-02 = 12.71 Mn Ha.
Total land used in 2001-02 = 29.93 Mn Ha.
Percentage of land used for paddy
=
or
2. c
12.71
× 100 = 42.5%
29.93
13
× 100 ~
− 43.3% approximately.
30
Productivity = Production/Area = Yield.
Yield of ragi in 2002-03 = 1327 kg/Ha.
Yield of ragi in 2001-02 = 1378 kg/Ha.
Percentage change in yield
1378 − 1327
× 100 ; 3.72%
1378
Calculation technique: One needs to calculate
51/1348. 10% of 1378 is 137.8 and hence 5% is
68.9. Thus 51 will be less than 5% and only one
option is left.
Now yield =
8. c
It can be visually figured out that the year would be
2000-01 therefore total production would be 35.52
Mn T.
9. a
In the year 2000-01, the production should be
2.53 + 2.58 i.e. the production should be a little more
than double the original production. Thus yield also
should be a little more than double of the given yield
i.e. a little more than 2 × 1329 = 2658. Only one
option is in that range.
10. d
Production of paddy in 2001-02 = 1602 × 12.71
= 20.36 Mn Ha.
Production of paddy in 2000-01 = 12.81 Mn T.
Increase in production
=
3. b
4. c
5. d
Barley. We do not need to solve it. Mere inspection
reveals the answer.
Production = Area × Yield.
If yield increases by 10% and the area remains the
same, then the production will also go up by 10%.
Hence production in 2003-04
= Production in 2002-03 × 1.1 = 2.43 × 1.1
= 2.43 + 0.243 = 2.67 Mn T.
Production = Area × Yield.
Production = 521 × 12.71/1000 = 6.62 Mn T.
1
.
Alternatively, we know 12.5% =
8
15.3 × 1000
= 2,508 kg/ma.
6.1
=
20.36 − 12.81
× 100 = 59% .
12.81
Alternatively, we can approximate as well.
Production of paddy in 2001-02 = 1600 × 12.71
100
1
= 20 Mn Ha as = 12.5
8
8 × 100
Production of paddy in 2000-01
= 12.81 Mn Ha ~ 12.5 Mn Ha.
= 1600 ×
Increase in production =
20 − 12.5
× 100 ~ 60%.
12.5
1× 100
= 6.5 Mn T.
8 × 1000
It would be higher than 6.5 Mn T as 12.717 > 12.5
So 521 ×
Page: 4
MBA
Test Prep
Solution Book-2
11. c
12. e
13. c
In such questions use options. Start with an intelligent
choice of option e.g. as maize appears in 3 of the
four choices, check trend for maize. As trend for
maize does not satisfy the required condition, safely
mark the fourth option i.e. (c).
The given data pertains to production and not sales.
We cannot assume that 100% of paddy produced,
was sold. Thus, data is inadequate to solve the
given problem.
Cost of production of barley = Rs. 1,250 per quintal.
If there is no loss of produce, then selling price for
25% profit = 1250 × 1.25 = Rs. 1562.5.
Loss in produce is 10%.
Thus, for every 100 kg produced, only 90 kg can be
sold. So for 25% profit, SP = (1562.5/0.9)
= Rs. 1,736 per quintal.
14. b
By rule of alligation, ratio of quantity available to
PDS to open market = 2 : 5.
Therefore, 2/7th or 28.57% is allocated for PDS.
15. d
Yield required = 1518 kg/Ha.
Amount of production = 8.06 Mn T.
8.06 × 1000
= 5.31 Mn Ha.
1518
Land should be less by (5.86 – 5.31) = 0.55 Mn Ha.
5. c
The number accounted for by Hyundai and Daewoo
= (10 + 25)% of 4 lakh = 35% of 4 lakh = 1.4 lakh.
6. a
Number of Santros = 75% of 25% of 4 lakh = 75000.
7. d
Number of Santros = 75000 (from question 6).
Number of Zen = 20% of 40% of 4 lakh = 32000.
Difference between Santro and Zen sales = 43000.
8. c
Sales of Maruti 800 in Mumbai
= 60% of 40% of 5% of total India sales
= 0.012 × 24 lakh = 28800.
9. b
Sales in 2004 = 1.15 × 4 lakh = 4.6 lakh.
Thus, HM sales = 15% of 4.6 lakh = 69000.
10. c
Maruti sales in 2003 = 40% of 4 lakh = 1.6 lakh.
Increase in Maruti sale in 2004 = 20% of 1.6 lakh
= 0.32 lakh.
Thus, Maruti sales in = 1.6 + 0.32 = 1.92 lakh and
total sales = 4 + 0.32 = 4.32 lakh.
Percentage share of Maruti = (1.92/4.32) × 100
= 44.44%.
11. d
Hyundai’s sales in 2003 = 25% of 4 lakh = 1 lakh.
Hyundai’s sales in 2004 = 90% of 1 lakh = 0.9 lakh.
Total sales in 2004 = 130% of 4 lakhs = 5.2 lakh.
Hyundai’s share of sales = (0.9/5.2) × 100 = 17.3%.
12. a
Total sales in 2004 = 5.2 lakh. (Refer question 11)
Daewoo’s sales in 2003 = 10% of 4 lakh = 0.4 lakh.
Daewoo’s share in 2004 = (0.4/5.2) × 100 = 7.7%.
13. b
Sales of Maruti in 2003 = 40% of 4 lakh = 1.6 lakh.
Increase in sales of Esteem in 2004 = 40000.
Thus, sales of Maruti in 2004 = 2 lakhs.
Total sales in 2004 = 1.2 × 4 lakh = 4.8 lakh.
Percentage share of Maruti in 2004
= (2/4.8) × 100 = 41.67%.
14. c
Sales of Esteem in 2003 = 16000.
Sales of Esteem in 2004 = 40000 + 16000 = 56000.
Sales of Maruti in 2004 = 2 lakh. (Refer question 13)
Esteem’s share in Maruti sales = (0.56/2) × 100
= 28%.
15. d
Sales of Opel in 2003 = 25% of 10% of 4 lakh
= 10000.
Thus, land required =
Practice exercise – A5
1. b
Number of Esteems = 10% of 40% of 2.5 lakh
= 4% of 2.5 lakh = 0.1 lakh.
2. c
60% of 40% of total car sales = 48000, i.e. 24% of
total sales = 48000.
Thus, total sales = 48000/0.24 = 2 lakh.
3. a
HM market share = 15%.
Zen market share = 20% of 40% = 8%.
Therefore, difference = (15 – 8)% = 7%.
= 7% of 2.5 lakh = 17500.
4. c
Number of Maruti cars sold = 40% of 2 lakh = 80000.
Sale of Maruti 800 = 48000.
Increase of Maruti 800 sold = 25% of 48000 = 12000.
New total sales of Maruti = 92000 = 0.92 lakh.
New total car sales = 200000 + 12000 = 2.12 lakh.
0.92
× 100 = 43.4% .
2.12
Thus, percentage points by which Maruti share has
increased = 43.4 – 40 = 3.4 points.
Percentage share of Maruti =
Explanations: Fundamentals of
Logical Reasoning & Data Interpretation
MBA
Test Prep
Page: 5
Practice exercise – A6
7. b
We are making table for convenience sake. You can avoid
making it. The following figures, based on the graph, have
been used for the calculations.
Year
Shareholders
wealth
Total
debt
Net
fixed
assets
Net
working
capital
1999
125
410
180
350
2000
280
350
250
375
2001
475
430
335
550
2002
750
675
550
745
2003
1150
800
760
100
SW growth rate in 2003 =
745 – 550
= 35.45%.
550
1150 – 750
= 53.33%.
750
Thus, NWC is less than SW by
53.33 –35.45
3.33
= 33%.
Alternatively, you can approximate, since choices
are far apart.
NWC growth rate in 2002 ≈
750 − 550 200 4
=
=
550
550 11
1
= 36.36% = 9.09%
11
Following abbreviations have been used.
SW = Shareholder’s wealth
NFA = Net fixed assets
NWC growth rate in 2002 =
SW growth rate in 2003 ≈
TD = Total debt
NWC = Net working capital
1150 − 750 400 8
=
=
750
750 15
= 53%.
1. a
2. d
Average of the total debt is 556.
Hence, the total debt is above the average for
2 years (2002 and 2003).
NWC in 1999 = 350, NWC in 2003 = 1000.
If NWC in 1999 = 100, then
NWC in 2003 =
3. d
1000
× 100 = 285.6
350
NFA growth rate in 2000 =
NWC is less than SW by =
8. e
Growth rate of NWC is less than that of SW for all
the years.
9. c
Growth rate of:
NFA in 2000-2002 = 120% and TD in 2000-2002
= 92%.
SW in 1999-2000 = 124% and SW in 2002-2003
= 53%.
TD in 2001-2002 = 56% and NWC in 2001-2002
= 35%.
TD in 2002-2003 = 23% and SW 2001-2002 = 58%.
10. a
While doing this visually, the only doubt is whether
for the year 2001 the NWC is above average or not.
Assume the average as the value in 2001 and use
the method of deviation. If sum of deviation is +,
average is above the value of 2001 or else other
way round. In this case the sum of deviation would
be –200 – 175 + 225 + 450 which will obviously be
positive. Thus average will be greater than that for
year 2001. Hence, (a) is correct.
250 – 180
= 38.9%.
180
550 – 335
= 64.2%.
335
Hence, the difference is 25.3%.
NFA growth rate in 2002 =
4. c
Total growth = 1150 – 125 = 1025.
1025
= 256.
4
Thus, SW in 2004 = SW in 2003 + 256
= 1150 + 256 = 1406.
Thus, average annual growth =
5. e
53 − 36 17 1
=
≈ = 33.3%.
53
53 3
Average annual growth rate
1150 –125
× 100 = 205%
125
(Refer question 4)
Growth rate is less than 205% for all the choices.
6. d
The word compounded has no relevance and the
item which has grown the maximum will have the
highest CAGR and average annual growth rate.
So shareholder’s wealth has grown the maximum.
Page: 6
MBA
Test Prep
Solution Book-2
Practice exercise – A7
8. c
If initially the amount of electricity = 1, then at the
end of period amount of electricity
= 1.085 × 1.08 × 1.04 × 1.065 × 1.063 = 1.3796 or
electricity will grow by 37.96%
If growth rate of electricity in 2000-01 is replaced
with growth rate of mining 1999-2000 (9.5%), then
electricity = 1.085 × 1.08 × 1.095 × 1.065 × 1.063
= 1.4526
or electricity will grow by 45.26%
Hence, difference = 45.26 – 37.96 = 7.3%
9. d
The highest total production would be in 2001-02,
because in the last year 2002-03, there is a negative
growth and thus is less than the figure in 2001-02.
10. c
The highest growth rate for the various years are:
1998-99-10%, 1999-2000-14%, 2000-01-7.5%
2001-02-6.5% and 2002-03-6.3%
Thus, figure at the end of the period
= 1.1 × 1.14 × 1.075 × 1.065 × 1.063 = 1.526
Hence, overall growth = 52.6%
For questions 1 to 10:
Again we are making the table, but you can avoid that.
The following figures, based on the graph, have been used
for the calculations. The table gives the percentage change
over the previous year or the growth rates.
Year
General
Mining
1998-99
8.5
7
Mfg. Elect.
8.5
8.5
FD GDP
10
4.2
1999-2000
13
9.5
14
8
8
5
2000-01
5.5
-2
6.5
4
7.5
4
2001-02
6.5
6
6.5
6.5
5.5
6
2002-03
4
-2
4.5
6.3
5.8
5
1. b
Growth rate of manufacturing in 2001-02 = 6.5%.
Thus, growth in manufacturing in 2001-02 = 6.5%
of 80 = 5.2
Mining data is redundant over here.
2. d
Average of GDP growth rate
=
3. b
4. e
5. c
4.2 + 5 + 4 + 6 + 5.5
= 5%
5
Add the growth rate of 5 years for all the items, this
sum will give the relative ranking of the growth
rates. This sum for mining and electricity is less
than that of fiscal deficit.
In question of this type where one has to eliminate
using options, choose which sector to start with
very intelligently. Choose that one which appears in
two options and thus with just one calculations,
atleast you can shortlist your choice to two options.
Practice exercise – A8
We have made the table for convenience sake, but you
should not make it.
Let production of :
A in the various years be given by A2000, A01, A02 and A03
B in the various years be given by B2000, B01, B02 and B03
C in the various years be given by C2000, C01, C02 and C03
D in the various years be given by D2000, D01, D02 and D03
The following table, based on the graph, gives the values of
the percentage share of production for A, B, C and D for the
four years
Year
A
B
C
D
The chart gives us the percentage change from the
previous year. With the GDP given, Fiscal deficit
can not be calculated because the data does not
give the relation between GDP and fiscal deficit.
2000
23%
37%
20%
20%
2001
23%
35%
22%
20%
2002
20%
30%
30%
20%
Add the percentage change for the four items from
1999-2000 to 2002-03, the item which has the
highest sum will end up with the highest figure.
2003
10%
30%
20%
40%
6. e
Cannot be determined because the data gives us
the percentage change only and we do not have
the actual values for any of the years.
7. b
GDP 1999-2000 = 3,09,000 × 1.05
= Rs. 3,24,450 crore
Fiscal deficit 1999-2000 = 11,200 × 1.10 × 1.08
= Rs. 13,305 crore
Fiscal deficit as percentage of GDP in 1999-2000
=
13305
13
=
= 4.1%
324
324450
Explanations: Fundamentals of
Logical Reasoning & Data Interpretation
Let total production in 2001 = p, thus total production in
2003 = 1.21p
Given C03 – A01 = 1320 thus 0.2 x 1.21p – 0.23p = 1320
thus p = 110,000 MT, and total production in MT in
2000 = 1,00,000
2001 = 1,10,000
2002 = 1,21,000
2003 = 1,33,100
1. a
C2000 = 20% of 1,00,000 = 20,000 MT.
C02 = 30% of 1,21,000 = 36,300 MT
Thus, 36300 = 20000(1 + r /100)2 ⇒ r = 35%
One could also have approximated it as follows:
The production increases from 20k to 36k i.e. a
increase of 80%. Thus CAGR has to be less than
40% and once could check with 35%
MBA
Test Prep
Page: 7
2. e
Given data pertains to production and not sales.
3. a
Product B
4. d
Cumulative production of B will still be the highest
because the ratio of production for the various
products for the various years is constant.
5. e
Cannot be determined because the price of A is not
known for the four years.
6. c
The total revenue from selling B and D
= 3k × 150 + 2k × 120 = 690k.
Thus total cost in producing B and D
690k
= 657k
1.05
Cost in producing D = 2k × 100 = 200k.
=
{Q cos t of one unit =
20% × Total03 – 23% × Total01 = 1320
20% × 0.95 × 1.1 × Total01 – 23% × Total01 = 1320
20% × 1.045 × Total01 – 23% × Total01 = 1320
This gives a negative production figure for 2001
and hence data is inconsistent.
10. e
Practice exercise – A9
1. b
Percentage change in the average price for
2000 = 50%
2001 = 16.66%
2002 = 14.28%
2003 = 37.5%
One does not need to calculate this.
One can just check it visually. The changes are 1,
0.5, 0.5, –1.5 and the base are 2, 3, 3.5, 4.5. The
year where change is maximum and base is least
will have the highest growth rate.
2. b
Production of Maa-roti in 2000 = 30,000 and
Production of Maa-roti in 2003 = 60,000
Average annual growth rate for 2000-2003
Rs.120
= Rs.100 }
1.20
⇒ Cost in producing B = 457k
457
= 152.3
3
2.3
×100 = 1.53%
⇒ Thus loss % on selling B =
150
⇒ CP of 1 unit of B =
7. b
Remember that in this question you cannot use the
total production in 2003 as 133,100 because this
was arrived at when there is a 10% increase in
production in each year. So all calculations have to
be done again for this question.
Total production of B in 4 years
= 37000 + 38500 + 36300 + 39930 = 1,50,630
Total sale of B in 4 years
= 37000 + 38500 + 36300 + 39930 × 0.5 = 1,31765
Sales Value of B = 131765 × 400 = Rs. 5,27,6000
Total cost of B = 100 + 150630 × 200 = Rs. 30126100
Profit = 74.9%
=
60,000 − 30,000
1
× 100 × = 33.3%
30,000
3
C03 – A01 = 1320
20% × Total03 – 23% × Total01 = 1320
3. b
40 × 2 + 30 × 3 + 50 × 3.5 + 40 × 4 + 60 × 2.5
220
= Rs. 2.97 lakh
20% × 1.5625 × Total01 – 23% × Total01 = 1320
=
(0.3125 – 0.23) × Total01 = 1320
Total01 = 16,000
Total production in 2003 = 16,000 × 1.5625 = 25000
Thus required difference = 10% of 25000 = 2500
8. c
9. e
All this problems asks is that what is the value of
x% such that -5% and x% successively amounts
to 21%. Thus 0.95 × k = 1.21
k = 1.2736
Thus growth in 2002 should be 27.36%
Though data of growth rate given originally is
changed, but since this question has nothing to do
with actual production amounts and we do not have
to use 1320 MT, the question is easy.
This is very similar to question number 7. The original
data of 10% increase every year is changed and
hence we cannot use the production figures found
earlier. With the new growth rates we have,
C03 – A01 = 1320
Page: 8
Average price per car
4. e
Cannot be determined because total cars produced
by the car industry is not known for any year.
5. b
Production of Maa-roti in 2001 = 50,000
Percentage share of Maa-roti =
50,000
× 100
200,000
= 25%
6. a
Production 2001 = Production 1996 × 1.08 × 1.15,
Production in 1999 =
200000
= 161030
1.08 × 1.15
7. e
The assumption that production = sales is not valid
unless it is specifically stated.
8. a
Cars sold in 2000 = 80% of 30,000 = 24,000
Price per car in 2000 = Rs. 3 lakh
Revenue in 2000 = 24,000 × 3 lakh = 72,000 lakh
MBA
Test Prep
Solution Book-2
9. c
10. c
Revenue in 2000 = 72,000 lakh (Refer question 8)
In 2001, the entire production of 2001 and 20%
2000 production is sold = 56,000
Revenue in 2001 = 56000 × 3.5 = 196000 lakh
Revenue 2000 + 2001 = 2,68,000 lakh
Revenue for each year is given below:
1999 : 40 × 2 = Rs. 80,000 lakh
2000 : 30 × 3 = Rs. 90,000 lakh
2001 : 50 × 3.5 = Rs.1,75,000 lakh
2002 : 40 × 4 = Rs.1,60,000 lakh
2003 : 60 × 2.5 = Rs.1,50,000 lakh
Thus, in 2001 Maa-roti has the highest growth.
11. a
2001. Calculate from question 10.
12. e
Again, we cannot assume that production = sales.
13. c
Average price per car for Maa-roti = Rs. 4 lakh
Average price per car of car industry = Rs. 4.5 lakh
Market share of Maa-roti = 62.5%
Let average price of rest of the car industry = p,
then 0.625 × 4 + 0.375 p = 4.5
Thus, p = Rs. 5.33 lakh
Recollect that in case of averages, oral calculations
are far better as follows : 62.5% is 5/8 and thus 8
items have an average price of 4.5 and 5 of them
have an average of 4. Thus these 5 were “given” a
total of 0.5 × 5 = 2.5 by rest 3 and thus rest 3 will
have an average of 4.5 +
14. e
FC = Fixed cost
SP = Selling price
2.5
= 4.5 + 0.83 = 5.33
3
VC = Variable cost/unit
N = Number of units sold
Break Even Point = BEP is the production level at
which the company is in a no profit or no loss
situation.
BEP = FC / (SP – VC), since the question does not
give us all the required data, we cannot determine
the BEP.
15. c
16. b
17. e
18. c
40,000 cars are produced in 1999
If 10% gets rejected, net production = 36,000
Turnover = 36,000 × 2 lakh = 72,000 lakh
= 720 crore
Number of Maa-roti Omni produced in 2001
= 15% of 50,000 = 7500
Since we do not have the average price per car for
Maa-roti 800 for the two years, we cannot determine
the ratio of sales revenue.
19. b
Maa-roti Esteem
Production 2001 = 20% of 50,000 = 10,000
Production 2002 = 22% of 40,000 = 8,800
Growth rate =
20. a
8800 − 10000
× 100 = −12%
10000
Refer to question 10 for data, simple annual growth
rate
=
150000 − 80000
1
× 100 × = 21.8%
80000
4
21. a
Refer to question 9 for data. For CAGR apply formula
for CI, A = P (1 + r/100)n
A = 1,50,000 lakh, P = 80,000 lakh and n = 4 time
periods, 150 = 80 (1 + r/100)4 thus r = 17%
22. e
Only the average price per car is given. The number
of cars produced is not given thus the average
price per car for 2001-2004 cannot be determined.
23. a
In 1999 Maa-roti produced 40,000 cars, which is
20% more than in 1998, thus production in 1998
=
40000
= 33333
1.2
Practice Exercise – A10
The following figures, based on the graph, have been used
for the calculations. Abbreviations used are :
Int. = Interest
Depn. = Depreciation
NP = Net Profit
OI = Other Income
Year
OPBDIT
Int.
D ep n .
NP
Tax
OI
1999
110
30
5
75
0
1
2000
285
80
20
165
20
2
2001
395
80
50
205
60
4
2002
520
110
80
285
85
10
2003
380
145
120
10
105
6
1. d
Operating profit after interest and taxes but before
depreciation = Net profit + Depreciation, this is
highest for 2002.
2. c
Refer to the table above, tax and depreciation have
witnessed a constant growth across the years.
Ratio of Maa-roti Gypsy produced in 2002 to 2001
5% of 40,000
200
= 10% of 50,000 = 500 = 2 : 5
Explanations: Fundamentals of
Logical Reasoning & Data Interpretation
MBA
Test Prep
Page: 9
3. b
4. b
NP has grown from 75 in 1999 to 165 in 2000,
hence percentage increase in NP for 1999-2000
= 120%.
OPBDIT has grown from 395 in 2001 to 560 in 2002,
hence percentage increase in OPBDIT for
2001-2002 = 41.8%
Thus, difference = 120 – 41.8 = 78.2%.
This set is DI type reasoning set, where the reasoning is
primarily used in solving individual questions.
1. d
Total bill amount:
(150 × 1 + 50 × 3.5 + 25 × 7 + 14 × 5 + 15 × 2) + 99
= 699
2. b
Total bill amount:
(150 × 1+ 50 × 3.5 + 25 × 7 + 14 × 5 + 15 × 2) + 99
+ 12% on (25 × 7) = 720
3. a
Since, in STD calls it’s not mentioned that the call
has been made to which operator, (like: Airtel, WLL
etc.) one may get tempted to mark option (d).
But upon solving, and considering all the possible
cases, we get
Interest grows from 30 in 1999 to 145 in 2003, thus
simple annual growth rate
145 – 30 1
=
× = 95.8%
30
4
5. e
Practice Exercise – A11
For CAGR, use the formula for compound interest
145 = 30(1 + r/100)4 thus r = 48.3%
Alternative method:
Total growth from 1999 to 2003 is 400%.
Now, let us take a convenient value, say 50%
(4 year)
Use a + b +
50 + 50 +
Rohan spent:
On STD: 18 × 1.5(Airtel, GSM) = Rs 27
Or
18 × 2 (WLL) = Rs 36
Local: 60% of 120 = 72 min (because Airtel and
GSM have same rates) @ 1 Re. min.
∴ Amount spent = 72 × 1 = Rs 72
40% of 120 = 48 min @ Rs. 2/min.
∴ Amount spent = 48 × 2 = Rs 96
Total Bill = (72 + 96 + 27 or 36)
= Rs 195 or Rs 204
ab
100
50 × 50
= 125
100
125 × 125
~ 400%
100
So CAGR is closer to 50%, hence the answer is 48%.
125 + 125 +
6. b
The company would have been placed in the MAT
category in 1999 because its tax for the year is
zero.
7. a
OPBDIT 2002 = 520 and OPBDIT 2003 = 380
OPBDIT 2003 as a percentage of OPBDIT 2002
=
380
= 73.1%
520
8.b
NP and OPBDIT increase till 2002, after which both
of them decrease.
9. a
Interest in 1999 = 30, then interest in 2003 = 145,
thus if interest in 1999 = 100, then interest in 2003
100
= 483
= 145 ×
30
10. c
The maximum depreciation was in 2003, hence
maximum assets were acquired in 2002.
Page: 10
Similarly for Mohan:
On STD: 30 × 1.5 = Rs 45 or 30 × 2 = 60
Local: 70 × 1 = Rs 70
30 × 2 = Rs 60
Minimum and Maximum amounts that Mohan would
have spent are Rs.175 and Rs.1907.
The maximum possible difference is
(Rs.204 – Rs.125) = Rs.59.
4. a
Normal tariff for 40 SMS/month would have been
= 40 × 1.5 = Rs 60
With SMS scheme: 40 × 0.60 + 35 = Rs.59
In fact, in order to benefit from the scheme, one
needs to send a minimum of 39 SMS.
So, option (a) is the correct choice.
5. d
Option (a): 30 × 1.5 + 55 + 99 (Price of Plan)
= Rs 199 (assuming local calls are made @ Re. 1)
Option (b): 16 × 1.5 + 76 + 99 (Price of Plan)
= Rs 199 (assuming local calls are made @ Re. 1)
Option (c): 10 × 1.5 + 10 × 2.5 + 10 × 2.5 = Rs 65
Now local calls (since not mentioned) may be
distributed like 5 × 2 (WLL/ Landline) + 25 × 1 (GSM/
Airtel) = Rs. 35
Total 65 + 35 + 99 (Price of Plan) = Rs 199
Therefore, bill amount of Rs 199 per month is
possible in each of the three options (a), (b) and
(c).
MBA
Test Prep
Solution Book-2
∴ Option (d)
Total bill = 8 × 1.5 + 4 × 3.0 + 7 × 3.5 + (55 × 1 or 55
× 2) + 99 = Rs.202.5 or Rs.257.5
Hence (d) is correct.
6. a
Checking options
Option (a): 8 ISD calls can be distributed as:
4 calls to Rest of the world and another 4 calls to
Gulf.
Bill amount = 4 × 40 + 4 × 10 + 99 = Rs 299
Option (b): Checking with the maximum rate of
both GSM and WLL, maximum possible bill amount
= 35 × 3 + 10 × 3.5 + 99 = Rs 239
7. b
Ear lie r
Ne w
Diffe r e nce
A irtel
1.5
1
0.5
GSM/CDMA
1.5
1
0.5
2
1
1
WLL/Landline
Therefore, to cover the additional charge of
Rs.20/month
20
= 40 calls to Airtel/GSM
One has to make
0.5
20
= 20 calls to WLL/Landline.
and
1
Hence, in
Option (a): No profit
Option (b): profit = (35 × 0.5 + 10 × 1) – 20
= Rs. 7.50
Option (c): profit = (50 × 0.5) – 20 = Rs. 5
In order to have no loss from the plan, user have to
balance out additional charge of Rs. 250 by making
calls.
For to be definitely sure, we will take that existing
STD rate which will give us minimum difference
when compared with new scheme
∴ STD of Airtel = Rs. 1.5/min
Difference = 1.5 – 1 = 0.50/min
250
= 500
0.5
Minimum 500 calls has to be made such that I am
definitely sure of having no loss from opting for this
plan.
∴ Minimum calls =
9. b
10. b
Rates in Delhi-Mathura segment:
Outgoing calls to
8. c
(b) Difference in the rates:
In STD = 1.5 – 0.5 = Rs 1/min
Amount saved = 1 × 14 = Rs 14
In local = 1.0 – 0.5 = 0.5/min
Amount saved = 0.5 × 18 = Rs 9
Total amount saved = 9 + 14 = Rs 23
Additional charge paid
= 5 × 5 = Rs 25
∴ Loss = Rs (25 – 23) = Rs 2
Option (b) is the correct choice.
Rest of the options need not be checked.
Checking options:
(a) Difference in the rates
In STD = 1.5 – 0.5 = Rs 1/min
Amount saved = 1 × 12 = Rs 12
In local = 1.0 – 0.5 = Rs 0.5/min
Amount saved = 0.5 × 8 = Rs 4
Total saved amount = 12 + 4 = Rs 16
Additional Charge levied @ Rs. 5/day for 3 days
= Rs 15
∴ One can gain Re. 1 in this calling pattern.
Explanations: Fundamentals of
Logical Reasoning & Data Interpretation
Difference in amount = 1 (Airtel local) – 0.5
= Rs 0.5/ call is saved
Additional Rent paid = Rs. 5 daily
To balance the additional charge, minimum no. of
5
= 10 calls
0.5
∴ In order to GAIN one should make atleast
10 + 1 = 11 calls
calls that should be made =
11. e
This was a tricky one.
There is no need to calculate anything.
Absolute profit will always remain the same.
Profit will always be :
[100 × 1 (free calls) + 100 × 1.5 (free SMS) – Rs. 200]
= Rs. 250 – Rs. 200 = Rs. 50
Note: Has it been said which will have “maximum
profit percentage”, then one has to calculate.
Be careful while reading the question.
12. a
This one is also a tricky question.
Total ISD calls made 12 min/month and 80% of the
bill came from the Rest of the world.
Let, calls to USA, Canada, Europe (Fixed Lines) be
of ‘a’ minutes duration, calls to GULF, Europe (Mobile),
SAARC be of ‘b’ minutes duration and calls to Rest
of the World be of ‘c’ minutes duration.
⇒ 7a + 10b + 40 c = Total Bill (in Rs.)
Or
7a + 10b = 20% of (7a + 10b + 40c)
1
(7a + 10b + 40c)
5
⇒ 4(7a + 10b) = 40c
...(i)
and also,
a + b + c = 12 (total duration) ...(ii)
Solving equations (i) & (ii), we get only one set as
possible solution. (0, 6, 6)
i.e.
a = 0 minute
b = 6 minutes
c = 6 minutes
∴ calls to USA, Canada, Europe (Fixed Lines) were
for minimum duration.
7a + 10b =
MBA
Test Prep
Page: 11
For questions 13 and 14: The outgoing call and SMS
charges are reduced to half.
13. b
4. c
Since, 20% drop in the total bill was because of
drop in the rates by 50%.
20
0.5
= 40% of the call-time during the Happy Hours.
⇒ He made 60% of the calls during the day and
40% during Happy Hours
∴ In order to get 20% drop, he needs to spend
Ratio =
14. e
The fact that the question exists means that only 1
of the 4 options can be right. Check each one till we
find a team which, based on its score after 5 games,
played can still reach the 2nd round.
After tabulating the points, we can see that only
Benfica has a chance to qualify for the next round.
Other teams given in the options will never qualify
for the next round.
Sure-shot Approach: Ideally the points for the
groups to which each option pertains should be
calculated and noted for future use. (1 at a time:
and not all at the start). People trying this question
mentally stand a good chance at getting it wrong.
60 3
=
40 2
Since, details of local calls made and SMS sent is
not available, percentage of local calls made during
the day cannot be determined.
Common Speed Breaker: Checking all 4 options
inspite of having found the right option, just to be
doubly sure, often costs students extra time in the
exam. Be sure of yourself!
Practice Exercise – A12
The set can be solved in two ways:
A.
Adding an extra column for points in Set A and
tabulating all points before proceeding. This
approach will involve repeating this step for atleast
60% of the teams again after question 6. This
approach, though time consuming, is likely to give
guaranteed results. For anyone who finds this set
tough to conceptualize, this may be the preferred
approach
B.
Take each question as and when it comes and tackle
it on merit.
5. b
Easy, the team which has the most money at this
stage, which is the team which has won all the
games it has played - Arsenal. This question is a
rare example of where scanning the data rather
than reviewing the options would give a faster
solution.
6. c
Add results / points for new instructions given and
then use options. Need to be done only for the 4
options, not for entire data set.
Anderlecht is yet to score a point so far in the
tournament and playing “away” in the last match of
the 1st round will result in a loss for them, thus
earning ‘zero’ point from the competition.
7. a
All home terms win their matches in their last first
round matches. Lyon , Inter Milan, Barcelona,
Liverpool and Arsenal have 16, 12, 13, 11 and 18
points respectively.
8. e
Cannot be determined because the scores from the
last set of matches of round 1 are not available
9. c
Can be counted and is equal to 77.
Short-cut: The total number of games won in the
last round was 16 and this was an exception as
there were no draws in this set, hence, the total
number of game won in all 6 rounds has to be
< 6*16 = 96. This rules out Option (d).
Option (a) is less than 16 (last set wins) and has to
be wrong.
Simple observation shows that 38 cannot be the
answer, so it has to be 77
10. c
Review the options.
Option (a): Even after losing the last match of their
group, Juventas and Bayern Munich, with 12 points
each will reach the next round, leaving behind club
Brugge with 9 points.
For questions 1 to 15:
1. c
(32 teams) × (6 Games Played) / 2 = 96 (it takes two
teams to play a match) matches will be played in
round 1, 8 matches in round 2, 4 in Quarter-finals,
2 in Semi-finals & 1 in the final
Total = 96 + 8 + 4 + 2 + 1 = 111
Short-cut: Answer has to be > 96 eliminating options
a & b instantly
Common-error – counting each match twice
forgetting that two teams are playing and hence not
dividing by 2
2. e
Will be either 3 or 4 depending on whether the
second round was a ‘home’ match or not.
3. b
All teams play atleast 3 away matches in the first
round and another 3 away matches in the quarterfinal, semi-final and final.
Remember, ‘neutral’ venue is also an away match
for any team.
Page: 12
MBA
Test Prep
Solution Book-2
Option (b): Rangers by virtue of winning the last
match of the group, will end with 9 points, Thus will
qualify from group H as a ‘runners-up’ behind Inter
Milan.
Option (c): Lille, after losing the last match of the
group (because Lille is playing away in the last
match) will end up with only 6 points. So, from
group D, Benfica and Villareal will quality with 8 and
10 points respectively.
So, option (c) is the right answer choice.
11 a
12. c
13. c
14. a
15. d
From the table given in the explanation of the
previous question, it is obvious that Juventas is the
only surviving team among the options given in this
question.
So option (a) is the correct answer.
As Arsenal has won the most number of games (6)
till this stage they are ahead of all teams on the
money earners list. Juventus have won 5 & have 0
draws, Liverpool have won 4 & have 2 draws and
are hence also behind. Rangers with 2 wins and 3
draw’s is not in the reckoning. The key in this
question is that the actual values of amount earned
need not be calculated.
Number of students in XI Arts = 720 (Given), thus
0.24N + 0.048N = 0.288N = 720, thus N = 2500
We will have the following:
LPS
Outside
Total
2000
400
2400
400
80
480
XI commerce
1000
200
1200
XI arts
600
120
720
Total standard X
2500
Total standard XI
XI science
1. b
As N = 2500.
2. a
2000 (solved above)
3. c
(0.4 + 0.08) × 2500 = 1200
4. b
Percentage of arts students in LJC standard XI
=
5. c
480 (solved above)
6. a
Number of science students in XI = 480
Number of science students in XI from LPS = 400
Percentage of science students in LJC standard XI
Again doesn’t need to be calculated, the two teams
records (wins, draws & losses) need to be
compared and are found to be identical
Options (c) & (d) are out as Arsenal has been
knocked out as seen in the last question, leaving
Juventus as an automatic choice and just a check
required between Liverpool & Rangers.
Since all ‘Draws’ took place in 1st round, comparing
the number of draws by Liverpool and Rangers,
we get Rangers had 3 draws, while Liverpool had
2 draws. So option (a) is the correct choice.
We have no data on quarter-final results, hence
this question only checks on the teams which have
definitely been knocked out and hence cannot be in
the final. AC Milan & Lyon have both been knocked
out, making this an easy choice.
who are from LPS =
Explanations: Fundamentals of
Logical Reasoning & Data Interpretation
400
= 83.33%
480
7. e
We do not know what percentage of the girls took
arts though we can find out the total number of girls.
8. c
10% of commerce stream = 10% of 1200 = 120
5% of the arts stream = 5% of 720 = 36, thus number
of students in XII arts = 720 + 120 – 36 = 804
9. a
5% of arts stream = 5% of 720 = 36
6% of commerce stream = 6% of 1200 = 72, thus
number of students in XII LJC = 2400 – 36 – 72
= 2292
10. c
8% of commerce stream = 8% of 1200 = 96,
10% of science stream = 10% of 480 = 48, thus
change in number of commerce students = –48
Percentage change in commerce students
= –48/ 1200 = –4%
11. e
Though we know the ratio, we do not have the ratio
of number of boys and girls in any stream or in total.
12. b
Number of girls in science = 50% of boys in science
Thus, girls form 33.33% of science = 480/3 = 160
Thus, number of girls in arts = 3/2 × 160 = 240
because the ratios Arts : Science is 3 : 2.
Practice exercise – A13
If N is the number of LPS students who took standard X
examination, then the number of students who passed the
examination is 0.8 N.
Number who joined the various streams:
Science = 0.2 × 0.8N = 0.16N
Arts = 0.3 × 0.8N = 0.24N
Commerce = 0.5 × 0.8N = 0.4N
Number of students joining from outside = 0.2 × 0.8N =
0.16N Ratio of these students in = Science : Arts : Commerce
= 2 : 3 : 5, thus number of these students in
Science = 0.2 × 0.16N = 0.032N
Arts = 0.3 × 0.16N = 0.048N
Commerce = 0.5 × 0.16N = 0.08N
720
= 30%
2400
MBA
Test Prep
Page: 13
13. b
Number that failed in XI class = 10% 2400 = 240
From V, we get
14. b
Number of boys in science stream = 320
(from question 12)
Number of boys in arts stream = 720 – 240 = 480
Thus, ratio = 320 : 480 = 2 : 3
Trunk
Box
25 pounds 5 pounds
20 pounds __
American
__
... (3)
From II, we get
15. b
Total number of students in standard XI after
introduction of home science = 2400 + 300 = 2700
Percentage of home science students in standard
XI = 300 / 2700 = 11.11%
Trunk
Box
25 pounds 5 pounds
20 pounds __ 10 pounds
American
... (4)
Logically, the left place in (4) will be occupied by 15 pounds
belonging to Indian as per IV
Logical Reasoning
Practice Exercise – B1
For questions 1 to 5: Note that from statements I, IV and VI
we can make out that Chandra lives in New Delhi. Now from
V, we can make out that Aparna is not the dancer; and from
II, since Chandra lives is New Delhi, she is also not the
dancer. Therefore, Bharti is the dancer and lives in Kolkata.
Also from VIII, Bharti is married to Ramsingh. Now since
Chandra lives in New Delhi and Bharti in Kolkata, it is obvious
that Aparna lives in Mumbai, and from IX, she (Aparna) is
married to Mansingh. Hence, the result is as follows.
Kolkata
New Delhi
Mumbai
Sisters
Bharti
(Dancer)
Chandra
Aparna
Brothers
Ram Singh
Bhim Singh
Man Singh
1. b
Trunk
Box
25 pounds 5 pounds 20 pounds 15 pounds 10 pounds
American Indian
... (5)
From III, Swede can occupy the second spot from left only
Trunk
Box
25 pounds 5 pounds 20 pounds 15 pounds 10 pounds ... (6)
Swede American Indian
From VII, we get
Trunk
Box
Crate/Suitcase Crate/Suitcase Carton
25 pounds 5 pounds 20 pounds
15 pounds
10 pounds
German Swede American
Indian
Belgian
Here 10 pounds weight will be owned by Belgian because
it is the one which is not owned by German as it is a carton.
Therefore, owned by Belgian and consequently 25 pounds
will be owned by German.
6. c
2. a
7. d
3. a
8. e
4. c
9. b
5. a
10. e
For questions 6 to 10:
According to the question from VI, we get
20 pounds __
__
__
__
American
... (1)
From VIII, we get
Trunk
25 pounds
Page: 14
__
20 pounds __
American
__
... (2)
MBA
Test Prep
Solution Book-2
For questions 11 to 16: According to the question, we
have
Persons
Campsites
Lakes
States
A
E
I
M
B
F
J
N
C
G
K
O
D
H
L
P
If we represent the data in the following way, we get
–, H , J, N
From (ii)
(1)
For questions 17 and 18: As given that the names of
brothers and sisters do not begin with the same letter and
Pinku and Gaurav are not Saroj or Sangeeta’s brothers,
Pinku cannot be the brother of Pooja and hence he is the
brother of Rakhi.
Now we have that Gaurav cannot be the brother of Saroj,
Sangeeta or Rakhi. Therefore, Gaurav is the brother of
Pooja. As given that Saroj is not Ratan’s sister and Rakhi
and Pooja can also not be the sister’s of Ratan (from above
conclusions), Ratan is the brother of Sangeeta. Anil will
have to be the brother of Saroj as this is the only valid
combination left. Therefore, we have this table finally.
B, –, K, – From (i) (2)
(4) From (v) A, –, –, O
Sister
Pinku
Rakhi
Gaurav
Pooja
Ratan
Sangeeta
Anil
Saroj
From (iv) (3)
D , F, –, –
We have I camping on P ... from (iii) ... (3)
Now, from (3), I and P have to be together. These I and P
cannot be with (1), (2) or (4) as they are already occupied.
Therefore, I and P will be there with (3).
So we have
–H J N
17. d
18. b
19. a
A
Brother
O
B
K
Relative speed of approach of the cyclists
= (15 + 15) = 30 miles per hour.
30
Therefore, they will meet after
= 1 hr.
30
Since the fly has travelled throughout this 1 hr
(regardless of direction) at a speed of 20 miles per
hour, total distance travelled by it is (1 × 20)
= 20 miles.
D F I P
Logically, as all the states are occupied except M,
M will come along with B and K. Therefore, we
have the final arrangement depicted by the table
given below.
Person
A
B
Campsite
C
D
H
F
L a ke
L
K
J
I
State
O
M
N
P
20. b
49
= 7 cigarettes.
7
∴ The duration of time he will take to smoke these
3
hr = 5.25 hr (i.e. 5 hr and
7 cigarettes = 7 ×
4
15 min). Now note that after he has smoked these 7
cigarettes, he will collect 7 more stubs (one from
each), from which he will be able to make another
3
cigarette. This will take him another
hr (45 min)
4
to smoke. Therefore, total time taken = 6 hr.
He has got =
And on the basis of this table we can solve rest of
the questions very easily.
11. b
12. a
13. d
14. a
15. d
Explanations: Fundamentals of
Logical Reasoning & Data Interpretation
16. b
MBA
Test Prep
Page: 15
For questions 9 to 13: On the basis of the given information,
we arrive at the following sitting plan that does not violate
any of the given conditions.
Practice Exercise – B2
For questions 1 to 3:
J
Note: Jayesh is to the right of Alok, i.e. J, A. Pramod is
between Bhagat and Subodh, i.e. B, P, S. Subodh is
between Jayesh and Pramod. So the sequence is:
Bhagat
Pramod
Subodh
1. a
Alok is at the extreme left end.
2. d
Subodh is in the middle.
3. a
Statement I is superfluous.
Jayesh
Alok
From III, we get
_
_
_
_
_
E or
B
B
E
_
_
_
_
_
From IV, we get
_
_
_
F
G
E
B
From V, we get
the final sequence
as
E
G
F
_
_
_
B
E
G
F
D
C
A
H
K
E
And on the basis of the above figure rest of the
questions are solved as follows:
9. e
K is seated between E and H.
10. c
Three persons H, L and J or G, I and E are seated
between K and F.
11. b
The three lady members are E, H and G.
12. c
J is to the immediate left of F.
13. a
Clearly, J is a male member.
For questions 14 and 15: According to the given question
from II, we get
B
or
F
G
I
For questions 4 to 8: According to the question from II, we
get
D
C
L
H isto ry
C ivics
... (1)
From III, we get
... (1)
Now on the basis of the sequence given in (1), we
can solve rest of the questions very easily.
G eog raph y
E ng lish
... (2)
4. e
5. a
6. c
7. d
8. e
Page: 16
MBA
Test Prep
Solution Book-2
From IV (1) and (2), we get
On the basis of the above table, rest of the questions
can be solved very easily.
16. c
17. a
18. d
19. b
20. d
H isto ry
Practice exercise – B3
C ivics
For questions 1 and 2:
Check out the numbers that have four factors and try to
understand, e.g. 6 – 1, 2, 3, 6.
Note that in case of 1, 2, 4, 8, x = 2 (not 4), which is prime.
Note that x and y will always be prime.
Therefore, x . y = N
Therefore, x . x . y = x . N
G eo gra ph y
E n glish
... (3)
E con om ics
Since history and civics cannot be at any other
place than this, according to the given conditions.
On the basis of this very arrangement, rest of the
questions can be solved very easily.
14. c
15. d
1. a
2. d
For questions 3 to 5:
r
Clearly, C gives us the clue that the science book is
placed at the bottom. Thus, we know that there are
three books between the civics and science books.
9
8
Clearly, history, civics and geography are the three
books kept above the English book. To deduce this,
no additional information is required.
r
8
9
r–9
r
r–8
For questions 16 to 20:
From the given information in the question:
From II, we get Dr Choudhary teaches zoology in Mumbai
University.
From III, we get Dr Natrajan is neither in Osmania nor in Delhi
University. Therefore, he will be either at Mumbai or Gujarat
University. Similarly, as he teaches neither geology nor
history, therefore, he must be teaching physics or botany.
... (1)
From IV,
Dr Zia → Physics but as he is not teaching in either Mumbai
or Osmania University, he must be teaching either in Delhi or
Gujarat University
... (2)
r
9–r
From V, we get Dr Joshi teaches history in Delhi University.
... (3)
From (1) and (2), we conclude that Dr Natarajan teaches
botany.
8
University
Subject
Dr Joshi
Delhi
History
Dr Davar
Osmania
Geology
Dr Natrajan
Gujarat
Botany
Dr Choudhary
Mumbai
Zoology
Dr Zia
Gujarat
Physics
Explanations: Fundamentals of
Logical Reasoning & Data Interpretation
r
8–r
And from (1), (2) and VI, we get both Natarajan and Zia
teach in Gujarat University. Finally, on summarisation we
can prepare the following table.
Names
r
9
(r – 8)2 + (r – 9)2 = r2
∴ r = 29, 5
3. b
4. c
MBA
Test Prep
5. b
Page: 17
For questions 6 and 7:
1000009 = 293 × 3413
You could get the factors using options.
Practice exercise – B4
For questions 1 to 4: The given information can be
tabulated as follows.
6. e
7. a
Department
For questions 8 to 11:
The prime numbers less than 45 are as follows:
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43.
So there are 14 prime numbers. Now, since the average of
the goals scored is also a prime number, for question (10),
we can readily eliminate choice (a). Now the sum of all the
prime numbers listed above is 281. So the sum of total goals
scored must be less than 281. Note that only choice (b), i.e.
23 gives a total number of goals of (23 × 11) = 253 which is
less than 281. Hence, average is 23.
Thus, 23 is not the number of goals scored by any player.
Of the remaining 13 numbers, whose sum is (281 – 23)
= 258, we have to delete two numbers, such that the sum of
the other 11 numbers is 253. That means the sum of those
two numbers, (which are to be deleted) is 5. Obviously,
those two numbers must be 2 and 3.
8. b
12. e
13. b
9. a
10. a
Now, since we know that the number of
compartments is less than the number of soldiers in
each compartment, there have to be 359
compartments.
For questions 14 and 15:
Obviously, the maximum number of such points possible
would be 3, where the points are the vertices of an
equilateral triangle.
If the points do not have to lie in a plane, then we can have
four such points, where they form the vertices of a trapezoid.
14. d
15. b
For questions 16 to 18:
Let the length be x, and the breadth y.
Since area = Perimeter, xy = 2(x + y).
The only values of x and y which satisfy this equation are
(x, y) = (4, 4) and (x, y) = (3, 6).
Therefore, there are two such rectangles, and only one
such square.
Therefore, if the rectangle is not a square, its dimensions
will be 6 by 3.
Page: 18
17. a
Male
Mechanical
Electrical
Female Male Female Male Female
Professor
A
E
H, I
L
O, P
S
Lecturer
B, C, D
F, G
J, K
M, N
Q, R
T
Further, following constraints are added.
(i)
(ii)
Male lecturer in
Civil Department
×
∴ (B, C, D)
× T… (Not allowed)
Female professor
× Male lecturer
∴ (E, L, S)
×
11. b
222221 = 619 * 359. However, we do not know
which is the number of compartments, and which
is the number of soldiers in each compartment.
Therefore, the answer cannot be determined due
to insufficient data.
16. b
Civil
Female lecturer in
Electrical Department
B, C, D, J, K, Q, R
(Not allowed)
As given that BCD cannot be with T and ELS cannot
be with BCDJKQR, then using these conditions we
can solve rest of the questions very easily.
(1) Choice (a) is not valid as no lecturer from
electrical department is there. Choice (b) is not
valid as no lecturer from civil department is
there. Choice (c) is not valid as no lecturer from
mechanical department is there. Therefore,
choice (d) is the only valid choice as it justifies
all the given conditions.
(2) According to the question. HIL cannot go with
BCDFG.
... (1)
The delegation must have minimum three
professors.
... (2)
Representation from every department must
also be there.
... (3)
Checking from the above conditions, we get
(a) is not valid as it violates (1).
(b) is not valid as A, H, D are males.
(c) is again not valid as it violates (1).
(d) obviously, it is the valid choice because it fulfills
all the conditions.
(3) According to the question, only civil and
electrical departments must be used.
..(1)
Committee must have two professors and two
lecturers.
...(2)
18. b
MBA
Test Prep
Solution Book-2
Explanations: Fundamentals of
Logical Reasoning & Data Interpretation
MBA
Test Prep
Page: 19
For questions 7 to 9: Simple! You are being asked about
the product of which two consecutive integers is equal to
product of three consecutive integers. You have two such
sets.
(i) 1 × 2 × 3 = 2 × 3 (This is not possible in this case.)
(ii) 5 × 6 × 7 = 14 × 15
For questions 17 and 18:
Doctor cannot be a woman. (As the two women,
Sujith’s mother and his wife are not blood relatives.) Son
cannot be the doctor because he is the youngest of them
and all people are his blood relatives.
⇒ Sujith is the doctor. Lawyer can be anyone except
Sujith and his mother (Sujith’s mother also because of
condition I).
7. e
17. a
10. d
18. c
8. c
9. e
2r
For questions 19 and 20:
The joker can be with Ajay or Sujith, since Sanjay cannot
have it.
Three queens can only be with Sanjay, as 3 queens plus a
minimum of 1 king add up to a minimum of 4 cards.
19. d
2r
2r
2r
20. b
Practice Exercise – B5
For questions 1 and 2:
Here, the best method to solve these questions is going by
choices. In (1) choice (b) is only valid choice as it is between
70 and 79 and is not a multiple of 6, whereas other choices
are not valid as they do not follow the given criteria.
In (2) obviously, none of the information is superfluous.
Therefore, choice (e) is the answer.
1. b
2. e
From III and IV, we get that Puneet arrived at the mansion
after midnight and because the detective arrived at the
mansion at midnight, therefore, Puneet is not the detective
... (i)
Now the detective can be either Aditya or Vijay.
From II, we get that the detective came at the second number
(then only II is justified) .
From III and V, we can conclude that Vijay was the earlier
arriver between Puneet and Vijay, therefore, he is not the
detective. Similarly, from (i) Puneet is not the detective.
Therefore, logically, it is Aditya who is the detective and
arrived at the mansion at the second number.
... (ii)
From IV, I and VI, we conclude that Vijay came first. Aditya
(the detective) came second and Puneet came third. Now
from VI Puneet was not the murderer and from I and IV it
was Aditya, the detective, who was the murderer . ... (iii)
Using (i), (ii) and (ii) rest of the questions can be solved
very easily.
4. b
Page: 20
For questions 11 to 13:
We have
P + Q + R + S = 45
1
and (P + 2) + (Q – 2) + 2R + S = 45
2
2(P + Q) = R + S
…(i)
…(ii)
... (iii)
Solving the equations, we have their ages as follows.
For questions 3 to 6:
According to the given conditions in the question.
3. c
In fact you will get a hexagon joining the centres of all the
external circles.
5. c
6. a
P
Q
R
S
?
?
10
20
11. b
12. c
13. a
For questions 14 to 18: If a room has an odd number of
visitors, it will be closed.
Any room number that is twice a perfect square will have
an odd number of visitors. The room with the largest such
number (twice a perfect square) will be the last room to
have an odd number of visitors.
Note that the 38th room with an open door will be the 38th
room whose number is not twice a perfect square, which
happens to be 88.
Note that if only occupants of 2 to 1000 do the task, then
only 1 person, i.e. the occupant of number 1000 will go to
room number 2000, (since 2000 is a multiple of 1000), and
this room will therefore be closed. (Since it is initially open.)
Number of room closed upto 1000 = 22
Since there are 22 twice a perfect square numbers.
Number of remaining room = 500
Number of twice a perfect number between 1000 and 2000
= 9.
All rooms with twice a perfect square between 1000 and
2000 will be open.
MBA
Test Prep
Solution Book-2
∴ Number of room closed = 22 + 500 – 9 = 513
As regards question 18, anyone who lives in a room with a
number greater than 1000 will obviously have to visit only
that particular room, as it will not have any multiple which is
not greater than 2000.
14. c
19. e
20. e
15. b
16. a
17. a
For question 6 to 10:
The best method to solve this question is to make a table and
fill the places according to the information given logically.
On analysing the information given in the question, we
arrive at the following table.
18. b
Student
Optional Subject
The only such number below 200 is 153.
M
Geography (given)
English (given)
Therefore, Q is 5.
N
Geography (given)
Biology (given)
You have two such numbers: 370 or 371.
Therefore, R may be 0 or 1.
O
Geography (given)
Physics (given)
P
English (given)
Geography
(given) Female
student (given)
Q
Chemistry (given)
Physics (given)
R
Physics (given)
Chemistry (given)
Practice Exercise – B6
For questions 1 to 5:
According to the given information in the question,
we get
O is a bank manager and is a male (as he is married to a lady
professor) ... (X)
Q is a medical representative and is a male (as he is the son
of M ... (Y)
N is a chartered accountant and is a female (as she is the
daughter-in-law of L) ... (Z)
As given in VI that businessman is married to the chartered
accountant, this would mean that the businessman cannot
be L (as N, chartered accountant, is the daughter-in-law of
L), O, Q and P (as they are having different professions,
other than business). Therefore, logically M is the
businessman with whom N is married. So M is businessman
(male). ... (A)
Given that P is an unmarried engineer.
6. d
O
(M ale/ Bank M anager)
M
P
(U nm arried Engineer)
2. c
8. e
9. a
10. d
Students
Choices
A
B
C
D
E
F
1
R
M
R
M
M
R
2
B
P
B
C
P
B
3
C
B
T
T
B
P
R ⇒ Reading
C ⇒ Cricket
P ⇒ Painting
B ⇒ Badminton
M ⇒ Music
T ⇒ Tennis
Note: The best way to solve this problem is
to fill the information given logically and
complete the table. On the basis of the given
information the third choice for both C and D
would be tennis.
L
(Fem ale/Professor)
11. c
(M ale/Businessm an)
Son
7. a
For questions 11 to 15:
The given information can be tabulated as follows.
Here as we know the profession of all the members except
L and only profession unclaimed is professorship, therefore,
L is a professor and because L is the grandmother of Q,
therefore, L is female ... (B)
From III, II and (B), we can very clearly conclude that O, the
male bank manager is married to lady professor L.
Using (A) and (V), we get that M is the son of L and O.
From (IV), we get that Q and P are kids of M ... (C)
So, all the conclusions using (X), (Y), (Z), (A), (B) and (C)
can be depicted with the help of this family tree with the
help of which rest of the question can be solved very easily.
1. d
Compulsory Subject
12. d
13. c
14. c
15. b
N
(D aughter-in-law /C hartered accountant)
Q
(Son/M edical R epresentative)
3. a
4. e
5. b
Explanations: Fundamentals of
Logical Reasoning & Data Interpretation
MBA
Test Prep
Page: 21
For questions 16 to 20: The given information can
be tabulated as follows:
Name
Subject
City
A
Philosophy
(given)
Hyderabad
... from (i)
B
Mathematics
(given)
Bangalore
... from (ii)
Economics (as
D does not
teach
economics
therefore, C
teaches it.)
Delhi (given)
D
History (as only
history is left)
Jaipur (given)
E
Geography
(given)
Lucknow
... from (iv)
18. c
20. b
C
16. b
17. d
19. d
Solving the two equations, we get
x = 5z and y = 19z.
Therefore, the ratio y : x = 19 : 5.
19 + 5 = 24 which means that there must be at least 24
ladoos as the cats have an integral number of ladoos. If
there are 30 ladoos, the monkey must have cheated the
cats of 6 ladoos (i.e. 30 – 6).
5. b
6. c
For questions 7 and 8: From the given data, we can make
the following table with the help of which rest of the
questions can be solved very easily.
Male (40)
Female (30)
Married
7
12
Unmarried
5
0
Married
8
3
Unmarried
20
15
Total
40
30
Above 25
Below 25
Practice Exercise – B7
For questions 1 and 2:
Five such pairs are possible. (10x + y) – (10y + x) = 36
⇒ x–y=4
∴ Numbers are 15 and 51, 26 and 62, 37 and 73, 48 and 84,
and 59 and 95.
7. b
8. d
For questions 3 and 4:
x = 10a + b, x2 = 100a2 + 20ab + b2
y = 10 b + a, y2 = 100b2 + 20ab + a2
⇒ In the square, middle number (ten’s place) will always
be the same.
122 = 144, 212 = 441
132 = 169, 312 = 961
For questions 9 and 10:
For every plate of ice cream, there are 2 employees.
For every plate of sambar, there are 3 employees.
For every plate of chicken, there are 4 employees.
∴ Number of employees must be a multiple of the
LCM of 2, 3 and 4, i.e. 12.
Only choice (d) of question 9 satisfies this condition.
Another way of looking at this is as follows.
Let the number of employees be x.
x
Therefore, the number of plates of ice cream =
.
2
3. b
Therefore, the number of plates of sambar =
1. e
2. b
x
.
3
x
Therefore, the number of plates of chicken =
.
4
4. b
For questions 5 and 6:
Initially:
When the white
cat gives z ladoos
to black cat
Black cat gives z
ladoos to white cat
Page: 22
Black cat
x
x+z
x–z
White cat
y
y–z
⇒ (y – z)
= 3 (x + z)
y+z
⇒ (y + z)
= 5 (x – z)
Therefore, x + x + x = 13 x = 65
2 3 4
12
Therefore, x = 60 (employees).
Number of ice cream dishes =
9. d
60
= 30.
2
10. a
MBA
Test Prep
Solution Book-2
For questions 11 and 12:
21 can be factorized as 3 × 7. Thus, to have 21 hand
shakes the combinations for number of males in the two
families can be as follows.
P
R
C a se 1
7
3
C a se 2
3
7
For questions 19 and 20:
Suppose the holiday lasted for x days.
∴ There were 11 nice morning.
∴ On x – 11 days it rained in the morning.
Similarly, on x –12 days it rained in the afternoon.
It never rained in the morning as well as afternoon.
∴ We have x – 11 + x – 12 = 13 or x = 18.
Therefore, the holiday lasted for 18 days.
It obviously never rained in the morning as well as in the
afternoon.
C a se 3
1
21
19. a
C a se 4
21
1
20. e
Practice Exercise – B8
Let there be x females in P family and y females in R family.
Then for the first case, number of kisses will be
7y + 3x + xy = 34.
Note that y = 2 and x = 4 satisfy the equation. So the total
number of females will be 6. This will be true for the second
case also. For case 3, total number of kisses will be equal to
1y + 21x + xy = 34.
1. a
Shikha is to the left of Reena and Manju is to her
right. Rita is between Reena and Manju. So the
order is: Shikha, Reena, Rita, Manju. In the
photograph, Rita will be second from left.
2. a
B is to the right of D. A is to the right of B. E is to the
right of A and left of C. So the order is: D, B, A, E, C.
Clearly, A is in the middle.
Note: That x = 0, y= 34 satisfy the equation. Therefore, the
number of females will be 34. This will be true for case 4
also. Thus, the number of females can be 34.
3. b
Q is to the left of R and to the right of P, i.e. P, Q, R.
O is to the right of N and to the left of P, i.e. N, O, P.
S is to the right of R and to the left of T, i.e. R, S, T.
So the order is: N, O, P, Q, R, S, T.
Clearly, Q is in the middle.
4. a
S is sitting next to P. So the order S, P or P, S is
followed. K is sitting next to R. So the order R, K is
followed because R is on the extreme left. T is
sitting not next to K or S (as given). So the
arrangement will be R, K, S, P, T. Clearly, P and K are
sitting adjacent to S.
5. a
Clearly, the order is Anuradha, Rashi, Monika,
Sulekha, Abha because Rashi is not adjacent to
Sulekha or Abha and Anuradha is not adjacent to
Sulekha. Therefore, Anuradha is adjacent to Rashi.
6. c
Putting the conditions given in the question, the
position of swimmers can be:
11. e
12. e
For questions 13 and 14:
1×2×3=2×3
5 × 6 × 7 = 14 × 15
The choices in question 14 give the answer.
13. b
14. c
For questions 15 and 16
ABCABC is divisible by 2 since it is even. Also, according to
the divisibility rule of 7, ABC – ABC
= 0 → The number is divisible by 7.
⇒ It is divisible by 14.
For 16, data is insufficient.
15. a
16. a
For questions 17 and 18:
1001 = 7 × 11 × 13
1
-year-old. Therefore, my
2
only possible age is 11 x 13 = 143 years. One of my greatgrandsons is 7-year-old and one of them is 1-year-old.
1
2
3
4
5
A
C/D
B
D/C
E
1
2
3
4
5
A
D
C
B
E
I am obviously more than 95
17. c
18. c
7. e
Positions of B and E violate rule 1.
Explanations: Fundamentals of
Logical Reasoning & Data Interpretation
MBA
Test Prep
Page: 23
8. a
There is only one possible arrangement, which
satisfies the rule, i.e.
1
2
3
4
5
B
D
A
C
E
For questions 16 to 20: According to the given question,
Fa rgo
C ad illa c
From I
... (i)
Fia t
9. d
D is to the left of C, i.e. C, D. B is to the right of E, i.e.
B, E. A is to the right of C, i.e. A, C. B is to the left of
D, i.e. D, B. From the above statements, the correct
order is: A, C, D, B, E. Clearly, D is sitting in the
centre.
From II
... (ii)
Fa rg o
Note: It is given that A, B, C, D and E are sitting
facing you. So your right and left will be considered
as left and right respectively.
C ad illac
From III, we get
Fiat
For questions 10 to 14: According to the question
From III, Indian is wearing a green cap and a Jacket ... (1)
From VI, Kurta is worn alongwith red cap and sits next to
Japanese
... (2)
From VIII, T-shirt with white cap combination is seated at
one end.
So from (1), (2), (3), VII and I we conclude that the Japanese
wear a shirt of yellow colour.
From IV, V, VI and VII, we conclude that the placement of
people will be like
(1)
(2)
(3)
(4)
German American Japanese Indian
B e dford
M aru ti
A m b assa d or
Fa rg o
C a dilla c
M erce de s [lo gica lly
it ha s to be he re on ly ]
From IV
Fiat
From (2) and IV, we arrive at the following table with the
help of which rest of the questions can be solved very
easily.
A m ba ssa do r
Nationality
German
American
Japanese
Indian
Clothes
T-shirt
Kurta
Shirt
Ja cke t
C aps
Whitecap
Red cap
Yellow
ca p
Green
ca p
10. d
11. c
15. a
From I, we get France, America, India ... (i)
From (II), we get India, Australia, Japan, China ... (ii)
Combining (i) and (ii), we get the correct sequence
as: France, America, India, Australia, Japan, China.
The two flags in the centre are of India and Australia.
Page: 24
12. c
13. c
14. c
Fa rgo
Hence, the sequence of cars is as follows:
Fiat, Bedford, Maruti, Ambassador, Fargo, Cadillac,
Mercedes.
16. d
Clearly, Maruti is in the third place and Mercedes in
the seventh, i.e. Mercedes is fourth to the right of
Maruti.
17. b
Clearly, Cadillac is in the sixth place, to the immediate
left of Mercedes, which is in the seventh place
(from the top).
18. d
On the sides of the Cadillac are the Fargo and the
Mercedes.
19. a
Clearly, Maruti is in third place (from top), and is to
the immediate left of the Ambassador, which is in
the fourth place.
20. c
To the right of Ambassador are Fargo, Cadillac and
Mercedes.
MBA
Test Prep
Solution Book-2
7. c
Practice Exercise – B9
For questions 1 to 5: On the basis of the given information
following graph can be constructed and from the graph, we
can easily find the answers of the rest of the questions.
Let A, M, P denote the sets of people liking apples,
mangoes and pineapples respectively.
n(M) = 20% of 10,000 = 2000
n(A ∩ M) = 5% of 10,000 = 500, n (M ∩ P )
= 3%of 10,000 = 300.
n(A ∩ M ∩ P) = 2% of 10,000 = 200
We have to find number of people belonging to the
shaded region in the following Venn diagram.
M
I
1 km
M
J
1 km
0 .5 km
K
L
P
0 .5 km
H
A
Number of people who likes only mangoes
1 km
G
D
= n ( A ′ ∩ M ∩ P′ )
1 km
1 km
B
0 .5 km
E
C
(
= n M ∩ ( A ∪ P )′
A
)
= n (M) – n (M ∩ ( A ∪ P ))
1. d
= n (M) – n (M ∩ A ) ∪ (M ∩ P )
2. a
= n (M) – n (M ∩ A ) + n (M ∩ P ) – n ((M ∩ A ) ∩ (M ∩ P ))
3. d
= n (M) – n (M ∩ A ) + n (M ∩ P ) – n (M ∩ A ∩ P )
= 200 – [500 + 300 – 200] = 1,400
4. c
5. d
6. a
8. d
Using the data given, the seven friends are sitting in
the following order.
EAGBDFC
Therefore, E was sitting fourth to the left of D.
9. c
Given information can be rewritten as
(Nidhi – Pavbhaji)
__ ×
... (i)
(Dhruv – Kriti)
__ ü
... (ii)
According to the given question, we have
At 12 noon (Waiting)
After announcement
(Waiting)
Left audience
At 12.50 p.m.
Required percentage =
Total
audience
180
Male
108
Female
72
108
72
36
72
(108 + 18)
126
36
36
90
36
90
250
% = 83.3%
× 100 =
108
3
Explanations: Fundamentals of
Logical Reasoning & Data Interpretation
(Parul – Movie)
__ ×
... (iii)
(If beach – Paanipuri) __ ü
... (iv)
(Nidhi – Beach)
__ ×
... (v)
(Circus – Pavbhaji)
__ ×
... (vi)
(Harsh – Parul)
__ ×
... (vii)
Clearly (ii) and (vii) implies Dhruv – Kriti, Harsh –
Nidhi, and Amit – Parul.
Also (iv) and (v) implies that Nidhi did not eat paani
- puri.
Also using (i), we conclude that Nidhi and Harsh
ate chaat. Now clearly only the following
combinations of place to visit and food items are
possible.
MBA
Test Prep
Page: 25
They are Movie – Pavbhaji or Chaat (using iv)
Beach – Paanipuri
Circus – Chaat, Paanipuri
Applying elimination method and using (iv) we get
beach – paanipuri, and using (vi) we get circus –
chaat as the true combination
and finally Movie – pavbhaji
Thus we have
Harsh – Nidhi – Circus – Chaat as one correct
combination.
Then the other combinations are:
Using (iii), we get Amit – Parul – Beach – Paanipuri
and obviously Dhruv – Kriti – Movie – Pavbhaji.
10. c
13. c
On the basis of the analysis of the information given,
the combinations of the movies and their type and
the number of prizes they bagged is as follows.
D
Love story
C
Horror
2
A
Action
1
B
Kids
3
4
5
6
7
8
9
–
×
Sunday
Physics
–
–
–
–
Physical
education
Now, using (VIII) there is a gap of two days between
computers and biology. This means that computers
examination is either on 2nd or 5th and biology is
either on 5th or 8th. If we take the second case, i.e.
computer 5th and biology 8th, then using (VII)
mathematics will be on 6th but this does not satisfy
the (V) condition given.
Now we consider the other case, i.e. computer 2nd
and biology 5th. Using (vii), we have mathematics
7th and using (v) chemistry 6th. The remaining
engineering drawing is on 8th, the only day left. So,
in tabular form we have
2
Computer
3
Holiday
Sunday
4
Physics
Monday
14. d
15. c
Biology
Tuesday
6
Chemistry
Wednesday
7
Mathematics
Thursday
8
Engineering draw ing
Friday
9
Physical education
Saturday
The combinations are as follows.
A
G
B
F
Beach
Rose
C
E
Movie
D
H
Chocolate
The combinations are as follows.
A
G
B
F
Beach
Rose
C
E
Movie
D
H
Chocolate
16. b
If we take first statement of I as false, then discount
is with saree and wristwatch is free with it. The
only condition where the logic of one statement is
false and other true is justified, is possible when
second statement of II is false and first statement of
III is false. On this basis, we conclude that cap is
free with bedsheet and T-shirt is free with shirt.
Hence, choice (b) is the correct answer.
17. d
The given information can be depicted with the help
of following diagram.
Saturday
5
None
So (c) is the answer.
We get the following table directly from the
information given.
2
3 (As it has won maximum awards)
Ten nis
8 80
4 30
6 90
S w im m ing
1 90
11. b
12. a
Y is intelligent in mathematics and physics. Also Y
isintelligent in chemistry and physics.
3 45
4 60
(d) is not possible as rubber and watches cannot
be exported together. (b) and (c) are not possible
as timber and garlic both are exported. So (a) is the
answer.
6 75
B illiards
460 + (430 – 190) + (345 – 190)
= 460 + 240 + 155 = 855
Page: 26
MBA
Test Prep
Solution Book-2
18. a
19. a
20. a
Members playing only tennis
= 880 – (430 + (460 – 190)) = 880 – (430 + 270)
= 880 – 700 = 180
Members playing only billiards
= 675 – (460 + (345 – 190)) = 675 – (460 + 155)
= 675 – 615 = 60
So 180 + 60 = 240
Since the given information is not sufficient (because
nothing is mentioned about the positioning of Vinay)
to definitely depict the exact seating arrangement,
‘none of these’ is the correct choice.
4. d
Option (a) has D and G both which is not allowed.
All others are allowed. Therefore, choice (a) is the
correct answer.
On the basis of the analysis of the given information,
we arrive at the following diagram with the help of
which question can be solved easily.
A nu
P allavi
Both B and C cannot be selected. So the other door
paint has to be A. So A is always selected.
R am ola
Practice Exercise – B10
M a nd ira
N ata sh a
1. d
All of D’s children are in Z. So (d) is the answer.
2. c
From the question, we get that
_ Sapna _ _ _ _ _ _ _
1 2
3 4 5 6 7 8 9
... (A)
Given that
(i) Megha, Sapna and Riya cannot sit at 1 or 9.
(ii) Beena and Megha does not have anybody sitting
adjacent to them.
(iii) There is only one empty chair between Megha
and Riya.
(iv) Charu is adjacent to both Jiya and Riya.
From (iv) it is very clear that Jiya - Charu - Riya or
Riya - Charu - Jiya will be the sitting arrangement
... (v)
From (v) and (iii), we get
Megha — Riya - Charu - Jiya ... (vi)
From (i) and (ii), we conclude that Beena and Megha
must have nobody adjacent to them. It means they
must have at least one place empty adjacement to
them. Now based on all the conditions (i), (ii), (iii),
(iv), (v) and (vi), we get the only possible and valid
arrangement as
1
X
8
9
X Beena
... (vii)
From (vi), we get Megha will sit at chair number 7.
3. e
2
3 4
5 6 7
Sapna Jiya Charu Riya X Megha
One of the possible sitting arrangements based on
given information is depicted by the diagram given
below.
Vinay
S he hul / M ilind
K om olika
5. b
Applying the rules given in the question, only two
combinations are possible.
A
Cricket
A
Cricket
B
Football
or
B
Cricket
C
Football
C
Cricket
So A always play cricket.
6. b
(a) is not feasible as it does not satisfy condition (i).
(c) is not feasible as it does not satisfy condition
(iii). Again option (d) does not satisfy condition (i).
The only feasible group is given by option (b).
7. c
Clearly, using (II) and (III) Poonam and Poornima do
not stay on 1st floor and Priya and Priyanka do not
stay on top floor.
Using (I), Priya does not stay on 1st floor.
So the only option for the girl staying on 1st floor is
Priyanka. Thus, Poornima stays on 2nd floor, so
Priya stays on 3rd floor and Poonam on the 4th
floor.
8. a
Using (II) and (VII), their full names are Sunil
Sachdeva, Rohit Sehwag and Sandeep Sharma.
Using (III) Mr. Sachdeva – Purse
Now, using (V) Mr. Sharma – Cosmetics and
Mr. Sehwag – Saree.
Thus, (IV) implies that Himani is Mrs. Sehwag.
Using (I) Dhwani is Mrs. Sachdeva and thus Vidhi is
Mrs. Sharma.
Thus can be tabulated as follows.
K ira n
A nil
R aje sh
Name
Surname
Gift
Wife’s name
Sunil
Sachdeva
Purse
Dhwani
Rohit
Sehwag
Saree
Himani
Sandeep
Sharma
Cosmetics
Vidhi
S he hul / M ilind
Explanations: Fundamentals of
Logical Reasoning & Data Interpretation
MBA
Test Prep
Page: 27
9. e
Let x be the number of employees at branch A.
Then, 320 – x is the number of employees at branch
B.
11. c
Gunjan and Piya have to be together.
∴ Sameer would not be there according to (ii) and
hence Priyanka would not be there according to
(v). So the four girls are Shikha, Aastha, Gunjan
and Piya. Hence, Anand and Vineet would not be
there according to (iv) and (iii). Hence one of either
Biswas and John will be there according to (vii).
Since Biswas and Aastha have to be together, the
member of the team are the ones given by choice
(c).
12. c
Clearly, using (III) the only salaries that satisfy this
statement is Rs. 6,000.
10
% of 320 = 32
100
At branch A, number of absentees = 20% of x
Total number of absentees =
At branch B, number of absentees = 7
9
% of
13
(320 – x)
Now,
⇒
20
100
x+
(320 – x) = 32
100
13 × 100
Q 6000 =
x
1
(320 – x) = 32
+
5 13
∴ Kishan’s salary is Rs. 6,000
... (1)
Using (IV) and (V) Ganesh’s salary is not Rs. 10,000
... (2)
Also Ganesh – EXCEL
Convergys – Rs. 10, 000
... (3)
Then (VI) and (I) implies that Shiv’s salary is either
Rs. 12,000 or Rs. 15,000 and Shyam’s salary is
either Rs. 6,000
or Rs. 7,500
... (4)
Using (VII) person working in Daksha gets
Rs. 15,000
... (5)
Using (1) and (4), Shyam’s salary is Rs. 7,500
∴ Shiv’s salary is Rs. 15,000 and hence he works
in Daksha [Using (5)].
Thus, Ravi’s salary is Rs. 10,000 and he works for
Convergys (Using 3)
Using (III), Kishan works in Global Vantedge and
Shyam in GE Capital.
⇒ x = 60
⇒ 320 – x = 260
But the number of females present on the day is
30% of the total employees present. But it is not
given that how many of those who were absent
are females. So, we cannot find the total number of
males at branch B.
10. d
Let us assume that they carried N chapattis.
Let N = 3x + 1
... (i)
He eats his share.
Number of chapattis left = 2x
Let 2x = 3y + 1
... (ii)
The second friend eats his share.
Number of chapattis left = 2y
Let 2y = 3z + 1
... (iii)
The third friend eats his share.
Number of chapattis left = 2z.
Let 2z = 3w + 1
... (iv)
All three of them get up and eat their shares.
3
× 10,000
5
Name
Call Centre's name
Salary (in Rs.)
81
65
W+
8
8
For even W – N will not be an integral number for
Ram
Convergys
10000
Shyam
GE Capital
7500
W = 1, N = 73
4
Kishan
Global Vantedge
6000
Shiv
Daksha
15000
Ganesh
EXC EL
12000
Using (i), (ii), (iii) and (iv) we get N =
77
W = 3, N =
2
235
4
W = 7, N = 79
∴ Minimum number of chapattis = 79
W = 5, N =
Page: 28
13. c
The information given can be written in a
compressed from as follows.
(i) Aman, Algebra, 2
(ii) Ankur, __ , 4, Scientist
(iii) __, Botany, 1, Teacher
(iv) Ankit, __, __, Businessman
(v) Amrita, Geometry, __, Architect
(vi) Mathematician and Physics expert stay on same
floor 2.
MBA
Test Prep
Solution Book-2
Using (i), (ii) and (iii). Aman is not a scientist or
teacher and Ankur is neither good at algebra nor at
botany.
Again (i), (iv) and (v) imply that Aman is not a
businessman. So, by elimination method, Aman is
an architect. Similarly, Ankur is either good at zoology
or physics. But using (vi), Ankur is a zoology expert.
One of the mathematician stays at 2nd floor and
that is Aman. So, Amrita is not the one staying on
2nd floor. Hence, by elimination she stays at 3rd
floor. Thus the other person staying on 2nd floor is
Ankit and he is the physics expert.
So using (iv), Ankit, Physics, 2, Businessman is the
correct combination.
DS Practice Exercises
5. e
Neither statements I and II alone nor taken together
can give remainder as remainder will vary according
to the numbers.
∴ Answer is (e).
6. e
Statement I does not have data to find the cost of
each album. Statement II gives the total amount spent
but we cannot find out the percentage of sales tax
per copy.
7. d
Combine both the statements to find the answer.
8. e
Since nothing about expenditure and the rate of tax
is mentioned, the salary cannot be calculated.
9. d
Let the total number of students appear in the
examination be x.
Statement I will give the information about passed
students.
Statement II is also not complete in itself.
Combining the statements I and II, we get
Practice exercise – C1
1. d
Statement I is very tempting as 4 = 4 × 1, but both
these numbers can be negative also. Hence, I is not
enough. Statement II says that both A and B are
positive. So both statements taken together solve
the problem.
2. a
Since all the 3 are odd and z – x = 4, they have to be
consecutive.
3. c
By statement I, a + 1 = a + a + 2
∴ a = –1
By statement II,
a(a + 2) = –1
∴ a = –1
∴ Answer is (c), i.e. it can be found out using either
statements alone.
4. d
50
51
x + 10 =
x
100
100
Hence, x can be calculated.
10. b
Statement II clearly gives that a can of beer is costlier.
Thus, (b).
11. b
CP =
12. d
Only statement I or II alone is not sufficient.
Combining statements I and II, we can calculate the
list price as (Rs. 1485 × 12) = Rs. 17820.
Hence, discount is Rs. 2,320.
13. a
From statement I, cost price = Rs.
From statement I, we get
a–2
a
a+2
–1
1
3
2
SP ⇒ SP = 1.5 CP
3
Profit percentage is 50.
15
and
12
6
4
Hence, profit can be worked out. So statement I is
sufficient. From statement II, we have no idea about
the selling price and hence the profit. So statement
II alone is insufficient.
selling price = Rs.
–3
–1
1
–5
–3
–1
A ll 3 case s
w e g e t ‘– 1 ’
So we do not know what the average will be from
statement I only.
Using only II we can say that the numbers are
–1, 1, 3 or 1, 3, 5.
Using both the only possibility is –1, 1, 3 and hence
we can find the average.
∴ Answer is (d).
Explanations: Fundamentals of
Logical Reasoning & Data Interpretation
14. e
Knowing the highest and the lowest scores tells us
nothing about the other scores. So statement I is
not enough. Statement II is very tempting but students
must realize that suppose 2 people take the GRE,
one scores 2300 and the other scores 1000. Surely
the average is not 2000. So statement II cannot
give us the answer. The two statements taken
together also cannot answer our question.
MBA
Test Prep
Page: 29
15. c
Statement I indicates that the number of 50-paisa
coins is 2 and the number of one-rupee coins is 3.
Statement II independently gives the same result.
16. d
A relation between a and c can be found using both
statements I and II, i.e.
From statement I, it can be found that the total of the
deposits = 75 × 5 = $375
18. c
d = Kt , u = 0 from statement I, 1205 = K.
2
1205
25
We need to determine D
⇒ 25 K =
10
7. e
Let the amount of money Prem has be P, and the
amount that Jagdish has be Q.
Statement I gives the inequality P ≥ J + 100.
Statement II gives the inequality P + J ≤ 500.
It is obvious that we cannot get the answer from
statement I alone or statement II alone or even from
statements I and II together.
8. e
Both the statements do not give any information
about the speed limit. Thus, (e).
9. e
It is not specified which tap is opened and which
one is closed, and what part of the tank was initially
full.
10. d
From statement I, we cannot find the hours.
–D
9
120.5
120.5
× 100 , D9 =
× 81
25
25
D10 – D9 =
Statement II repeats statement I. Hence, the question
cannot be solved.
a+c a
= +1
c
c
17. a
D10 =
6. e
x
y
+
= 1 which enables
5
3
us to get the answer. By using both the statements
together we get the answer.
From statement II, we get
120.5
120.5
(100 – 81) =
(19)
25
25
Note:
You do not have to calculate2 the actual answer.
From statement II, 490.4 = Kt
11. b
1
Statement II gives time required by B as × 10
2
490.4
100
Proceeding as above, D10 – D9 can be determined.
2
K . (10 ) ⇒ K =
19. e
20. d
= 5 days. Hence, statement II alone is sufficient.
There are more unknowns than the number of
equations. Hence, both statements I and II are not
sufficient.
12. e
The statements given do not give any information
about the number of rooms to be vacuumed.
Both statements are required.
13. b
From statement I, we can only get an equation in
terms of x and y, but not the value of x and y.
From statement II, we can get the value of x as 0.
Practice exercise – C2
1. a
Statement I gives only the relative efficiencies and
no idea of time taken is there.
So statement I is insufficient.
x
2
+ 18 = x , where x is
3
2
the initial amount of money he took to the mall.
Using only statement I,
2. a
a+b a
a
= + 1 : − 1
a−b b
b
3. e
Both the statements give the same information.
4. e
10 students in statement II may not be the remaining
beyond the 30 given in statement I.
5. e
Using both statements I and II, we still cannot say
anything about the number of students in the class.
Page: 30
14. a
2pq + pq2
=2+q
pq
15. b
Statement (I) gives us an inequality which is not
enough to answer the question. Statement II
indicates that g is greater than h because
irrespective of the sign of the integer, the integer
whose cube is greater will obviously be the greater
one. Therefore, statement II alone is enough.
16. d
From the given condition, a6 = b6
But since the power is even we do not know
whether a = b. But from I and II we can conclude
that both ‘a’ and ‘b’ are +ve. Thus, a = b
∴ a3 – b3 = 0
∴ Answer is (d).
MBA
Test Prep
Solution Book-2
17. d
From I, we have r = 0 or r = 3. So it is not sufficient.
From II, we have r = (–1) or r = 3.
So it is also not enough. The two statements taken
together give r = 3.
18. d
Statement I says that there could be 25 or more
books. Statement II says that there could be 25 or
less books. The two statements taken together give
us the answer as 25 books.
19. a
x2 + 2xy + y2 = (x + y)(x + y)
20. e
From I and II, we get 2 equation in two variable. But
they are dependent. Thus, eventually we get one
equation in two variables. Which has infinite solution
set. Thus, (e).
Practice exercise – C3
1. b
Statement I can be factorized as [x – 3][x – 1] = 0
giving two possibilities, statement II is also a quadratic
equation but when factorised it yields (x – 1)2 = 0
2. b
Statement I does not give any information, since if
‘x’ is –ve and ‘y’ is +ve, then also x2y is +ve and if x
is +ve and y is +ve, then also x2y is +ve.
Statement II tells us either both x and y are +ve or
x
is +ve.
both –ve. In either case
y
3. d
4. e
5. b
From statement II, we get x2 < 1
x2 – 1 < 0
(x + 1)(x – 1) < 0
–1
6. e
2x > 6 implies x > 3, while 3x < 10.9 implies
x < 3.6333.
Hence, the answer cannot be conclusively
determined.
7. d
From statement II, we get x as positive.
Hence, using this in the first statement we can infer
that x > y.
8. d
Since the value of x2 – 7x + 12 either increases or
decreases depending upon the domain of x, both
statements I and II are required to answer the
question.
9. a
Statement I implies that x can be between 1 and 2 or
–1 and –2.
Statement II implies that x is between 1 and 2.
10. a
Statement I is true only if a = b = c = 0
11. d
From statement I, we can say
1
= 4 >1
0.25
From statement I, we have x² – 7x ≥ 8, means
x² – 7x – 8 ≥ 0. So (x – 8)(x + 1) ≥ 0. So x ≤ –1.
And x ≥ 8. And combining with statement II, we get
x = 8.
Both the conditions are not sufficient to say if
xy > 1.
Since if x = 0.25 and y = 10
∴ xy = 2.5 > 1 and if x = 0.025
y = 1.2
xy < 1
∴ Answer is (e).
From statement I, we have
1
x
1
= 0.5 < 1
2
Thus, statement II is needed where x < 1
∴ Answer is (d).
12. e
<1
Consider x = 4
1
∴
= 0.25 < 1; for x > 1
4
Now consider x = –4
1
= − 0.25 < 1 ; for x < 1
−4
So even if ‘x’ is > 1 or x < 1, statement I is satisfied.
Thus, we cannot say from that if x > 1 or x < 1
∴
Explanations: Fundamentals of
Logical Reasoning & Data Interpretation
+1
Thus, we can say that x lies between +1 and –1.
Thus, x is not greater than 1. Thus, (b).
13. a
m
= n + p cannot be said if n > p.
n
From statement II, n > m, i.e. n = km, where k > 1
n
m
Again it cannot be said if n > p. Combining,
=
p
n
1
∴
= n + p, i.e. n + p < 1
x
It cannot be determined if n > p.
From statement I,
From statement I,
a = 40°
∴ a + b + c = 180°
∴ b + c = 140°
And x + b + c + y = 360°
∴ x + y = 360 – (b + c) = 360 – 140 = 220
∴ The answer is (a).
MBA
Test Prep
Page: 31
Page: 32
MBA
Test Prep
Solution Book-2
Explanations: Fundamentals of
Logical Reasoning & Data Interpretation
Page: 33
From option (d), it can be seen that fact (v) is
worthless as far as the data about weights go.
Also, facts (i) and (ii) do not confirm D as the lightest
among all.
Therefore, option (c) consists of sufficient facts
and hence, is our answer.
For questions 4 to 6:
Another new type of question set. Basically, instructions
would have given you the idea that the question is not to be
answered but the information required to find the answer to
the question must be singled out from the sea of data.
We analyze the facts one by one in a tabular format.
Fact #
(i)
(ii)
(iii)
Height
C>D
D>B
A>B
(iv)
-
(v)
SUMMING
UP
4. a
A
C>B
D
A
C > >B
D
Weight
C>D
B>C
B>A
C
A
>D
-
B>
C
>D
A
Here, we are interested in the weights of B and D
relative to each other. We see that the first two
facts tell us that B > C and C > D. Thus, both of them
combined have told us that B > C > D and hence, B
> D. Note that even (iii) and (iv) combined would
have directed us to the result
(B > A > D). But in the options, we had
a. (i), (ii)
B>C>D
b. (ii), (iii) No information about D
c. (i), (iv) All information (about weights) in fact
(i) is already contained in fact (iv).
d. (ii), (v) Fact (v) does not tell anything about
weights and (ii) alone is insufficient.
So, only option (a) contains all the facts required to
answer the question. The answer to the question
asked is "YES", but we have nothing to do with that
answer.
5. e
Here, a sequential procedure was to be followed.
First, we had to find who is the shortest individual.
Then, we had to find who is the heaviest individual.
Finding both of them can confirm if the same individual
is referred or not. Hence, from facts (i), (ii), (iii) and
(iv) we get to know that B is the shortest as well as
heaviest. So, option (e) is correct.
6. c
From option (a), it becomes clear that B or C cannot
be the lightest (because they are heavier than
somebody). But lightest could be either A or D.
From option (b), information revealed is that C or A
is not the lightest. But still, we cannot be sure about
D because information about B is still concealed.
From option (c), we come to know that neither B
nor C nor A is the lightest. Hence, D has to be the
lightest.
Page: 34
In questions 7 to 11, the task was not to identify the option
containing sufficient facts, but to tell the minimum sufficient
number of facts required. In this scenario, redundancy and
intersection of the information in the facts must also be kept
in mind while marking the answer. In these questions, the
ultimate result acquired through the synthesis of all the
information given by all the hints taken together would be
employed.
7. a
In the table, we can see that C is the tallest. Height
order of C can be computed by fact (v) alone, which
directly tells that C is the tallest. Minimum 1 fact
required. Option (a).
8. c
From the table, B is the heaviest. To know that B is
the heaviest, we must make sure that there is at
least one person heavier than each of A, C and D.
For that purpose, we require facts
[(ii) AND (iii) AND { (i) or (iv) } ]. Minimum 3 facts
required. Option (c).
9. a
We can see in the table that D is not the shortest. To
confirm this, we need just one information that could
tell us that there is at least one person shorter than
D. Fact (ii) tells us just that. Only one fact required.
Option (a).
10. b
From the table, D is the lightest. To know that D is
the lightest, we must make sure that there is at least
one person lighter than each of A, B and C. For that
purpose, we require facts
[(iv) AND { (ii) or (iii) } ]. Minimum 2 facts required.
Option (b).
11. c
B is the shortest. To know that B is the shortest, we
must make sure that there is at least one person
shorter than each of A, C and D. Only facts [(ii) AND
(iii) AND { (i) or (v) } ] can confirm the premise that B
is the first in ascending order of heights. Minimum 3
facts required. Option (c).
MBA
Test Prep
Solution Book-2
For questions 12 to 20:
This is a typical kind of set similar to that featured in CAT
2005. The questions asked you to decipher the data and
then ascertain the dependency of the two statements
specific to each question. Hence, it must be understood that
these kinds of questions ask for a bit more than forming
cases or inferring a pictorial / tabular simplification of the
information. The data can be synthesized as:
Five ladies
Colour of sarees
Mrs. Laali
Red
Mrs. Neelima
Blue
Similarly, number of affairs of L is greater than that of
someone else. So, it can never be minimum (6 for orange).
Information 3: In Adorable colony, H has the maximum number
of affairs.
Interpretation: From the table, we can see that the maximum
number of affairs in
Adorable colony is 4, which are those of the lady in Red and
the lady in Green. Hence, H is wearing either Red or Green.
Now, we must move on to the questions.
12. c
Straight from the information 3, we deduced that H
is in red or green. So, none of the two statements is
definitely true. However, exactly one of them is
true. Hence, option (c)
13. a
Option (a): Statement I tells that green is the colour
of the saree worn by K. And since, from information
I, if total affairs of K = 10, total affairs of N will be 9,
that means N is wearing blue coloured saree. So
statement I confirms statement II. Note that we
assumed statement I to be true and concluded the
truth of statement II. Hence, this option is correct.
Option (b): If statement II is assumed to be true, then
according to information 1, i.e. K = 1+N, K has 10
affairs. Hence, K wears green saree which makes
statement I true. So option (b) is not correct, because
it concludes statement I to be false.
Option (c): If statement I is assumed to be false,
then K doesn’t wear green saree. Hence, she wears
either red or black saree. That means, number of
her affairs = 7. Thus, number of affairs of N = 6
(orange saree). So statement II is also false. Hence,
option (c) is not correct, because it concludes
statement II to be true.
Option (d): None of the two statements is definitely
true because there are other cases possible.
14. a
2 affairs in Affection colony means either red or
green. And Mrs. Hariyaali is in one of red or green
sarees (from information 3). Hence, statement I is
true. 2 affairs in Adorable colony means either blue
or orange. And from information 1, we have seen
that K cannot wear blue or orange. Hence, statement
II is also true.
15. d
S cannot be green and L cannot be orange from
information 2. Hence, only statement II is true.
16. c
Information 2: S has lesser affairs than L
Interpretation: S < L. First things first. S cannot be
maximum(Green) and L cannot be minimum(Orange). That
is because number of affairs of S is less than someone
else’s. Hence, it cannot be maximum (10 for green).
We know that K wears the red or green or black
coloured saree only. Hence, blue is not possible.
Green coloured saree for K is possible. So from the
options, we see that option (c), which says that
statement II is true and statement I could be false, is
the correct option.
Now information 1, 2 and 3 must be combined for
necessary deductions. If N is not wearing orange,
she must be wearing blue. And N(blue) => K(green).
This way, H (green or red) has to wear red. And
since S < L, S (orange) and L(black) is the only
combination possible for N wearing blue.
Explanations: Fundamentals of
Logical Reasoning & Data Interpretation
MBA
Test Prep
Mrs. Hariyaali
Green
Mrs. Kesari
Orange
Mrs. Shyama
Black
However, the colour of the saree worn by an individual
lady in particular is not known.
Only possible source of information to come to any kind of
conclusion regarding the colour of saree each lady wears
is the following table along with three additional information
given in the question.
Number of extra-marital affairs
Colony
Colour of the saree
Red
Blue
Green
Orange
Black
Affection
2
4
2
3
1
Adorable
4
2
4
2
3
Fondness
1
3
4
1
3
Total
7
9
10
6
7
Since time is what we fall short of while attempting such
crucial questions in any exam, we must turn as economical
for that resource as possible. Therefore, Mrs. Laali will be
addressed as L, Mrs. Neelima as N, Mrs. Hariyaali as H, Mrs.
Kesari as K and Mrs. Shyama as S from now on.
Information 1: K is having 1 affair more than N
Interpretation: Number of affairs of K has to be exactly one
more than that of N.
The possibilities from the table above are
N = 6 and K = 7 or
N = 9 and K = 10
There is no other possibility. So it is clear that: (i) N wears the orange saree or blue saree
(ii) K wears the red or green or black saree
Also,
N (blue) ⇒ K (green) and N (orange) => K (black / red)
Page: 35
Now, we assume that N is wearing orange. Hence,
K will be wearing black or red. When K wears red,
H (green or red) wears green and hence S wears
black and L wears blue (because S < L). When K
wears black, H wears green or red and accordingly,
S wears red or blue obeying S < L.
The exhaustive list of possibilities
Red (7)
Blue (9)
H
N
K
S
L
K
L
H
N
S
S
L
H
N
K
H
S
L
N
K
17. a
18. c
19. d
Green (10)
Orange (6)
20. b
Black (7)
Option (a): If statement I is true, H doesn’t wear
green. That means, H wears red. Correspondingly,
S could wear orange or blue, and not necessarily
orange.
Option (b): H not wearing green means H wears
red. In that case, S can wear blue or orange. But
when it is given that S wears orange, we can see
from the table that H wears red for sure (and hence
not green). So given that statement II is true, we
can infer that statement I must be true. So, this is
the right option choice.
Option (c): If statement I is false, it means that H
wears green. Then S wears black or red but never
orange. Hence if statement I is false, statement II
would necessarily be false and not true.
Option (d): If statement II is false, S wears black or
red or blue saree. Hence, H may or may not wear
the green saree.
Minimum number of affairs = 6 (orange). Neelima
not wearing orange means she wears blue. In that
case, Shyama will wear orange (see table) and
she would be having only 1 and not 3 affairs in
Fondness colony. Neelima wearing orange means
Shyama could be wearing red or blue or black. 3
affairs in Fondness colony is true for lady wearing
blue saree and lady wearing black saree. Shyama
could have 1 affair in Fondness colony if she
happens to wear the red saree. Hence, statement II
could be true if statement I is false.
LRDI Practice Exercises
1. b
We know that S cannot wear green. And K cannot
be orange because K can be red or green or black
only. Hence irrespective of the conditions, both
statement I and statement II are definitely false.
There are just two hotels having grade ‘A’ and
exceeding 13,450 customers, viz. Taj Residency
and Mandar International.
2. c
Just count the hotels with grade ‘A’ or ‘B’ and having
more than 11,500 customers per year.
3. b
There are just five hotels with their names starting
with alphabet ‘M’ or ‘R’. Of these Madhuban Deluxe
has the highest customers per year and which is
greater than 1,500, the second highest. So even
with increased customers for other hotels,
Madhuban Deluxe will have the highest customers
per year.
4. b
Same explanation as in question 3.
5. b
In metro, valuation is > 10 times the equity. In A
circles, valuation is more than 4 times but less than
5 times. In B and C circles, valuation is more than 5
times, so valuation is least for A circles.
6. c
Combined share of metro and A circles
= Rs. 1,48,350 million
Total valuation of all circles = Rs. 2,02,900 million
Thus, percentage share of metro + A circles
Option (a): If we take statement II to be true, L has
maximum (4) affairs in Affection colony. So she
wears blue saree. In that case, H has the maximum
number of affairs (green saree). Hence if statement
II is true, statement I is also true. This option claims
that statement I is false. Hence, not the right option.
Option (b): If statement I is taken to be false, H
doesn’t have maximum affairs and therefore, she
doesn’t wear green. So she wears red. It implies
that L could wear black or green. Therefore, L can
never have maximum affairs in Affection colony
(which is true for lady wearing blue only). So
statement II is certainly false. Hence, this option is
not correct.
Option (c): Both the statements are not
independently true, as can be seen from the table.
We do have cases when H does not have maximum
number of affairs. Hence, this option is not valid.
Option (d): H having maximum affairs means H is
wearing green (10 affairs). In that case, L definitely
wears blue (as can be seen from the table) and
hence she has the maximum affairs in Affection
colony (4 affairs).
So, statement II is necessarily true if, statement I is
true.
So, this is the right option choice.
Page: 36
Practice exercise – D1
=
7. c
148350
< 74%, i.e. 73.18%
202900
Total losses written off is Rs. 1,000 crore = Rs.
10,000 millions. Thus, the increase in valuation for
Metro and A circles is Rs. 4,444.4 million and Rs.
5,555.5 million respectively. Valuation to equity ratio
has to be equal for A and B circles by reducing the
MBA
Test Prep
Solution Book-2
equity of A circles, thus
72656 × 7875
= Rs.12,858 million.
x=
44500
Thus, percentage decrease in equity of A
12858
= 1 −
= 16%
15300
8. e
Total valuation for
Delhi : 81250 × 0.42 = 34,125
Maharashtra : 67100 × 0.25 = 16,775
Karnataka : 67100 × 0.22 = 14,762
⇒ Delhi : Maharashtra : Karnataka
= 34.1 : 16.7 : 14.7
9. d
By observation, we can say that E has registered
maximum increase.
10. b
(i)
11. e
12. e
13. b
Percentage change
44500
67100 + 5556
=
or
7875
x
Price of share A on day 1 = Rs. 198
Price of 100 shares on day 1 = 198 × 100
= Rs. 19,800
If we encash 50 shares of A on day 5, we get
201 × 5 = Rs. 10,050
And on day 6 = 203 × 50 = Rs. 10,150
Gain in case of share A is (10050 + 10150) –
19800 = 20200 – 19800 = Rs. 400
(ii) Investment of share B on day 1 = 1012 × 50
Total investment = Rs. 50,600
Total investment on day 1 on share C = 52 × 50
= Rs. 2,600
(iii) Investment of share B on day 1 = 1012 × 100
If 50 shares of B and C are encashed on Days
5 and 6, then amount encashed = (1034 + 1067)
× 50 – 1012 × 100 = Rs. 3850
Amount of share B on day 6 = 1067 × 50 = Rs. 3,350
Amount of share C on day 6 = 56 × 50 = Rs. 2,800
Total return = 53350 + 2800 = Rs. 56,150
Gain = 56150 – 53200 = Rs. 2,950
So, (ii) is better than (i).
We can observe from the table that the percentage
increase for stock E is more than 20%.
Stock C has shown the least increase in absolute
terms but in percentage terms stock D is the least.
Volatility is not defined whether it is in percentage
terms or in absolute or something else.
So none of these is the correct choice.
Day 4 = 199 × 20 + 1030 × 30 + 54 × 25 + 414 × 15
+ 62 × 10 = 3980 + 30900 + 1350 + 6210 + 620
Day 4 = 43060
Day 5 = 201 × 20 + 1034 × 30 + 55 × 25 + 417 × 15
+ 63 × 10 = 43300
Explanations: Fundamentals of
Logical Reasoning & Data Interpretation
14. e
=
Day 5 − Day 4 43300 − 43060
=
Day 4
43060
=
28050
× 100 = 0.55%
43060
Percentage increase in gross assets from 2001-02
800
to 2005-06 =
− 1 × 100 = 515 .4%
130
15. b
Total disbursements from 2002-03 to 2004-05
= 560 – 65 = Rs. 495 crore
16. b
Maximum percentage increase in gross profit is for
33
1995-96 = 21 − 1 × 100 = 57.1%
For questions 17 to 21:
17. b
Route
Carrier
No. of
Fare
Fuel (Rs.)
tickets
3
Rs. 14100
2820
I
Jet Airw ays
II
Indian Airlines
4
Rs. 13640
III
Air Deccan
2
Rs. 3510
2728
912
Rs. 6460
18. b
Total expenses = Fare + Time not used in working
Wage = Rs. 200 / hr
⇒ By air = Rs. 2,375 + Rs. 200 × 2 = Rs. 2,775 and
By train = Rs. 2,286 + Rs. 200 × 9 = Rs. 4,086
So, profit = Rs. 4,086 – Rs. 2,775 = Rs. 1,311 (Since
total expenses are lower for sending him by Air
Deccan).
19. d
Neither remuneration nor number of employees is
known.
20. c
Check the options, (c) is maximum.
21. d
The last digit can be found only if we know which
number is higher. Hence, both the statements are
needed.
22. d
Overall winning percentage is the weighted average
of percentage won in the 2 years, with the number
of matches being the weights.
Thus, we need both the statements.
MBA
Test Prep
Page: 37
23. c
24. c
If he was dot on time for the class as per his watch,
7. d
We can write the equation as 3rd (r + d) = 0, r can
8. d
Shares are bought and not sold, so no tax is paid,
since tax is paid only when shares are sold.
9. a
Shastri = 7 ×
6
= 14
3
Akram = 10 ×
6
= 20
3
be − 3 or r can be 0. Hence using either of the
statements alone, one can say that r = − 3.
25. d
Combining (I) and (II):
Time taken by Shyam = 2 hr
Ratio of speed = 2 : 3
So time taken by Ram
2
=
3
Razzaq = 12 ×
20
t
t = 45 min
So Ram reaches at 11.40 a.m. + 45 min = 12.25 p.m.
Practice exercise – D2
1. a
Binny = 9 ×
10. c
18
1 1 1
10 + 3 + 5
= 1− +
+ = 1−
= 1−
30
30
3 10 6
Shastri =
6
4
2
= 1−
=
=
10 10 5
Binny =
3. d
32
= 16
2
It seems, either Maninder or Shastri is the answer.
Maninder =
Skin = 1 = 10%
10
Proteins = 16%
⇒
4. c
12
=4
3
Quantity of water in a body of 50 kg = 70% of 50 kg
70
× 50 = 35 kg
100
10
× 100 = 62.5%
16
Percentage requirement of proteins and other dry
elements = 30%
6
= 27
2
31 32
<
3
3
Azad =
=
6
= 72
1
32
> 10
3
Akram =
Part of body made of neither bones nor skin nor
Muscles
11. d
2. e
P
ratio has not decreased by more than 20%. So
E
no share is bought. So (0, 0)
1
he would have been 22
min late.
2
21
12
= 1.7
= 1.9 , Shastri =
11
7
12. e
We cannot find the runs scored as we do not know
the extras.
13. e
Number of literate females in India in 1991
= 39.4% of 430 million = 169.4 million
14. e
Cannot be determined because we do not have the
population (the weights for average) for the states.
15. e
Number of illiterates in Maharashtra in 2001
= 36.9% of 19 million = 7.01 million
16. e
Because the infant population of states is not given.
Hence, the data is insufficient.
o
30
× 360 = 108°
Therefore, the required angle =
100
5. b
In value term, Godrej appreciation has been highest
= 200
6. a
Very clearly, Dabur, which has become almost 3
times in 2008 from 2006.
Page: 38
For questions 17 to 21:
17. e
Some people can be lawyers as well as women
Also passage of the bill is not defined.
∴ We cannot find out the answer.
18. a
Average age =
MBA
Test Prep
63 × 550 + 58 × 250
= 61.43.
550 + 250
Solution Book-2
19. c
20. b
Minimum case is when all the Rajya Sabha female
members are Hindus, i.e. 42.
So, minimum number of Hindu males is 64.
Percentage of Congress(I) members =
=
21. b
4. b
Aggregate deposits for the year ended March 1999
= (717271 – 111861) = Rs. 6,05,410.
Hence, percentage increase in deposits
111861
=
= 18.5%
605410
379
= 47.37%
800
From statement I, we have Y > – |X|, which means
Y is less than or greater than X. So statement I by
itself cannot answer the question. From statement
II, we have Y > |X| which means Y is greater than X.
So it is sufficient.
The sum of a, b and c cannot be found from
statements I and II individually or together.
23. d
Statement I alone is not sufficient as the total number
of guests is unknown.
Statement II alone is not sufficient as the proportion
of single-scoop and double-scoop is not known.
Combining statements I and II, we can find the
solution.
25. a
Japan
267 + 112
550 + 250
22. e
24. c
3. a
5. b
366003
× 100 = 51%
717271
6. e
800000 − 717271
× 100 = 11.5%
717271
7. b
480000 − 366003
× 100 = 31%
366003
8. c
Statement I indicates that n is divisible by 1, the 2
prime numbers and their product. Hence, there are
4 different positive integers. From statement II, the
factors of 8 are 1, 2, 4 and 8 itself, again 4 different
positive integers.
Using statement (I):
a × b = 16 ⇒ a = 4, b = 4
a = 2, b = 8
a = 8, b = 2
In all the above cases, the last digit of (5 × ab)ab will
be zero.
So statement (I) alone is sufficient.
Statement (II) alone is not sufficient.
9. d
2. b
Increase in total population = 548 – 439 = 109
million
Increase in Hindu population = 453 – 366 = 87
million
Increase in Hindu population in total population
increased =
356
× 10 ≈ 10%
3694
123
= 512
Population of Hungary =
0.24
622.72 – 549.8
× 100
549.8
= 13.3%
Practice exercise – D3
Percentage =
Hindu population in 1971
= 1.2 × 0.78 × 665.3 = 622.72
Percentage increase =
10. a
1. e
For a credit deposit ratio of 60%, if deposit is
Rs. 8,00,000 crore, then deposit should be Rs.
4,80,000 crore.
87
≈ 80%
109
Population of Hindu, Jain and others in 1941
= 424 million
Population of Hindu, Jain and others in 1961
= 641 million.
Ratio =
424 2
≈
641 3
549.8 (550)
≈
≈ 660 million
453.4
450
2
Population of Pakistan =
285
= 109
2.62
11. e
Hindu in 1971= 549.8 ×
Others in 1971 =
So difference = 403
Ratio =
Explanations: Fundamentals of
Logical Reasoning & Data Interpretation
75.5 × 75.5 (75)2
≈
= 94 million
61.4
60
660 7
≈
94
1
MBA
Test Prep
Page: 39
12. a
Clear from Data.
13. a
Clear from Data.
14. a
R.D. =
604 604
−
= 37.75 − 30.2 = 7.55
16
20
J.K. =
503 503
−
= 25.15 − 22.86 = 2.29
20
22
M.K. =
902 902
−
= 43 − 39.21 = 3.79
21
23
23. e
Combining (I) and (II):
The oldest institute could be any one of the three,
i.e. Career Launcher, Erudite, or IMS.
Hence, it cannot be determined.
24. a
The cyclicity of 2 is 4.
Using statement (I), xy can be 43, 49, 83, or 89.
So last digit of (72)xy could be 2 or 8.
Using statement (II), xy can be 46 or 86.
So the last digit of (72)xy has to be 4.
25. d
Combining (I) and (II):
c=x
a=x–1
b = 2x
d = 2x + 2
832 832
−
= 39.62 −34.67 ≈ 5
20
27
Clearly R.D. has the highest difference.
∴ Option (a) is the correct choice.
A.S. =
15. d
a x −1
c
x
=
and =
b
2x
d 2x + 2
ad = (x – 1)(2x + 2) = 2x2 – 2
bc = 2x (x) = 2x2
So
Calculating the strike rate of the given bowlers we
get
(a) 3.2 for Harmission
(b) 4.68 for Shane Warne
(c) 4.26 for M. Muralidharan
(d) 2.9 for Pollock
As ad < bc,
a c
<
b d
Practice exercise – D4
For questions 16 to 20:
1. b
Total amount spent on ad
= Ad spent on (Print + TV + Others)
In 1998 = 6824, in 2000 = 8988
Average annual growth rate of ad spent
16. c
Except Reliance Growth, Birla Dividend Yield Plus
and Templeton India Growth all others fit in the
condition. Hence option (c) is the correct choice.
17. a
Reliance Growth is maximum at 0.58.
18. a
Templeton India Growth is 2nd best at 0.52.
19. a
HSBC Equity – 0.09
Franklin India Blue-chip – 0.14
20. c
Barring top – 3, all others are eligible.
4745
− 1 = 24.2%
Print =
3820
21. d
Combining both the statements, we can determine
that S is standing at the middle position.
3542
− 1 = 44%
TV =
2460
22. d
Combining (I) and (II), we get the following sitting
arrangement.
701
Others =
− 1 = 28.8%
544
8988 100
= 15.8%
=
− 1 ×
2
6824
2. b
3. e
Stuti
Amount spent on TV as a percentage of total ad
2952
= 38.4%
spent for 1999 =
7687
S h ikh a
R itu
Growth rate of ad spent is the highest for TV as
shown:
4. b
5. c
Decreases continuously.
512.02
× 100 = 21.6%
2365.23
B a sis th
S u kan ta
M and ar
Page: 40
MBA
Test Prep
Solution Book-2
6. b
Exports of pearls in September 1998 have fallen
by US $0.26 million (0.54-0.28) from the value in
September 1997.
Percentage decrease =
7. e
0.28
16.63 × 100 = 1.68%
8. e
2365
2860 × 100 = 82.7%
9. e
10. c
0.26
= 48.1%
0.54
(210 – 220) + (225 – 230) + (232 – 215) +
(215 – 205) + (215 – 248) + (226 – 247) +
(245 – 239) + (250 – 223) + (217 – 248) = – 40
Export earnings of jewellery + leather = 32% of
230
food grain imports = 46% of 225.
Percentage of food grain imports that could not be
(0.32 × 230)
paid for = 1 −
× 100 = 28.88%
(0.46 × 225)
11. b
In June, cotton exports = 27% of (100 – 7.25)% of
215 = 0.27 × 0.9275 × 215 = Rs. 53.8 million
12. e
Chemical import = 0.23 × 225 = Rs. 51.75 million
Export of jewellery = 0.18 × 230 = Rs. 41.4 million
Difference to be covered by spices = Rs.10.35
million
Spice exports = 0.07 × 230 = Rs.16.1 million
Thus, percentage of export of spices spent in
10.35
× 100 = 64.28%
chemical imports =
16.1
13. c
For questions 14 to 16:
Total matches played = 50
Total number of Losses = 44 + Y
11 + Z
2
Total matches played = number of loss + number of ties
Total number of wins = Total number of losses
11 + Z
So, 50 = 44 + Y +
2
⇒ 2Y + Z = 1
Since Y and Z are non-negative integers.
Z = 1, Y = 0
So, X = 19.
Total number of Ties =
14. c
15. a
16. e
For questions 17 to 19:
17. b
Total oranges produced in May = 31 × 5 = 155
Checking the options,
(1) 26th June
Total oranges needed by Kitto = 1 + 2 + 3 + ... + 26
= 351
Total oranges produced till (and on) 26th June
= 155 + 26 × 8 = 155 + 208 = 363
So, 12 oranges will be left.
(2) 27th June, Kitto will be in need of 27 oranges.
Taking a clue from option (1), maximum available
orange
= 12 + 8 = 20.
So, it will run short of oranges on 27th June.
18. b
Total oranges produced in May = 31 × 8 = 248
Checking the options,
(1) 1st July
Total oranges needed till (and on) 1st July
= 1 + 2 + ... + 31 = 496
Total oranges produced till (and on) 1st July
= 62 × 8 = 496
On 2nd July, kitto will obviously run start of oranges.
So, option (b) is the answer.
19. e
Total oranges needed till 1st July
= 1 + 2 + … + 15 + 15 × 15 = 345
Total oranges produced till 1st July = 31 × 5 + 30 × 8
= 395
So oranges left on 1st July = 50.
20. d
Combining I and II: There are 25 students, one girl
and 24 boys in the class. Santosh’s rank is 24th in
the class and since the girl is not 25th, she has to
be above Santosh. Therefore, Santosh’s rank
among the 24 boys is 23rd. So the answer is (d).
Export earning from cotton in April in US dollar
(0.27 × 220)
= $1.41 million
42
Import of chemicals would cost 0.23 × 210
= Rs. 48.3 million
For this to match $1.41 million, the exchange rate
=
48.3
= 34.25
has to be
1.41
Explanations: Fundamentals of
Logical Reasoning & Data Interpretation
MBA
Test Prep
Page: 41
21. d
Combining I and II, the code for ‘love you’ is ‘mike
sika’ (from I). If we combine I with II, ‘you’ has to be
‘mika’. So ‘love’ has to be ‘sika’. The answer is (d).
22. e
All that we can get after combining the information
given in the two statements is that ‘God is’ coded as
‘mau pau’, but we cannot find out exact code of
God.
23. c
Statement I:
Z has to be second.
X has to be third.
Y has to be first.
Statement II:
Y finished second.
Z has to be third.
X has to be first.
Thus, both the statements individually give the
answer.
24. e
5. a
Total number of outside university students
= 3 + 2 + 4 + 1 + 2 + 3 = 15
Total number of outside state students
= 2 + 2 + 4 + 1 + 2 + 3 = 14
Total number of OBC students
= 10 + 12 + 11 + 10 + 11 + 13 = 67
Total number of college III students:
General category = 35% of 500 = 175
Others = 3 + 4 + 4 + 5 + 11 + 18 + 1 = 46
Hence,
6. d
In college III, total number of handicapped, outside
university, NRI and OBC students = 23 and total
number of outside state, SC/ST and industrysponsored students = 23.
7. b
In college IV, students admitted through quotas
= 38.
Students admitted through general quota
= 14% of 500 = 70
Combining I and II: We known that A and C are males,
and B and D are females. But we do not know the
gender of E who could be a male or female.
Hence,
25. e
Combining I and II:
Even after combining both the statements, we cannot
determine who is the tallest. It could be either Rohit
or Manish. When we say that Rohit is not taller than
Manish, it means Rohit could be of the same or
lesser height than Manish.
(15 + 14 + 67 )
× 100 = 43.4%
(175 + 46 )
38
× 100 = 54.28%
70
8. d
None of the above: the figures given are for increase
in GDP.
9. a
1 × (1.025)4 × (1.018)4 = 1.1854
10. b
It is clear from the table.
11. e
Since we cannot take average value of GDP
because of possibility of different GDP figures for
different regions within a group.
= 393 : 331 = 1.2 : 1
12. e
Cannot determined because data of each region is
not given.
2. d
Among all the crops grown in Bihar, MP and
Maharashtra, Jowar has the highest production.
13. e
3. c
Production of wheat by Punjab as a percentage of
Required cost
= (240 × 1.5) + (1200 × 3) + (840 × 1.8)
= Rs. 5,472
14. a
Required average
Practice exercise – D5
1. d
In 1985, production of rice and wheat is 393 lakh
tonne and 331 lakh tonnes respectively.
Hence, the ratio of production of rice to wheat
total production of India =
4. c
103
= 31.1%
331
Total number of OBC students
= 10 + 12 + 11 + 10 + 11 + 13 = 67
Total number of industry-sponsored students
=2+1+1+2+1+2=9
Total number of SC/ST students
= 15 + 16 + 18 + 19 + 11 + 10 = 89
Page: 42
=
or
15. d
48 × 1.5 + 48 × 1.6 + 48 × 2.4 + 48 × 2 + 48 × 1
( 48 + 48 + 48 + 48 + 48)
1.5 + 1.6 + 2.4 + 2 + 1
= Rs. 1.7
5
The cost of transportation for each alternative
are:
(1) 300 × 2.8 = Rs. 840
(2) 200 × 3
= Rs. 600
(3) 300 × 1.5 = Rs. 450
(4) 400 × 1.1 = Rs. 440
Obviously, the cost of transportation of
alternative (d) is the least.
MBA
Test Prep
Solution Book-2
16. d
Required percentage
1200 − 240
× 100 % = 400 % .
=
240
Alternative method:
In stead of using the formula for percentage growth
=
24. d
Combining I and II, we must have 6 silver and 2
bronze coins. No other possibility is there.
25. e
Combining I and II, the only information we get is that
C has four sisters. Now R has four siblings but
how many brothers and sisters is not known.
Hence, the number of uncles for K cannot be
determined. So the answer is (e).
Final − Initial
× 100; we can use;
Initial
Final
100 ×
− 1 ;
Initial
as it saves one mathematical operation of
subtraction.
Practice exercise – D6
1. e
≈ 9.2 : 28.2 ≈ 1 : 3
2. c
Total sales =
3. b
Answer is ‘The Decan Herald in 1999’.
‘Others’ is not included because it is not a
newspaper. ‘Other’ here includes newspapers
which are not mentioned.
4. b
Deccan Herald gained Rs. 5.34 crore.
5. a
Jadeja’s total = 164.
Tendulkar’s average = 57.5
Hence, the ratio is 1 : 2.86.
14
220 × 7 × 100
=
= 5.6%
250 × 1.1× 100 25
6. a
Tendulkar’s points are 8 + 3 + 4 + 4 + 7 = 26.
Dravid’s points are 4 + 10 + 10 = 24.
Others are below this.
19. e
200% increase in 4 years. (From 3 crore to 9 crore
or 50% average annual growth rate.)
7. a
Since the total of 5 batsmen is more than 226 runs.
20. c
Ratio =
165
= 55 . (Do not write data for every
3
year, try to do it mentally)
This is maximum in 1996-97.
8. d
Against England =
1200
− 1 × 100 = (5 − 1) × 100 = 400%
So
240
17. b
1997-98, nearly 50% increase. (Can be figured out
by looking at the slope)
18. a
Company turnover in 1997-98 = Rs. 250 crore.
Total industry turnover in 1997-98 = 250 ×
Total industry turnover in 1998-99 =
100
7
250 × 100
× 1. 1
7
1.32 × 100
= 17.4 crore
7.6
Company’s share in 1998-99 = Rs. 220 crore
∴ Company’s share =
21. d
22.
c
23. e
18
= 18
1
31
= 7.75
Australia =
3
14
= 3.5
Sri Lanka =
4
Africa =
Statements I and II alone are not sufficient to
determine Ravi’s age. From statement I, we get
Ravi’s age but from I, we get Ravi + 10 = 2
(Ram + 10) and from statement II, Ram = 5 years old.
Therefore, Ravi = 30 –10 = 20-year-old.
From statement I, 4 boys and 3 girls,
i.e. every boy has 3 brothers and 3 sisters.
From statement II,
every girl has 3 brothers and 3 sisters
∴ The answer is (c).
‘x’, ‘y’, ‘z’ be the lengths of the ropes in the increasing
order of length.
Then from statement I, y + z = 10
and from statement II, x + y = 9
2 equations and 3 variables. Thus, cannot get ‘x’.
∴ The answer is (e).
Explanations: Fundamentals of
Logical Reasoning & Data Interpretation
38
= 19
2
71
× 200 ~
− 8 × 200 ~
− 1600 crore
9
9. e
Sales =
10. b
From a figure of Rs. 90 crore in 1996-97, sales
touch Rs. 175 crore in 1999-2000. Since sales have
less than doubled, so percentage increase < 100%.
It has to be around 95%.
11. e
We do not have data on market shares in 1997-98.
MBA
Test Prep
Page: 43
12. a
40% of others = 40% of 5% = 2%, which is
1
of
4.5
18. c
Since after the investment of Rs. 30,000, for every
investment of Rs. 10,000, return from dairy is Rs.
900 whereas for the same amount, return from
shares is Rs. 1,100.
19. b
If equal amount is invested, than net return =
Rs. 4,200, which is maximum in all the cases.
20. e
It can be observed that if he invests
Rs. 20,000 — Agriculture
Rs. 10,000 — Dairy and
Rs. 20,000 — Poultry
Total return = Rs. 2,600 + Rs. 1,400 + Rs. 2,750
= Rs. 6,750
So, the percentage return = 13.5%
21. b
Statement I gives us superfluous data. To find out
the volume of the glass, we need its height and
diameter. Hence, statement II alone is sufficient.
22. e
No information has been given about the area or
any sides of the square base of pyramid, we are
only provided with distance from the square base
from IInd statement. Hence we cannot determine
the attitude of the pyramid by using the statements.
23. e
You do not know the shape of the bathroom floor.
Thus, the exact number of tiles cannot be
determined.
24. e
We still do not know the height of the room.
25. a
Using statement I, we get
70
= k × π × ( 4.52 − r 2 ) × t
k × π × 4. 5 2 × t ×
100
Solving this we can find r.
k is constant of proportionality to convert volume
into weight. Thus answer is (a).
Zenith home PC market share.
1
2
of 200 = × 200
4.5
9
= Rs. 44.44 crore
∴ Wipro home PC sales =
13. d
Among all the products, E has maximum sale as
well profitability. So obviously E will be having the
highest profit.
Sale = 60 lakh
Profit = 20%
60
= 50 lakh
1 .2
Profit = 50 × 20% = 10 lakh
Cost =
14. c
40 10
40
lakh
×
=
1.1 100 11
Profit earned by E = 10 lakh
Profit earned by B =
(Pr ofit )B
40 × 100 400
So (Pr ofit ) × 100 = 11 × 10 = 11 = 36.36%
E
15. c
Profit earned by A =
30 20
×
= 5 lakh
1.2 100
Profit earned by B =
40
lakh = 3.6 lakh
11
20
15
300
Profit earned by C =
lakh
×
=
1.15 100 115
= 2.6 lakh
50
7.5 300
lakh
×
=
1.075 100
86
= 3.5 lakh
Profit earned by E = 10 lakh
Total profit = 24.7 lakh
Profit earned by D =
16. c
The percentage share of the profits of A and
E=
(
Practice exercise – D7
1. a
15
× 100 = 60.63%
24.74
Percentage change in exports of automobiles
excluding HCVs in 1993-94 to 1990-91
=
70 + 83 + 5 – 20 – 44
× 100
4234
=
94
× 100 = 2.2%
4234
For questions 17 to 20:
17. e
It can be observed that shares give a flat 11% return.
Now, making all the possible investment from Rs.
10,000 – Rs. 1,00,000, return in dairy does not
become equal to the return obtained by shares.
Short cut: Check it through options.
Page: 44
)
2. e
We know the total number of two-wheelers in
1990-91 and 1993-94. But we do not know the
number of two-wheelers which were exported to
EU in 1990-91 and 1993-94.
Hence, data is insufficient.
MBA
Test Prep
Solution Book-2
3. d
The total revenue of government through duty on
exports = 33000 × 1.2 × 721 × 10 2 = 285 × 10 7
= Rs. 285 crore
4. d
Revenue earned by LCVs = Rs. 1012 × 5.5 crore
= Rs. 5,566 crore
Revenue earned by LMCs = Rs. 823 × 4.2 crore
= Rs. 3,456.6 crore
Total revenue in US dollars
5566 + 3456.6
crore = $ 21.2 million
=
42.5
5. d
Cost of production (or) wage bill
= Man-hour worked × Hourly wages × Days worked
Now it is very clear from the table that year 2000
has the highest value for all the three parameters.
6. a
It is clear that the least man-hours worked is in
1970 = 1300 × 12 = 15600 man-hours × 250 days
7. c
11. d
Computed special price
News − s tan d price for each
12. c
EP (3 years) = 648 + (24 × 3) – 60 = 660
660
EP
=
= Rs. 18.33
issue 36
13. e
100% =
1,300 workers by working 12 hr per day can
produce 550 tonnes. Hence, by working 24 hr per
day they can produce a maximum of 1,100 tonnes.
Hence, they can meet targets of 1980 but not of
1985.
Cannot be determined as only market share is
given.
15. d
HP’s sales of peripherals =
Hence, total market =
357 × 100
30.4
= Rs. 1174 crore
16. a
Revenues from consultancy services
= 8% of 6945 crore = 555.6
357 × 100
= 1,155
30.9
Hence, consultancy services as percentage of
Revenues from PC servers =
Special price
per issue
Percentage decrease
from previous term
1
22
—
2
20
Approx. 9%
3
18
10%
PC servers =
4
16
11.11%
5
13.5
15.6%
EP
= Rs. 22
Issue
EP for 5 years = 810 + 0 – (12 × 25) = 510 for 60
issues.
Family A yields =
15
× 14400 = Rs. 600
360
⇒ Ratio = 480 : 600 = 4 : 5
18. e
Family A spends =
B spends =
48
× 4800 = Rs. 640
360
39
× 7200 = Rs. 780
360
⇒ Required percentage =
EP
= Rs. 8.5
Issue
Difference = Rs. 13.5
Explanations: Fundamentals of
Logical Reasoning & Data Interpretation
18
× 9600 = Rs. 480
360
B yields =
9. e
Effective price (EP) = 480 + 48 – 0
= 528 for 24 issues
555.6
× 100 = 48.1%.
1155
For questions 17 to 20: In the pie diagram, the expenses
are given in total amount of Rs. 3,600.
17. b
10. b
2500
7
= Rs. 357 crore
HP’s share of peripheral market = 30.4%
For questions 9 to 12:
Term
1250 × 100 ~
− 70 × 100 = 7000 crore
18
14. e
Percentage increase in production is always higher
than the increase in number of workers.
8. b
1250 = 18%, then
MBA
Test Prep
640
× 100 = 82%
780
Page: 45
19. b
For family A, light =
18
× 4800 = 240
360
15
× 7200 = 300
36
⇒ Difference = 300 – 240 = Rs. 60 (least)
Practice exercise – D8
1. e
20. d
21. b
22. c
96
× 100
360
110
× 100
Percentage of expenditure of B on food =
360
⇒ Ratio = 96 : 110 = 48 : 55
Percentage of expenditure of A on food =
The factors of 630 = 2 × 3 × 3 × 5 × 7. By using the
statement II, the sum of the 2 numbers is 153.
Therefore, out of the factors of 630,
(3 × 3 × 7 = 63) + (3 × 3 × 2 × 5 = 90) add up to give
153. Hence, the 2 numbers are 63, 90 and we can
find the absolute difference between them.
Hence, by using statement II only we get the answer.
Hence, (b).
From statement I,
Required ratio =
swing =
2. e
No relation between voters turnout (votes polled)
and percentage of swing are given.
3. c
(iii) adds up to more than 200 seats.
(ii) gives negative value for Socialists.
4. d
For republicans to get 100 seats the vote swing
must be 8.75% which is not an integer.
5. d
The maximum observed value of ‘s’ is 10. Sam must
have answered 10 questions at least on one day.
The minimum he answers on any day is 5. From
Monday to Thursday there are 14 queries from the
previous week and 10 from that week. Hence in all
there are 24 queries. So there is no way he could
have answered 10 queries on any day from Monday
to Thursday. So he must have answered the
maximum (10) number of queries either on Friday or
on Saturday.
6. b
The queries received on Saturday have to be all
answered on Saturday. On Thursday Sam has to
answer atleast 1 query that was received that day
and on Friday Sam has to answer at least three
queries that were received on that day. All the other
queries can definitely be answered with atleast a
day’s delay.
Hence the minimum number that has to be answered
on the day of receipt is
(4) + (5) + (1) + (7 + 3 + 1) = 21.
For Example, they can plan their schedule in the
110
X=Y
100
90 Y
90 110
=
×
100 X 100 100
Statement I alone is sufficient.
10 X 10 10
Y
=
×
100 11 100
Similarly, statement II alone is sufficient.
From statement II,
23. d
Combining I and II, we must have 6 silver and
2 bronze coins. No other possibility is there.
24. d
Statement I is necessary and statement II is also
needed as if we take:
following manner:
x 1
= <1
y 2
c = 98
∴
x − 98 −97 97 1
=
=
>
y − 98 −96 96 2
∴
x− c 1
>
y−c 2
This defeats the given case. Thus, both statements
I and II are needed. ∴ Answer is (d).
25. e
The IInd statement tells us that triangle QPR is a right
angled triangle but the information is given only of
perpendicular distance QR, no information is given
about the base, so even by combining both the
statements we cannot get the length of PR.
It cannot be answered using both the statements.
∴ Answer is (e)
Page: 46
If the Republicans win 97 seats, percentage of
(97 − 65)
=8
4
Democrats win 90 – 8(3) = 66 seats.
For family B, light =
Mon
8 by
DI
Anu
6 by
LR
Dudi
Tue
Wed
18 by
Anu
6 by
Dudi
18 by
Anu
6 by
Dudi
Thu
10 by
Anu
6 by
Dudi
QA
2 by
Tiru
15 by 15 by 15 by
Tiru
Tiru
Tiru
VA
5 by
Sam
5 by
Sam
5 by
Sam
Fri
4 by Anu
1 by Dudi
Sat
(4)* + 2
by Anu
(5)* + 7
by dudi
6 by Tiru, 5
(1)* + 9
by Dudi and
by Tiru
4 by Anu
4 + (1)* (3)* + 4 by
by Sam Sam
(7)* + 3
by Sam
( )* indicates queries received on the same day.
MBA
Test Prep
Solution Book-2
7. d
8. e
9. c
Now from D there are again 6 ways by which A can
reach his office Ao
If all conditions have to be satisfied on each day
atleast 21 queries have to be answered.
The number of incoming queries on M, W, Th, Fri,
Sat are all less than 21, hence there must definitely
be some carry over from the previous day in the
week for each of these days.
Route
DH-EO-CO-DO-AO
DH-EO-DO-CO-AO
DH-CO-EO-DO-AO
DH-CO-DO-EO-AO
DH-DO-CO-EO-AO
DH-DO-EO-CO-AO
Number of queries of the beginning of the week
= 103.
Number of queries received during the week = 107.
Hence number of queries answered
= (103 + 107) – 20 = 190
In all DI has (24 + 40) = 64 queries to be handled.
Similarly LR has 37 queries to be handled.
QA has 72 queries to be handled.
VA has 37 queries to be handled.
Therefore minimum distance that A can travel is
40 + 52 = 92 kms.
11. b
Now range for dudi is 6 ≤ d ≤ 12; so at least on one
day he answered 12 queries, so minimum number
of queries answered by Dudi in the week
= 6 × 5 + 12 = 42.
42 ≤ Number of queries that can be answered by
Tiru in a week ≤ 77
Similarly, 58 ≤ Anu ≤ 98 and 35 ≤ Sam ≤ 55.
If Dudi answers the LR queries than he has to move
to other sections viz. either DI or QA.
So, atmost queries of three sections could be
handled by one single respondent.
The three friends whose houses are at a minimum
possible distance from the house of A are C, D and
E.
There are 6 ways by which A can visit C, D and E.
Route
AH-EH-CH-DH
AH-CH-DH-EH
AH-CH-EH-DH
AH-DH-CH-EH
AH-DH-EH-CH
AH-EH-DH-CH
12 kms
13 kms
18 kms
12 kms
17 kms
19 kms
12. a
BH
→ DH
→ DO
→BO
15km
12km
15km
@Rs.1/km
@Rs.3/km
@Rs.2/km
+
15
+
36
= Rs. 81
30
13. b
For questions 10 to 13:
10. d
AH-AO
BH-BO
CH-CO
DH-DO
EH-EO
FH-FO
Now combining each of the above distances with
the distances given in table 2. i.e. Distance of AH
from BH, CH, DH, EH and FH and so on, we find that
for C, E and F the distances of their house from
their respective offices is not less than the distance
from the houses of any of his friends.
Dudi in a week ≤ 66.
As explained above, minimum number of queries
answered by Tiru in the week = 25.
25 ≤ Number of queries that can be answered by
Distance (kms)
54
53
52
59
60
52
Distance (kms)
48
44
40
45
39
46
Now, DH – E0 < EH – E0 and CH – E0
DH – C0 < EH – C0 and CH – C0
DH – D0 < EH – D0 and CH – D0
and for AH – CH – EH – DH the distance travelled is
40 Kms.
Explanations: Fundamentals of
Logical Reasoning & Data Interpretation
The shortest route can be found through iteration
12km
13km
14km
16km
14km
13km
D→E→F→B→C→A→D
In all it takes 82 kms of travelling (or) Rs. 82/-
For questions 14 to 17: The following values have been
taken from the graphs for calculation
Year
1995
1996
1997
1998
1999
2000
Value (Rs Cr)
400
310
150
200
590
750
Quantity (Mn kg)
205
175
130
150
265
420
14. c
Value per kilogram for:
1995 = Rs. 19.50
1996 = Rs. 17.71
1997 = Rs. 11.53
1999 = Rs. 22.26
Thus, it is the lowest for 1997.
MBA
Test Prep
Page: 47
15. c
16. e
Increase in value of jute exports for
1996 = –22.5%
1998 = 33.3%
1999 = 195%
2000 = 27.11%
The graph shows that the steepest incline is from
1998 to 1999.
Hence, percentage change is maximum in 1999.
24. d
Statement I indicates the time taken by Ravi to eat
24 eggs. Statement II indicates the time taken by
Harish to finish eating 24 eggs. Together they give
the answer.
25. e
From statement I, we get that there are balls of
2 colours. This gives the probability of selection of
Refer solution to question 14.
2000 = 17.85
1998 = 13.33
4
. However, it has no
11
information about the total number of balls.
From statement II, we do not get any information
about the total number of balls.
the ball of other colour as
Alternative method:
Average Price =
19.50 + 17.71 + 11.53 + 13.33 + 22.26 + 17.85
6
= 17.75
Practice exercise – D9
1. c
17. b
The price per kilogram has decreased in 1996, 1997
and 2000, has been worked out in questions14 and
16.
100
42137
Students please note that to calculate the exact
value of this expression, we need calculator. Since
options given are not very close to each other, we
can approximate the values.
And using approximations, we get the value of the
100
2650
=
required ratio = (68600 – 42000) ×
42000
42
= 63%
= (68718 – 42137) ×
Alternative method:
For 1995 cannot be determined, but otherwise
growth is the highest in 1999 from 13.33 to 22.26.
18. e
1.25 × 10–3 times the world’s fresh cut flower exports
= Rs. 100 lakh
So the value of the world export of fresh cut flowers
=
19. c
100
1.25 × 10
21. c
22. b
23. e
lakh;
100 × 10 3
1 . 25
lakh = 8 × 104 lakh
2. c
20% of ‘Others’ = 4% of world’s export of cut
flowers in 1990-91
= 17 lakh. So 4% of x = 17 lakh.
⇒x=
20. e
−3
17
× 100 = 4.25 crore
4
Netherlands may be exporting flowers other than
cut flowers.
Using statement I and II we can independently find
out the unknown side of rectangle which comes
out to be ‘6’. So perimeter can be calculated by
making use of both statements independently.
∴ Ans (c)
From statement I, a + b = 3k but nothing can be said
about C. Therefore, statement I is not sufficient.
From statement II, c = 3p
⇒ 3(a + b) + c = 3(a + b) + 3p = 3(a + b + p) = 3q
⇒ 3(a + b) + c is divisible by 3.
Hence, statement II alone is sufficient.
The volume is unknown and the cost of each
individual element except zinc is unknown. Cost of
copper cannot be found.
Page: 48
Required percentage growth
Boo ks
1975
1980
Percentage
g row th
P rim a ry
4 21 37
6 87 18
6 6%
S e co nd a ry
8 82 0
2 01 77
1 28 %
H ig h er
secon da ry
6 53 03
8 21 75
2 6%
G ra du ate
le vel
2 53 43
3 66 97
4 4%
Hence, percentage growth is least for higher
secondary books, viz. 26%.
3. b
Again referring to the above table, we can see that
the percentage growth rate is maximum for
secondary level books, viz. 125%.
4. d
It can be seen from the given table that though
primary level books have shown a consistent
growth, it has declined in 1978. On the other hand,
even secondary and higher secondary level books
have shown a consistent increase except for 1977
when it had declined. But the graduate level books
have shown a consistent growth over the period.
MBA
Test Prep
Solution Book-2
7. d
For questions 5 to 8: Let us refer these hawkers as J, R,
S and P respectively in place of Jayram, Rajaram, Sitaram
and Peariram. From information (II), we know that P is selling
Chatpata. From information (I), we gather that R do not sell
either Fruits or Plastic Toys. So R must be selling
Newspapers. So J and S are selling Fruits and Plastic Toys,
not in any particular order.
Regarding their revenue earned, we know that
J + R + S + P = 1200.
But from information (IV), S = 250.
Therefore J + R + P = 950, or R + P = (950 – J)
…(i)
The amount to be paid to the hooligans by J and S
are Rs. 35 and Rs. 62.50 respectively. In order to
maximise their total ‘take-home’ revenue, the amount
to be paid by R and P should be considered. We can
observe that P has to pay a lesser percentage as
compared to R. So we maximise the revenue earned
by P at Rs. 249, (corresponding revenue earned by
R is Rs. 351), to get the minimum amount to be paid
to the hooligans. We can compile the result in the
following table.
Hawkers
Jayram Rajaram Sitaram Peariram
Revenue
350
351
250
249
earned (Rs.)
Total
1200
From information (III), we can say that
R+P
+ 50 = J
2
…(ii)
Solving (i) and (ii), we get that J = 350 and (R + P) = 600
From information (I), we know that R > 300. But if, say
R = 301, then P = 299, which is greater than 250. But from
information (II), P earned the minimum revenue. So, P should
earn less than S (Rs. 250).
So, we can say that R > 350 and P < 250. We can compile
the following table with this derived conclusion.
Hawkers
Jayram
Rajaram
Sitaram
Peariram
Items
Fruits/
Plastic
Toys
News
papers
Fruits/
Plastic
Toys
Chatpata
Revenue
earned (Rs.)
350
>350
250
<250
5. b
6. e
According to the question, S + P = R.
We know that S = 250 and P = (600 – R)
Solving we get, R = 425. Since Rajaram sold
Newspapers, the answer is (b).
Amount to be
paid to the
hooligans
(Rs.)
35
70.2
62.5
37.35
205.05
Take-home'
revenue (Rs.)
315
280.8
187.5
211.65
994.95
So, (d) is the right answer.
8. c
As integral number of Chatpatas’ were sold so (600
- Rajaram’s salary) should be divisible by 7.
Option (c) i.e. (600 –494) = Rs.106 is not divisible
by 7. So Rs.494 cannot be the revenue earned by
Rajaram on that day.
9. b
180 − 140
= 6.67 million tonnes (Approximately)
6
10. a
In 1997, kharif crop production is
110
× 99 = 108.9 million tonnes
100
And rabi crop production in the same year is
We know that either J or S is the Fruit-seller. J sold
items worth Rs. 350 and S sold items worth Rs.
250. If one orange was sold at Rs. 2.50 per piece,
then either 140 or 100 oranges were sold on that
day by one of these hawkers.
116
× 81 = 93.96 million tonnes
100
∴ Total production increases by
108.9 + 93.96 − 180
× 100 = 12.7%
180
11. c
Given that 83% equals 110. So 100% should
110
× 100 = 132.5 million tonnes(Approx.)
83
So quantity of food grains India should have imported
= 132.5 – 110 = 22.5 million tonnes (approximately)
equal
Explanations: Fundamentals of
Logical Reasoning & Data Interpretation
MBA
Test Prep
Page: 49
For questions 12 to 16:
14. c
The maximum number of times when all the three on
the podium can be from Ferrari is 6. For example, in
all races from 4 to 9 the 2nd position is occupied by
Ferrari.
The minimum could be 0, when Ferrari occupies
2nd position in races say 1 to 3 and 10 to 12.
15. c
Total number of points obtained by Ferrari team
= 9 × 10 + 6 × 9 + 9 × 8 = 216
McLaren team = 5 × 10 + 2 × 9 + 5 × 8 = 108
Renault team = 3 × 10 + 0 × 9 + 3 × 8 = 54
Let’s assume the remaining podium positions were
won by a single team; Y, then
Y = [18 – (9 + 5 + 3)] × 10 + [18 –(6 + 2 + 0)] × 9 +
[18 – (9 + 5 + 3)] × 8
= 1 × 10 + 10 × 9 + 1 × 8 = 108
∴ The minimum difference in points between the
top two teams = 216 – 108 = 108
16. a
From the table it can be seen that all the three can
never be simultaneously on the podium.
2
rd of the races i.e. 12 races.
3
Hence in (9 + 9) – 12 = 6 races it held both the 1st and 3rd
positions.
Ferrari was on the podium in
McLaren was on the podium in
2
rd of the races i.e. 12
3
races.
Hence since 5 + 5 + 2 = 12 in no race did it hold more than 1
position.
2
th of the races i.e. 4 races.
9
Hence it could have held the position (1, 3); (1, 3); (1); (3).
Based on this the following chart could be drawn:
Renault was on the podium in
Race
Ist
18
R
IInd
IIIrd
17
R
R
16
R
M
15
Others
M
R
14
M
R
13
M
Others
12
M
F
11
M
F
10
M
F
9
F
F
8
F
F
7
F
F
6
F
F
5
F
F
4
F
F
3
F
M
2
F
M
1
F
M
The above chart represents one of the options in the races.
Every other option would also satisfy the same condition.
Ferrari in the 2nd position can range anywhere from race 1
to race 12 but not beyond it (since Ferrari held the podium in
only 12 races)
12. c
13. b
From the table we can see that there would be 6
races when both Ferrari and McLaren would
definitely be on the podium.
Renault had podium positions in 4 races.
It won 6 podium positions.
Hence it is possible only in the following cases:
(1, 3), (1, 3), (1), (3). Hence the answer is (b).
Page: 50
For questions 17 to 20: Suppose that the selling price is
Rs. 100.
17. a
25% (of 30% (of 20%)) = 1.5%.
18. c
6 lacs = 0.6 million
19. a
Appraisal cost = 6 million
⇒ Product Testing cost = 3 million.
20. b
Error cost = 20% of quality cost
=
1
× (20% of total cos t )
5
=
1 1
× = 4% of total cost
5 5
21. c
Statements I and II both individually give a relation
between the capacities of water barrels A and B.
22. c
From statement I, we easily get the answer.
Since we know both the distance and speed.
From statement II, x + 50 →
x→ t
xt = (x + 50)
t
2
t
= 45
2
∴ xt = 45
45
+ 25t = 45
2
∴ t = 54 min
Thus, we can find time from statement II also
independently. Thus, (b).
MBA
Test Prep
Solution Book-2
23. e
From statement I,
1 1 1
+ =
A C 8
Case I:
Vc
R
From statement II, 1 + 1 = 1 . Here we have two
B C 6
equations and three unknowns. Hence, we need
one more equation.
24. a
Statement I clearly shows that x is either zero or
negative, i.e. not positive. We do not get any
information from statement II. Thus, (a).
25. c
From statement I, the SP of a pair can be obtained
while the CP can be obtained from statement II.
Practice exercise – D10
1. d
All are true.
2. e
New demand is 114% of 175000 = 199500 pairs.
So demand supply gap
= 199500 – 175000 = 24500 pairs
3. a
Actual demand in 1995-96 is 110% of 190
= 209000 pairs
Actual production in 1995-96 is 85% of 190
= 161.5 thousand pairs
So (Demand – Supply) = 209 – 161.5 = 47.5 × 1000
= 47500 pairs
S
If both C & D are moving in the anticlockwise
direction, then
… (ii)
Vc – Vb = ± 40
Vb + Vd = 60
… (iii)
Adding (i) and (ii)
Va + Vc = 90 or 10 … (iv)
Adding (iii) and (iv)
Va + Vb + Vc + Vd = 150 or 70
Therefore, the average of the speeds of these cars
is:
150
70
0r
i.e. 37.5 mph or 17.5 mph
4
4
Case II:
R
If B’s speed is 20 mph.
Then A’s speed = 30 mph
C’s speed = 20 mph
D’s speed = 80 mph
Ratio of their speeds = 3 : 2 : 2 : 8
At that instant A, B, C and D passed through the
points P, Q, R and S. As they passed these points,
all of them were moving parallel to each other. Let
their speeds be Va, Vb, Vc and Vd respectively.
At that instant A & B were moving in the same
directions (both clockwise), so we must have.
Va + Vb = 50
…(i)
Similarly C and D also moved in the same directions
(either both in the clockwise direction or both in the
anticlockwise direction) and their speeds with
respect to B are 40 mph and 60 mph respectively.
Explanations: Fundamentals of
Logical Reasoning & Data Interpretation
Vd
F ig . C & D , b oth anti-c lock w ise
Q
Vb
5. a
P
Vb
For Questions 4 to 7:
4. c
Va
Q
Vd
Vc
Va
P
S
If both C & D are moving in the clockwise direction,
then
Vd – Vb = +60
…(v)
…(vi)
Vb + Vc = 40
Solving, we get the value of the average of the
speeds of the four cars i.e. 37.5 mph.
Hence (a) is the only possible choice.
MBA
Test Prep
Page: 51
6. b
C observes that D is moving at a speed of 20mph.
As C and D are moving in the same direction we
must have:
Vc + Vd = 20
From this equation, the maximum values of either
Vc OR Vd can be 20mph only irrespective of the
direction of C & D
...(i)
Again, there are two cases:
Case I: Both C and D drive in the clockwise direction.
…(ii)
Vb – Vc = +40
Vb + Vd = 60
…(iii)
Case II: Both C and D drive in the anticlockwise
direction.
…(iv)
Vb – Vd = 60
Vb + Vc = 40
…(v)
Case II is impossible.
As all the speeds are either positive or zero,
equations (ii) and (iii) must be changed as following:
Vb – Vc = 40 or Vb = Vc + 40 …(vi)
{since Vc ≤ 20}
Vb + Vd = 60 or Vb = 60 – Vd …(vii)
{since Vd ≤ 20}
From both equations, we get
40 ≤ Vb ≤ 60.
Out of the given options, only option (ii) lies in the
given range. Hence (b) is the correct choice.
7. d
As E is stationary, the speeds observed by E, are
the actual speeds of the cars. Let both the drivers
C and D were driving with a speed of V mph. As C
and D have the same speed, we need not find the
solution under different cases as we did in the
previous problems. We can write the following
equations:
…(i)
Va — Vb = ± 50
…(ii)
Vb + V = 60
V — Vb = ± 40
…(iii)
Adding (ii) and (iii) we get:
…(iv)
V = Vc = Vd = 50 or 10
Subtracting (iii) from (ii) we get:
…(v)
Vb = 10 or 50
Putting both these values of Vb in (i), we get
Va = 50 ± 50 or 10 ± 50 i.e.
Va = 100, 0 or 60, – 40
As Va represents speed, it cannot be negative,
hence Va = –40 is discarded.
…(vi).
Va = 100, 0 or 60
From (iv), (v) and (vi) we observe that Vb and Va
do not have unique values rather they have more
than one allowed value. As a result (Va, Vb, Vc,
Vd.) can have any one of the following combinations
of values:
Page: 52
Case I: (Va, Vb, Vc, Vd.) have values (0,50, 10,10)
OR
Case II: (Va, Vb, Vc, Vd.) have values
(100,50, 10,10)
OR
Case III: (Va, Vb, Vc, Vd.) have values
(60, 10, 50,50)
Options (a), (b) and (c) are not correct. If we check
the options, only option (d) is correct. As V = 10
b
and V = 60 (more than double) is a possible
a
combination of speeds.
8. b
IT has the lowest acquisition value US $21.6 m.
9. b
For 2001 - 03, average acquisition
$ 1.6 b
= $ 13.3 m
120
So 3 companies were acquired at lower than
average cost - Expert Information Services,
Alpharma, Dashiqiao.
cost =
10. d
The Australian companies were acquired at a higher
value as compared to the others,
Total = US $ 59.5 m
Next is USA = US $ 39. 8 m, then UK and then China.
11. e
London Stock Exchange has more than 15
companies listed. But how many exactly is not
known. Also, a company may be listed in more than
one international stock exchange.
So we have no data to confirm that these more than
25 companies are all different.
12. c
Top 3 acquisition value = US $ 386.4 m
The total acquisition value for top 9 acquisitions in
the period = US $ 461.5 m.
Hence ratio = 5 : 6
For questions 13 to 15:
The following table can be made:
(all figures in rupee crores)
2004
2005
Revenue Profit % Profit Revenue Profit % Profit
A
4600
15
600
6720
12
720
B
3900
30
900
5000
25
1000
C
2700
35
700
3840
28
840
D
6600
10
600
5000
25
1000
E
7700
10
700
8800
10
800
By the given information, A is Dixons and D is Eletropaulo.
13. a
Here B is Coleco. The second highest profit in 2004
is Rs.700 crore. So both Alta Vista and Bultaco is
the answer.
MBA
Test Prep
Solution Book-2
14. b
15. d
Here C is Alta Vista. The highest profit earned by
any Company in 2005 is Rs. 1000 crore.
So, from the choices, Coleco and Electropaulo is
the answer.
For questions 18 to 20:
Designation
Manager
Assistant
Manager
Executive
Salary
15x
12x
8x
Name
Anny
Bobby
Citra
Expenditure
5y
4y
3y
% increase in profit over 2004 of A:
120
× 100 = 20%
600
% increase in profit over 2004 of B:
=
100
× 100 = 11.11%
900
% increase in profit over 2004 of C:
=
18. d
140
× 100 = 20%
700
% increase in profit over 2004 of D:
Anny does not draw the maximum and the Executive
does not spend the minimum. Hence, Anny is not
the Manager and Citra is not the Executive.
Anny
=
Manager
Assistant
Manager
400
× 100 = 66.67%
600
% increase in profit over 2004 of E:
=
X1 − X5
X1 − X6
16. b
17. b
X 2 − X5
X2 − X6
X2 − X 4
X2 − X 4
×
Now if we analyse the options,
Option (a) is wrong, since Citra cannot be the
Executive. Hence, our conclusion can never be
reached.
Option (b) says that Bobby is the Manager. Since
Citra cannot be the Executive, she has to be the
Assistant Manager and therefore our conclusion is
contradicted.
Option (c) says that Citra is not the Assistant
Manager. That means Citra is the Manager. Now,
either of Anny or Bobby can be the Assistant
Manager. Hence, the information in this option is not
sufficient for us.
Option (d) tells us that Bobby is the Executive. Since,
Anny cannot be the manager, she must be the
Assistant Manager. So this option is the correct
choice.
For questions 16 and 17:
A total of 5 different arrangements is possible
X1 − X3 X2 − X5 X4 − X6
X1 − X4
Citra
Executive
100
× 100 = 14.28%
700
Here E is Bultaco.
So C is either Coleco or Alta Vista.
Hence, the data is insufficient.
X1 − X4
Bobby
×
X3 − X 6
X3 − X5
X3 − X 6
X3 − X5
From the various possible arrangement shown
above, there are two possible arrangements for
(X1, X4).
19. c
If the pairs (X1, X3), (X1, X5), (X1, X6), (X2, X6)
or (X4, X6) are given, then all other pairs can be
determined,
while if the pairs (X1, X4), (X2,X4), (X2,X5), (X3,X5)
or (X3, X6) are given, then all other pairs can’t be
determined.
Therefore required probability
Total number of favourable cases
5
=
=
Total number of possible cases
10
Designation
Manager
Name
Anny
Assistant
Manager
Citra
Salary
15x
12x
8x
Expenditure
5y
3y
4y
Savings
15x – 5y
12x – 3y
8x – 4y
Manager spends the maximum. Hence, Anny is the
Manager.
Since Bobby is the executive, hence Citra must be
the Assistant Manager
Conclusion says that Citra saves the maximum.
Savings = Salary – Expenditure
Executive
Bobby
Bobby’s savings = 8x – 4y
Bobby earns less than Citra and spends more than
her.
Explanations: Fundamentals of
Logical Reasoning & Data Interpretation
MBA
Test Prep
Page: 53
So, her savings are certainly lesser than that of
Citra. Since, according to the conclusion drawn,
Citra saves the maximum, we have
(12x – 3y) > (15x – 5y) ⇔ 2y > 3x
Therefore, if with the help of certain additional
information, we can establish 2y > 3x, that will be
the right answer choice.
We have logically deduced that Anny is the Manager
and Citra is the Assistant Manager. So, options (b)
and (d) can be out rightly rejected.
In option (a), Anny’s savings is twice her
expenditure. Therefore, Anny’s salary must be three
times her expenditure.
⇒ 15x = 3 × 5y
⇒ x = y and hence Citra is not saving the maximum.
In option (c), Citra’s salary is twice her savings.
Hence, her salary is also twice her expenditure.
⇒ 12x = 2 × 3y
⇒ y = 2x
⇒ 2y > 3x and hence this information is correct as
well as sufficient to lead us to the stated conclusion.
So, option (c) is the right choice.
20. a
Therefore, Anny is the Assistant Manager and thus,
Bobby is the Manager.
Designation
Name
Salary
Expenditure
Manager
Assistant Manager Executive
Bobby
Anny
Citra
15x
12x
8x
4y
5y
3y
29
29
16
Savings
x or
y
x or 4y
4x or 3y
3
4
3
Conclusion says that savings of Executive is Rs.
12,000 per month. Hence, 4x = 3y = 12000. Assuming
the stated conclusion to be true, we can compile
the following table.
Designation
Manager
Manager
Salary
15x
Assistant
Manager
12x
Name
Bobby
Anny
Citra
45000
36000
24000
Expenditure
16000
20000
12000
Savings
29000
16000
12000
Option (a) says that Executive spends Rs. 12000
per month. This enables us to find the stated
conclusion.
Option (b):Bobby is the Manager but monthly salary
of Citra, if Rs. 36000 per month would mean that 8x
= 36000 and hence, x = 4500 (misleading)
Option (c): Anny is the Assistant Manager but if she
saves Rs. 20000 per month, 4y = 20000 and hence,
y = 5000 (misleading)
Option (d): Manager’s salary = 15x
Anny’s expenditure = 5y
It gives x = 1600 (misleading)
So, option (a) is the correct choice.
Executive
8x
Citra’s expenditure = 3y
Case I:
Citra is the Manager
⇒ 3y =
1
× 15x
2
⇒ 2y = 5x
From the first condition the number can be
97
79
88
From the second condition we get
11
22
33
:
99
∴ From I and II combined we get number as 88.
∴ Answer is (d).
22. a
Statement I implies that y is odd since the only way
for the average of x and y to be a whole number is
that x + y must be even. Statement II is not sufficient
since either y or y + 1 must be even.
Hence, x + y + 1 must be even.
23. e
Since the ball that has been picked up is not
identified, the probability cannot be found.
1
× 8x
2
3y = 4x
If Citra happens to be the Manager,
Then, Executive’s savings is either equal to
(8x – 4y) or (8x – 5y), which can never be negative.
But, if Citra is the Manager, then 2y = 5x.
Therefore, both (8x – 4y) and (8x – 5y) become
negative quantity, which is not possible.
Hence, Citra has to be the Executive.
∴ Citra’s savings = 8x – 3y = 6y – 3y = 3y
This saving is 25% less than the savings of the
Assistant Manager.
Hence, Assistant Manager must be saving 4y.
⇒ 12x – Assistant Manager’s expenditure = 4y
⇒ 9y – Assistant Manager’s expenditure = 4y
⇒ Assistant Manager’s expenditure = 5y
Page: 54
21. d
Case II:
Citra is the Executive
3y =
Executive
Salary
Citra’s savings = Citra’s expenditure = 1 × Citra’s
2
salary
Citra’s savings is 25% less than that of Assistant
Manager. Therefore, Citra is not the Assistant
Manager.
Designation
Assistant
Manager
MBA
Test Prep
Solution Book-2
24. e
Statement I gives probability of A winning but here
we do not know the number of contestants.
Statement II provides the information that there are
only 2 contestants but we do not know the probability
of a tie. Thus, answer is (e).
25. e
In statement I, 2500 includes families who have
neither of the devices and no data is given about
such families. Hence, even statement II is not useful.
5. a
Practice exercise – D11
1. b
The most obvious one is rank number 5. Since H
has scored less than each of A, B, C and J in all the
three sections, its total also has to be less than
these four students. By the same reasoning, total
marks of H have to be more than D, E, F, G and I.
Thus H is ranked 5th.
Alternative method:
‘A’ sales has been more than all other manufacturer
in all the years except 1995.
6. d
Among, A, B, C and J, J could be at any of the ranks
1 to 4 i.e. J could be greater or less than each of A,
B and C. Hence one cannot identify who ranks 1 to
4.
By same logic between E, F, G and I, student E
could be at any position from rank 7 to 10. Thus the
student at rank 7 to 10 also cannot be identified.
2. a
The minimum difference between marks scored by
F and C in quant can be 10 and this will happen only
when the difference between consecutively ranked
students (F & D; D & G; G & H; H & B; B & C) is 2.
(Observe that the sum total of the numbers in the
denominator would be higher than ‘five times 410’
and hence, the required percentage would be a
little less than 20%.)
7. c
It can be easily observed that, since the difference
is the highest in case of manufacturer C, the largest
percentage growth would naturally occur for him,
as the base is the smallest.
8. e
Required ratio = (400 ÷ 500) = 0.8
9. d
The figure for the highest sales of scooters over
the period shown is 520 units which occurs in 1999
in case of the manufacturer A.
For questions 10 to 12:
Only those cities that have three roads emanating from
them can be start/end city.
10. b
11. b
The minimum marks that A can score is 19 in Quant,
20 in EU and 22 in DI i.e. a total of 61. The maximum
marks that A can score is 25 in Quant, 23 in EU and
25 in DI i.e. a total of 73. Thus the required difference
is 12 marks.
4. e
As explained in answer explanation to first question
of this set, the student at rank number 7 cannot be
found as E could be ranked anything from 7 to 10.
Explanations: Fundamentals of
Logical Reasoning & Data Interpretation
Three routes starting from 2 and 4 each.
12. d
13. c
3. e
Required share
410
=
× 100 ≈ 19.5%
480
410
390
380
440
+
+
+
+
D’s is definitely more than the total of F and I (as D
has scored more in each of the sections compared
to these students). Also D cannot be less than G
(worst case scenario for D, as compared to E, is
Quant -5, EU +4 and DI +2). Similarly D cannot be
less than E. Thus 6th rank will be D.
The other close option will be rank 10. However
one cannot conclude that F will come last. E and F
both could be tied for the rank 9 or 10. Consider
worst case scenario for E and best case scenario
for F. Thus E will score 8 more than F in EU and 2
more than F in DI. But F can make up for this
difference by scoring 10 more than E in quant. Thus
for rank 10, one cannot identify the student.
The total sales over the period shown for different
manufactures are given as follows:
(a) For A = 440 + 480 + 470 + 500 + 520 + 510
= 2920 × 1000 units
(b) For B = 400 + 410 + 415 + 415 + 420 + 430
= 2490 × 1000 units
(c) For C = 380 + 390 + 390 + 400 + 420 + 495
= 2475 × 1000 units
(d) For D = 360 + 380 + 400 + 415 + 440 + 500
= 2495 × 1000 units
(e) For E = 480 + 440 + 440 + 420 + 425 + 435
= 2640 × 1000 units
Thus, the sales of the manufacturer A is the highest.
Total number of line employees with 3 to 5 years of
employment = 140
Out of the above, the number that attended training
on financial management = 80
∴ The number of line employees with 3 to 5 years
employment who did not attend training on financial
management is (140 – 80), i.e. 60.
MBA
Test Prep
Page: 55
14. b
15. a
Number of employees with less than 3 years of
employment who attended training on decisionmaking alone = (40 – 10) + (30 – 15) = 45
Similarly, the number of employees with less than
3 years of employment who attended training on
financial management alone
= (30 – 10) + (20 – 15) = 25
Answer = 45 + 25 = 70.
Number of line employees with more than 5 years
of employment who attended at least one
programme = 50 + 40 – 30 = 60
Number of staff employees with more than 5 years
of employment who attended at least one
programme = 40 + 50 – 20 = 70
Total number of employees with more than 5 years
of employment who attended at least one
programme = 60 + 70 = 130
Percentage of employment with more than 5 years
of employment who did not attend either workshop
=
Statement (A) indicates that the chief guest spoke
longer than 25 min but does not say how much
longer than 25 min. So statement (B) indicates that
the chief guest spoke for lesser than 35 min.
Hence, we can infer from statement (B) alone that
the chief guest did not speak for more than 45 min.
24. e
It is not possible to answer the question. Although
(c) might seem to be a tempting offer, it is not,
because we do not know whether Rashmi will go
shopping if Ranjana goes.
25. d
Both the statements are required, since from the
statement (A), we get
A +B+C
B+C
or 3A > A + B + C or A >
3
2
From the statement (B), we get
A>
A +B
or 2B > A + B or B > A
2
Combining both the statements, we get
B>
(200 + 160) − 130
230
× 100 =
× 100 = 64
200 + 160
360
16. e
(50 + 40) + (40 + 50) – (30 + 20) = 130
17. d
Middle management use highest percentage of
spread sheet, which is 53 out of 150
=
23. b
B+C
< A C
∴ ‘B’ weighs most.
Alternately: Statement (A) says there has to be at
least one of B or C or both B and C less than A.
Statement (B) says B is more than A.
So combining we can say the order has to be
C, A, B.
53
× 100 = 35.3%
150
18. e
Option (e). Answer is 202.
19. b
Top management have second highest proficiency
in spread sheet. Just check the second highest
value from the four of those given in the table.
Practice exercise – D12
20. a
42 + 23 = 65
1. b
21. d
From statement (A), he cannot watch a news
channel at night.
From statement (B), he can watch an entertainment
channel only in the evening and night (combining
with statement A).
Also note, he watches only one type of channel in
each of 4 parts of the day. Hence (d).
22. b
Statement (A) is irrelevant. It only says Pakistan is
the eventual winner. But nothing is said about points.
Statement (B), explains the point system.
Based on this, the maximum points possible to be
scored by the winning team is (3 × 3 matches) = 9
points. Hence (b)
Page: 56
79.37 − 68.75 80 − 70 1
−%
= = 14.28%
70
7
68.75
2. e
Demand = 1.15 × 102 = 117 MMT
Gap = 117 – 98 = 19 MMT
3. e
It can be checked by multiplying the All India supply
figures by 0.3 and comparing it with southern India
supply figures. No year satisfies that.
4. b
Supply = 22 × 1.3 = 28.6 MMT
Demand = 28 × 1.4 = 39.2 MMT
Deficit = 10.6 MMT
MBA
Test Prep
Solution Book-2
For questions 5 to 8:
Each of the participants received at least one vote in round
1. If the minimum number of votes received is 2, then 4 of the
11 votes are accounted for, since two of the contestants
were tied for the last place in Round 1. Payal has received
4 votes. Taking these votes into consideration, 8 votes are
accounted for. The remaining 3 votes can be divided among
the other two participants as either (3 + 0) or (2 + 1), both of
which are not possible. (3 + 0) is not possible because
each participant has received at least 1 vote and (2 + 1) is
not possible because we have considered the lowest number of votes as 2.
If the minimum number of votes received by two participants is 3 each, then including Payal’s 4 votes, 10 out of 11
votes would have been accounted for and therefore the
remaining two participants cannot receive at least 1 vote
each.
Therefore, the only possible combination is when two
participants receive 1 vote each (the minimum), Payal
receives 4 votes while the other 2 participants receive 3
and 2 votes respectively.
Further, one of the participants has received ‘0’ votes in
round 2.
i.
That participant cannot be Priti because she has received 1 vote in round 2.
ii.
That participant cannot be Priyanka because Mr. Biyani
has voted for her in round 2.
iii. That participant cannot be Payal because the judge
who voted for Poonam in round 1 voted for her in
round 2
iv. That participant cannot be Pooja because 50% of the
judges who
voted for Payal in round 1 voted for
Pooja in round 2.
v. Therefore, it is Poonam who got ‘0’ votes in round 2.
Further, it is given that the judge who voted for Poonam in
round 1 voted for Payal in round 2. Therefore, Poonam
would have got 1 vote in round 1. Also, Payal would have
got 3 votes in round 2. (50% of votes from earlier round and
1 vote of the judge who voted for Poonam in round 1).
The remaining 8 votes were divided between Pooja and
Priyanka. For this to be possible and Priyanka to be joint
second with another person, the only possible combination
in round 2 can be:
Priti : 1 vote, Payal : 3 votes, Priyanka : 3 votes, Pooja : 5
votes
From condition II, Pooja got 2 additional votes in round 2.
Therefore Pooja would have got 3 votes in round 1. Priyanka
got 1 additional vote of Mr. Biyani in round 2 and ended with
3 votes. Therefore Priyanka would have got 2 votes in
round 1.
Explanations: Fundamentals of
Logical Reasoning & Data Interpretation
The number of votes received after the first 2 rounds were
as follows:
Pooja
Payal
Priti
B'lore
Delhi
B'lore
Priyanka
Poonam
Round 1
3
4
Round 2
5
3
1
2
1
1
3
0
Since Poonam was did not contest in round 3, Priyanka
must be the other girl from Delhi. Further, the total number of
votes in Round 3 is 13. (Poonam will also vote). From
condition IV, the total votes won by Pooja and Priti (two girls
from Bangalore) will be 7 while Payal and Priyanka (two
girls from Delhi) together secured 6 votes in Round 3.
5. b
If Priyanka received 2 votes in round 3, then Payal
would have received 4 votes in round 3, because
both of them together received 6 votes in round 3.
6. e
Pooja was the person with the highest votes at the
end of round 2.
7. d
If Priti received 3 votes in round 3 and Payal received 50% of the remaining votes which is 5, then
Pooja received 4 votes in round 3 while Priyanka
received 1 vote in round 3. Therefore, option (d) is
definitely true i.e. Priyanka received the minimum
number of votes in round 3.
8. d
Priti and Poonam are the 2 contestants who received the minimum number of votes (1 each) in
round 1.
For questions 9 to 11:
Iqbal is watched by Bimal only. Beckham watches 3 movies
— Hence, does not watch Iqbal which is screened at plaza.
Movie
Salaam Namaste Iqbal
Movie Hall
Satyam Chanakya
Priya
Plaza
Bunty
Ö
Ö
´
´
Babli
Ö
Ö
´
´
Bonny
Ö/´
Ö/´
Ö/´
´
Bimal
Ö
Ö
Ö
Ö
Beckham
Ö
Ö
Ö
´
9. c
10. e
11. e
MBA
Test Prep
Page: 57
12. a
Net revenue by the sale of scooters
=
17. d
48
× 23784.1 = 11416.368 million
100
60
× 1307211
100
= 784326.6
Number of scooter units sold =
11416.37 × 106
784326.6
= Rs. 14,555.63 ≈ Rs. 14,500.
Revenue per scooter =
Alternative method: Net income of scooters
50
~ 12000
= 24000 ×
100
but it has to be less than 12,000, because both the
approximate values are higher. So let us take
11,500 million.
60
= 780000
Total scooters sold = 1300000 ×
100
4928
× 100 = 22
1999-2000 =
22400
18. b
In 1999-2000, the value of manufactured articles
and raw materials export = Rs. (22400 – 4928)
= Rs. 17472 crore
Since export in manufactured goods is twice that
of raw materials, Rs. 17,472 has to be divided in the
ratio 2 : 1, viz. export of manufactured goods = Rs.
11,648 crore and raw materials = Rs. 5,824 crore.
Hence, the difference between raw materials and
food = Rs. (5824 – 4928) crore = Rs. 896 crore.
19. d
In 2000-01, the combined percentage of
manufactured articles and raw materials = 77 and
this is in the ratio 4 : 3.
Hence, percentage of manufactured articles export
= 44% and that of raw materials export = 33%
Hence, the value of manufactured = 0.44 × 25800
= Rs. 11352 crore and the value of raw materials
= Rs. 8514 crore.
Hence, percentage of difference between the value
of raw materials in-between 1999-2000 and
2000-01
11500 × 106
~ 14500
Net realisation =
780000
13. a
Take total sales volume as 100, then motorcycle
sales volume = 22
25% increase = (25 × 22) ÷ 100 = 5.5
This 5.5 is the decrease in scooter sales volume
⇒ Percentage decrease in sales volume
= (5.5 × 100) ÷ 60 = 9.1%
14. d
If others is excluded, then of a total of 100, net
revenue left = 89
Of this, 3 is from mopeds
Hence, percentage of mopeds
3
× 100 = 3.4 % approximately
=
89
15. c
Net profit
20 48 + 17 25 21 + 3
=
+
× 23784.1million
100 100 100 100
= Rs.
Net revenue per vehicle would be the highest for
three-wheelers. (The ratio of percentage revenue
and percentage volume sales is greater than 1,
only for three-wheelers.)
The change in the value of exports from 1999-2000
to 2000-01 = Rs. (11648 – 11352) crore
= Rs. 296 crore.
21. e
Combining (I) and (II):
The only information we have is about the sitting
positions of the highest and the lowest aged
persons. Nothing can be said about the
occupation of chairs 2, 3 and 4. Hence, we
cannot find the age of C.
22. a
Statement (I) alone:
A → Q
C → S
B →
E→
D → T
B does not visit T, and E obviously cannot visit T.
So D will surely visit city T.
23. e
Statement A is not sufficient to answer the given
question. Statement B indicates the days on which
the club needs to be run but no information is
provided regarding saturday and sunday.
For questions 17 to 20: From the data that is given, we
can find the following data (the explanation of how the
following values were arrived at is given after the table):
Item
1999-2000 2000-01
Food (percentage)
22%
23%
Food (value)
4928
5934
Manufactured articles
11648
11352
Raw material
5824
8514
Total value of exports Rs. in crore
22400
25800
Page: 58
(8514 − 5824)
= 31.6%
8514
20. a
= (0.13 + 0.06) × 23,784.1 = Rs. 4,500 million
16. a
Food related exports in 2000-01
= 0.23 × 25800 = 5934
So food related exports in 1999-00
= (5934 – 1006) = 4928
Hence, percentage of food related exports in
MBA
Test Prep
Solution Book-2
24. e
Statement A gives the number of users in December
but does not indicate the volume. Statement B
indicates some idea on the volume reduction and
the number of maximum users. But, even together,
they do not give any answer to the volume of water
in the pool in the particular month of December.
For questions 7 to 10: The following table can be drawn
as per the information given.
P
A
From statement A,
R a hul
4
G au rav
2
C
D
Practice exercise – D13
1. a
Interest earned = 7201 – 12 × 5 × 100 = Rs. 1,201
2. d
Interest on 10 year 17409 − 12000 5409 ~
=
−9:2
=
Interest on 5 year
7201 − 6000
1201
3. b
Interest on 250 monthly investment for 20 year
= 130991 – (12 × 20 × 250) = Rs. 70,991
Interest on 500 monthly investment for 10 year
= 87047 – 60000 = Rs. 27,047
How much less would be earned
= 70991 – 27047 = Rs. 43,944
Alternative method:
The total amount invested in both the cases will be
same as half the amount is paid for twice the time.
Thus difference in earnings will be the difference in
amounts, i.e. 130991 – 87047 = Rs. 43,944
4. b
5. b
6. d
Very clearly the growth has been minimum in
2
= 2.74%
1993-94 =
73
Ö
Ö
S
T
2
Ö
Ö
1
E
Ö
Ö
Ö
Ö
Ö
Since S cannot be supplied by 4 vendors (Q is being supplied
by 4 vendors and each component is supplied by different
number of vendors), hence S has to be supplied by all 5.
Thus C who supplies just 1 component, supplies S. Further
since Q is supplied by 4 vendors, A and B have to supply it
as well. Thus table now looks like:
Kamal
∴ Obviously, Rahul is older than Gaurav by 2 years.
Statement B does not give any information.
∴ The answer is (b).
R
Ö
B
25. b
Q
4
A
P
(2+)
Ö
Q
4
Ö
B
Ö
Ö
´
´
C
1
R
Ö
S
5
Ö
T
2
Ö
´
Ö
D
Ö
Ö
E
Ö
Ö
´
Ö
Since P is supplied by at least 2, R has to be supplied by just
1 vendor and P has to be supplied by 3 vendors. Nothing
row wise, D is the only vendor who can supply 2 parts (as
all other supply more than 2). Thus table now looks as:
A
B
P
3
Ö
Q
4
Ö
R
1
Ö
S
5
Ö
T
2
Ö
Ö
´
Ö
C
1
´
´
´
Ö
´
D
2
´
Ö
´
Ö
´
Ö
´
Ö
Ö
E
Now it is obvious that P is supplied by E and thus E supplies
4 components. Only A can supply all 5 components and
hence, B supplies 3. Thus final table is:
P
3
Ö
Q
4
Ö
R
1
Ö
S
5
Ö
T
2
Ö
Profit of banks in 1991-92 = 3% of 630 = 18.9
Profit of mutual fund in 1992-93 = 5% of 710 = 35.5
Difference = 16.6
A
5
B
3
Ö
Ö
´
Ö
´
Profit to sales is maximum in 1994-95. It is 10%,
otherwise in all the year, it is less than that.
C
1
´
´
´
Ö
´
D
2
´
Ö
´
Ö
´
E
4
Ö
Ö
´
Ö
Ö
Therefore,
Explanations: Fundamentals of
Logical Reasoning & Data Interpretation
7. a
MBA
Test Prep
8. c
9. d
10. a
Page: 59
For questions 11 and 12:
11. b
12. b
13. b
16. c
Given: B > L
…(i)
To be concluded: I > R
In option (a), B > I and L > R. Combining it with (i), B
> L > R and B > I. Therefore, conclusion cannot be
confirmed for sure.
In option (b), B < I and L > R. Combining it with (i),
I > B > L > R. Therefore, conclusion has been
confirmed.
In option (c), B > I and L < R. Even if we combine it
with (i), our conclusion cannot be confirmed.
In option (d), B < I and L < R.
Combining it with (i), L < B < I and L < R.
Therefore, conclusion cannot be confirmed.
Here, x2 > x
⇒ x2 – x > 0
⇒ x(x – 1) > 0
⇒ either x > 1 or x < 0
Our conclusion, which states that x < 0 would be
true only when we come to know that other
possibility, i.e. x > 1 is not the case.
For that matter, if we know that x is a real number
less than 1, we can say that x is not greater than 1
and hence, x < 0. So option (b) would be the correct
choice.
If you look at the trend, area under irrigation has
been increasing over the year in minor as well as
major. Only in 1974-75, in case of minor, it has
decreased which suggests that some minor area
has came under major.
For questions 17 to 20: Australia won all 5 and Holland
lost all 5. For India or Germany, 7 points can be scored as
{W, W, D, L, L} or {W, D, D, D, D}. But since Australia defeated
both of them, the only possibility is {W, W, D, L, L}.
For Pakistan also, there has to be at least one W and at least
one L
⇒ Possibility = {W, D, L, L, L}
Total W = Total L
⇒ South Korea won 3 matches and lost one.
Table is:
P W L D Points
India (Ind)
5
2
2
1
Germany (Ger)
5
2
2
1
7
Australia (Aus)
5
5
0 0
15
Holland (Holl)
5
0
5 0
0
South Korea (SK) 5
3
1 1
10
1
3
Pakistan (Pak)
3.42
0.59
−
~ 0.17 – 0.13
1970-71 =
16.9 + 2.9 3.1 + 1.4
5
1974-75 =
5 .2
0.78
−
~ 0.19 – 0.11 = 0.08
27.1 7.2
5.49 0.84
−
1975-76 =
~ 0.19 – 0.12 = 0.07
28
7.1
So, it is highest in 1974-75.
14. c
Or ratio of gross cropped area to consumption of
fertilizers should be highest
Win
Ö
Loss
´
Draw
´
Holl
´
Ö
´
Pak
Holl
Ind + SK + Aus
Ger
SK
Holl + Pak + Ger
Aus
Ind
Ger
Ind + Holl
SK + Aus
Pak
Ind
Pak + Holl
Ger + Aus
SK
The matches between India and Germany as well as South
Korea and Germany did not end in a draw. Hence, Pakistan
and Germany match must have been a draw.
⇒ India defeated Holland and Pakistan only.
17. b
The result between India and Germany must be a
win in favour of Germany.
1970-71 =
174.8
174.8
=
1.11 + 3.42 + 0.59 5.12
18. a
10
173.1
;
5.51
177
6.06
19. e
7 + 7 + 15 + 10 + 4 = 43
1971-72 =
20. c
There were 2 matches which added in a draw in
the whole tournament.
21. c
Statement A indicates Bunty gets the middle most
sequence. Statement B indicates Bunty is the last
one in sequence.
Hence each statement individually is sufficient.
22. d
Statement A gives the number of hours Mr. Sharma
walks on weekdays and Statement B indicates the
number of hours he walks on Sunday.
1972-73 =
187.8
180.4
1974-75 =
;
1975-76 =
7.71
8.22
So, obviously, 1970-71 is answer.
15. e
4
Aus
= 0.04
4.22 0.73
−
1973-74 =
~ 0.18 – 0.12 = 0.06
23.4
6. 1
1
7
In 1972-73 = 23.20 + 32.77 = 55.97
In 1973-74 = 24.00 + 34.20 = 58.20
More = 2.23 (Remember these figures are cumulative
figures)
Page: 60
MBA
Test Prep
Solution Book-2
23. d
From statement (A),
since ‘D’ always lies, B > W.
From statement (B),
since ‘c’ always speak truth, c = White hat.
∴ From statements (A) and (B), ‘A’ and ‘B’ have to
wear black hats.
∴ The answer is (d).
Persons
A
B
C
D
Hat
B
B
W
W
24. e
Both the statements are not sufficient since ‘A’, ‘B’,
‘C’ positions can be any of these.
15
25
A
A
C
15
B
C
25
B
25
15
A
B
C
And many such positions are possible.
The answer is (e).
25. d
Both the statements together give the length
XY = 10 cm.
Practice exercise – D14
For questions 1 to 5: It is given that each member represents one of the five schools. Hence, the only logic which needs
to be understood is that if a particular school has won 1 red medal, the individual member winning 2 red medals cannot
belong to that school. On this inference, it could be seen that Qadir, Rishi and Yousuf have 2 red medals each. Thus, they
cannot be from Lucknow school or Ahmedabad school, because their respective cumulative red medal count is less than
2. On a similar reasoning, we can limit the possibilities
The following table illustrate the possible linkage between the individual and the schools the represent.
M e dals
M e m be r s Re d Blue
Pos s ible School
White Gr e e n De lhi Indor e Luck now Kolk ata Ahm e dabad
Puneet
1
1
1
1
Qadir
2
0
0
1
×
×
Rishi
2
0
0
0
×
×
Satyam
0
1
3
0
Tarun
0
1
1
2
×
×
×
×
Umesh
0
1
2
2
V ipul
0
1
0
1
Wasim
0
0
2
1
Xavier
0
2
1
0
Y ousuf
2
0
1
0
Zaheer
1
1
1
0
×
×
×
×
×
×
×
×
×
×
×
×
Till here, it can be seen that Satyam could be from Delhi school or Indore school. Let us assume that Satyam is from Indore
school. Thus, his blue and white medals account for the entire tally of blue (1) and white (3) medals of Indore school. The
remaining 2 Green and 2 Red medals of Indore school must have been won by other member(s) representing Indore
school. These other member(s) surely cannot win any Blue or White medal.
For any other member of Indore school, blue and white medal count should be 0 (only Qadir and Rishi). But even both of
them combined cannot equal the green medal count of Indore school. Hence, this case is not possible and our assumption
that Satyam is from Indore school is wrong. Therefore, Satyam must be from Delhi school.
Explanations: Fundamentals of
Logical Reasoning & Data Interpretation
MBA
Test Prep
Page: 61
Putting Satyam to Delhi school, the remaining medal tally of all the schools is as follows
Schools
Medals
Red
Blue
White
Delhi School*
3
1
1
Green
1
Indore School
2
1
3
2
Lucknow School
1
3
2
1
Kolkata School
2
1
1
2
Ahmedabad School
0
1
2
2
*(without the medals won by Satyam)
Now Xavier, who has won 2 blue medals, must be from Lucknow school. The remaining medal tally is
Medals
Schools
Red
Blue
White
Green
3
1
1
1
Delhi School
Indore School
2
1
3
2
Lucknow School*
1
1
1
1
Kolkata School
2
1
1
2
Ahmedabad School
0
1
2
2
*(without the medal won by Xavier)
Now, Lucknow’s one remaining red medal could come from Puneet or Zaheer. But Zaheer has won 1 each of red, blue and
white medals and hasn’t won any green medal. If Zaheer is from Lucknow school, it won’t get just one green medal from
some other member, because none of the members has won just one Green medal.
Hence, Puneet is from Lucknow school. This completes the member list of Lucknow school.
The remaining medal tally:
Medals
Schools
Red
Blue
White
Green
Delhi School
3
1
1
1
Indore School
2
1
3
2
Lucknow School*
0
0
0
0
Kolkata School
2
1
1
2
Ahmedabad School
0
1
2
2
*(without medal won by Xavier & Puneet)
Delhi school, which has got 3 red medals, must have got 1 red medal from a member and 2 red medals from another
member. Only remaining member having 1 red medal is Zaheer. He must be from Delhi school. Remaining medal tally:
Schools
Medals
Red
Blue
White
Delhi School*
2
0
0
Green
1
Indore School
2
1
3
2
Lucknow School
0
0
0
0
Kolkata School
2
1
1
2
Ahmedabad School
0
1
2
2
*(without medal won by Satyam & Zaheer)
Page: 62
MBA
Test Prep
Solution Book-2
Eliminating the ones whose schools have been confirmed and forming the table once again,
M e dals
Pos s ible School
M e m be r s Re d Blue White Gr e e n De lhi Indor e Luck now Kolk ata Ahm e dabad
Qadir
2
0
0
1
×
×
Rishi
2
0
0
0
×
×
Tarun
0
1
1
2
×
×
Umesh
0
1
2
2
×
×
V ipul
0
1
0
1
×
Wasim
0
0
2
1
×
Y ousuf
2
0
1
0
×
×
×
×
Only Qadir or Rishi could be from Delhi school, because remaining blue and white medals for Delhi are 0. If Rishi is from Delhi
school, one remaining green medal for this school would not be possible. Hence, Qadir is from Delhi school. This completes
the member list for Delhi school also.
The possible schools for the remaining members would further reduce:
Medals
Schools
Red
Blue
White
Green
Delhi School
0
0
0
0
Indore School
2
1
3
2
Lucknow School
0
0
0
0
Kolkata School
2
1
1
2
Ahmedabad School
0
1
2
2
M e dals
Pos s ible School
M e m be r s Re d Blue White Gr e e n De lhi Indor e Luck now
Rishi
2
0
0
0
×
×
Tarun
0
1
1
2
×
×
Umesh
0
1
2
2
×
×
V ipul
0
1
0
1
×
×
Wasim
0
0
2
1
×
×
Y ousuf
2
0
1
0
×
×
Kolk ata
Ahm e dabad
×
×
×
×
If Tarun is from Indore school, remaining red and white medal tally for Indore school is 2,2 respectively. It could be seen that
none of the remaining members have such a medal count. Even after combining the medal tally of two or more students,
Red-White 2-2, combination cannot be formed. Hence, Tarun is not from Indore school. Tarun is not from Ahmedabad
School either, because in that case only single White medal remains for Ahmedabad School, which is not available to any
Student. Instead, he is from Kolkata school. Other member from Kolkata school could be Rishi only. This makes Yousuf a
member of Indore school, because he is the only one remaining with 2 red medals.
The possible schools for the remaining members would further reduce as follows:
Schools
Medals
Red
Blue
White
Green
Delhi School
0
0
0
0
Indore School
0
1
2
2
Lucknow School
0
0
0
0
Kolkata School
0
0
0
0
Ahmedabad School
0
1
2
2
Explanations: Fundamentals of
Logical Reasoning & Data Interpretation
MBA
Test Prep
Page: 63
Medals
Members
Umesh
Possible School
Red
Blue
White
Green
Delhi
0
1
2
2
Vipul
0
1
0
Wasim
0
0
2
Indore
Lucknow
Kolkata
×
×
×
1
×
×
×
1
×
×
×
Ahmedabad
The remaining medal tally of Indore school and Ahmedabad school are identical. Also, the total medals of Vipul and Wasim
are identical to those of Umesh alone. Therefore, either Umesh is from Ahmedabad school and (Vipul + Wasim) are from
Indore school or vice-versa.
In summary,
Delhi School
Satyam, Zaheer, Qadir
Indore School
Yousuf, (Vipul+Wasim)/Umesh
Lucknow School
Xavier, Puneet
Kolkata School
Tarun, Rishi
Ahmedabad School
Umesh/(Vipul+Wasim)
1. d
Option (a) is necessarily correct. Option (b) is also
necessarily correct. Option (c) is necessarily
incorrect.
Option (d) could be correct, if Vipul and Wasim are
the members of Indore school. Option (e) is
necessarily correct.
2. e
Either Indore or Ahmedabad. Hence, cannot be
determined
3. b
Three representative members.
4. a
Among the options, only Puneet is from Lucknow
school
Page: 64
5. e
Option (a) is inconsistent because Satyam is not
representing Indore school.
Option (b) is redundant because it is already known
that Vipul and Wasim are fellows.
Option (c) is consistent, but doesn’t give any data
which could tell us the schools of Umesh, Vipul and
Wasim.
Option (d) is also consistent, but not sufficient.
Option (e) clearly tells that 3 members represented
Indore school. This implies that Umesh is a member
of Ahmedabad school and (Vipul+Wasim) are
members of Indore school.
MBA
Test Prep
Solution Book-2
For Questions 6 to 10: Aggregate number of cars manufactured by all the companies across the twelve months of a year
= 1200.
Since aggregate number of cars manufactured by all the companies is same in each month, therefore number of cars
1200
manufactured in each month =
=100.
12
In January, Mitsubishi manufactured 7 cars and Renault manufactured 37 cars. The 6 remaining companies have to
manufacture at least 7 + 2 = 9 cars. But total cars in January = 100.
Hence out of those 6 remaining companies, minimum cars any company could manufacture = 9 and maximum cars any
company could manufacture = 11.
Similarly, with the help of the given information, we can find the range of the number of cars manufactured by each of the
companies in the 12 months.
Mercedes
Jan
9-11
Feb
7-26
Mar
32
April
8-19
May
27
June
7-20
July
8-23
Aug
10 -18
Sep
7-23
Oct
3
Nov
10-11
Dec
10
Mazda
9-11
7-26
9-16
8-19
6-22
7-20
8-23
10 -18
5
5-33
10-11
32
Nissan
9-11
7-26
9-16
8-19
4
40
8-23
10 -18
28
5-33
8
10
Mitsubishi
7
7-26
9-16
8-19
6-22
7-20
8-23
10 -18
7-23
38
10-11
10
Porsche
9-11
7-26
9-16
8-19
6-22
5
31
8
7-23
5-33
31
10
Ferrari
9-11
7-26
9-16
35
6-22
7-20
8-23
24
7-23
5-33
10-11
8
Renault
37
5
9-16
6
6-22
7-20
8-23
10 -18
7-23
5-33
10-11
10
Honda
9-11
34
7
8-19
6-22
7-20
6
10 -18
7-23
5-33
10-11
10
Total
100
100
100
100
100
100
100
100
100
100
100
100
The maximum and minimum number of cars manufactured by different brands alongwith their optimum index (O.I.) and
average index (A. I.) is collated in the following table.
Maximum
Minimum
Sum
O.I.
A.I.
Ratio
(O.I./A.I.)
Mercedes
223
138
361
85
180.5
Mazda
236
116
352
120
176
Nissan
236
146
382
90
191
Mitsubishi
233
127
360
106
180
Porsche
235
136
371
99
185.5
Ferrari
252
135
387
117
193.5
Renault
224
120
344
104
172
Honda
214
119
333
95
166.5
0.470
0.68
0.471
0.588
0.53
0.60
0.604
0.57
6. c
Maximum possible number of cars manufactured by Mitsubishi across all the 12 months = (7 + 26 + 16 + 19 + 22
+ 20 + 23 + 18 + 23 + 38 + 11 + 10) = 233.
7. e
Mazda is the company for which the Optimum Index is highest.
8. c
Statement B and D are incorrect.
9. d
The ratio of O.I. to A.I. is least for Mercedes.
10. c
The minimum number of cars that a company possibly manufactured throughout the year is second largest for
Mercedes.
Explanations: Fundamentals of
Logical Reasoning & Data Interpretation
MBA
Test Prep
Page: 65
Maximum number of New Users who logged into
the SIS exactly two times = 49 – 11 = 38.
Total number of hits on the SIS on Thursday is 105
and total number of students who logged into the
SIS on Thursday is 52.
So a maximum of 52 New Users can possibly log
into the SIS exactly two times on Thursday, this is
possible when only one new User logged into the
SIS exactly thrice on Thursday.
Required answer = 38 + 51 = 89.
For questions 11 to 15:
11. c
Sunday: 131 = 25 × 5 + 2 × 3.
Minimum possible number of students who logged
into the website = 27.
Monday: 49 = 9 × 5 + 1 × 4.
Minimum possible number of students who logged
into the website = 10.
Tuesday: 92 = 18 × 5 + 1 × 2.
Minimum possible number of students who logged
into the website = 19.
14. d
Since on Sunday and Thursday no Old User logged
into the SIS, total number of students that logged
into the SIS on the remaining five days of the week
is 19 + 33 + 61 + 37 + 25 = 175.
So to minimize the number of New Users that logged
into the SIS, we need to maximize the number of Old
Users who logged into the SIS on exactly three
days.
175 = 57 × 3 + 2 × 2
Therefore at least 59 Old Users.
15. b
175 = 86 × 2 + 1 × 3. Maximum possible number of
Old Users that logged into the SIS is 87.
Therefore, number of Old Users that logged into the
SIS lies between 59 and 87 (both inclusive).
Wednesday: 157 = 31 × 5 + 1 × 2.
Minimum possible number of students who logged
into the website = 32.
Thursday: 105 = 21 × 5.
Minimum possible number of students who logged
into the website = 21.
Friday: 81 = 15 × 5 + 1 × 4 + 1 × 2.
Minimum possible number of students who logged
into the website = 17.
Saturday: 63 = 12 × 5 + 1 × 3.
Minimum possible number of students who logged
into the website = 13.
Therefore during this particular week the minimum
possible number of students who logged into the
website = 27 + 10 + 19 + 32 + 21 + 17 + 13 = 139.
12. b
Maximum number of students who logged into the
SIS on Sunday = (64 × 2 + 1 × 3) = 65
Similarly, maximum number of students who logged
into the SIS on Monday, Tuesday, Wednesday,
Thursday, Friday and Saturday is 24, 46, 78, 52, 40
and 31 respectively.
Required difference between the maximum and the
minimum possible number of students who logged
into the SIS on Sunday, Monday, Tuesday,
Wednesday, Thursday, Friday and Saturday is 38,
13, 27, 46, 31, 23 and 18 respectively.
Therefore, on 4 days the required difference is not
less than 27.
13. a
Total number of hits on the SIS on Sunday is 131
and total number of students who logged into the
SIS on Sunday is 49.
To maximize the number of New Users who logged
into the SIS exactly two times we need to minimize
the aggregate number of New Users who logged
into the SIS at least three times.
Minimum number of New Users who logged into the
SIS at least three times is when 11 New Users
logged into the SIS exactly five times.
Page: 66
For questions 16 to 20:
Let a, b, c, d, e, f, g, h, i and j be the number of questions in
sections I, II, III, IV, V, VI, VII, VIII, IX and X respectively.
From the data given in TABLE 1, we can draw the following
conclusions:
a+b
2
= 28,
a+c
2
= 21,
b+c
2
= 28,
a+d
2
= 26,
a+e
2
= 24
a+f
a+g
a+h
a+i
a+ j
= 35,
= 38,
= 30,
= 32,
= 28
2
2
2
2
2
Solving the above we get the following values
a = 21, b = 35, c = 21, d = 31, e = 27
f = 49, g = 55, h = 39, i = 43, j = 35
Similarly the number of questions solved correctly in each
of the given 10 sections can be easily calculated.
The following table provides information about the number
of questions in each section, number of questions solved
correctly in each section and the difference of the number
of questions and the number of questions solved correctly
in each section.
Sections
Total
I
II III IV V VI VII VIII IX X Total
21 35 21 31 27 49 55 39 43 35 356
Correct
7 21 15 21 23 35 41 33 29 27 252
Total – Correct 14 14 6 10 4 14 14 6 14 8 104
Absolute Value
of [Correct – 7
(Total – Correct)]
MBA
Test Prep
7
9 11 19 11 27 27 15 19
—
Solution Book-2
16. d
17. d
In section V, the difference between the number
of questions and the number of questions solved
correctly is the least.
Number of questions solved incorrectly and number
of questions left unattempted in section VI can be
3 and 11 respectively. Hence option (1) is possible.
Number of questions solved incorrectly and number
of questions left unattempted in section VI can be
6 and 8 respectively. Hence option (2) is possible.
Number of questions solved incorrectly and number
of questions left unattempted in section X can be 3
and 5 respectively. Hence option (3) is possible.
If option (a) is possible, then option (e) is also
3 3 + 11 14
possible Q =
=
11
11
11
Option (d) is not possible and hence the correct
choice.
18. b
The number of questions in sections VI, VII, VIII, IX
and X is not less than the number of questions in
section II. Hence there are 5 such sections.
19. c
In seven sections namely I, II, III, IV, V, VI and IX.
20. b
Maximum possible number of questions solved
incorrectly in sections I, II, III, IV, V, VI, VII, VIII, IX
and X is 1, 5, 3, 5, 5, 8, 10, 8, 7 and 6 respectively.
So, total number of questions solved incorrectly
= 58.
Therefore, number of question left unattempted
= 104 – 58 = 46.
Minimum possible marks obtained by the student in
the entrance test = 252 × 4 – (58 × 2 + 46 × 1) = 846
21. c
Since 909909 are divisible by 9 and 567 is also
divisible by 9.
∴ x is also divisible by 9.
22. e
Number of books is not known. Thus, by using
both statements also we do not get the answer.
Practice exercise – D15
For questions 1 to 5:
From information B and E, it is clear that surname of Asha is
Singh and she has ordered one San Marino Imbottiti and one
Sicilian Fugitive pizza. From information A, at least one of
them must be in Solo size. So, possible total bill amounts of
Asha Singh are Rs. 246, Rs. 372, Rs. 266 and Rs. 396 only.
From information E, it is also evident that Esha’s surname is
either Hazra or Malhotra.
Let us assume that Esha’s surname is Malhotra. Then, from
information C, the pizza, which is common between Esha
and Isha, did not start with ‘M’. So the single Montreal Ham
must have been ordered by Esha and from information F, this
is a Solo size pizza. From information E, Esha Malhotra’s bill
amount is Rs. 127 less than Asha. So this amount can be Rs.
119, Rs. 245, Rs. 139 and Rs. 269. Deduct Rs. 87 from these
amounts, which Esha Malhotra must have spent on Solo
size Montreal Ham. You will get Rs. 32, Rs. 158, Rs. 52 and
Rs. 182 respectively, which must be the price of the other
pizza Esha ordered. Out of these only Rs. 158 can be found
out in the menu for a double size Raffaele’s Bianco. But, the
pizza, which is common pizza between Esha and Isha, must
have been ordered at least twice. So Raffaele’s Bianco
cannot be the common pizza between Esha and Isha. So,
Raffaele’s Bianco must have been ordered by Usha and not
by Esha.
Therefore, our assumption, that Esha’s surname is Malhotra
is wrong. Esha’s surname is Hazra. She must have ordered
Hawaiian Honeymoon in Solo size and one other pizza in
Double or Family size. Again, the common pizza between
Esha and Isha, must have been ordered at least twice. So
Raffaele’s Bianco cannot be the common pizza between
Esha and Isha. So, Usha must have ordered Raffaele’s
Bianco. From information F, Usha must have ordered
Raffaele’s Bianco in any size other than Double. Now, Usha’s
surname is either Gupta or Malhotra.
24. d
Statement I: (c + d)(c – d) = 5.
This data is not sufficient independently.
Combining with (II) we get c + d.
Let us assume that Usha’s surname is Gupta. Then she must
have ordered Godfather’s Choice in Double size. So, she
must have ordered Raffaele’s Bianco in Solo size. So,
possible total bill amounts of Usha Gupta are Rs. 214 only.
From information E, Usha Gupta’s bill amount is Rs. 81 more
than Esha. So Esha’s bill amount can be Rs. 133 only. Deduct
Rs. 87 from these amounts, which Esha Hazra must have
spent on Solo size Hawaiian Honeymoon. The remaining
amount is Rs. 46, which must be the price of the other pizza
Esha ordered. There is no pizza with this price tag in the
menu.
25. c
Using statement I, the orthocentre lies outside the
triangle only in the case of an obtuse triangle.
Thus, ABC is not an equilateral triangle.
Using statement II, only in the case of an
equilateral triangle does the circumcentre
coincide with the orthocentre. Thus, ∆ABC is an
equilateral triangle.
Therefore, our assumption, that Usha’s surname is Gupta is
wrong. Usha’s surname is Malhotra and Isha’s surname is
Gupta. Usha must have ordered Raffaele’s Bianco in Solo
size and Montreal Ham in Double. So her total bill is Rs. 245.
That means Asha’s total bill is Rs. 372. So she must have
ordered San Marino Imbottiti in Solo size and one Sicilian
Fugitive pizza in Family size.
23. c
Solving the given data
x +y x − y y x 1 x y 1
+
= + = x + = + y
x
y
x y
y x
y
x
Thus, both statements can give answers
independently.
Explanations: Fundamentals of
Logical Reasoning & Data Interpretation
MBA
Test Prep
Page: 67
Now Isha Gupta must have ordered one Godfather’s Choice.
Let us assume this was in Solo size. She has ordered
another pizza common with Esha. We know, that Esha Hazra
must have ordered one Hawaiian Honeymoon in Solo size
and one other pizza in Double or Family size. The price of
Godfather’s Choice is Rs. 12 less than Hawaiian Honeymoon
in Solo size. But, from information E, Ms. Gupta’s total bill
was Rs. 81 more than Esha. So, the other pizza must have
created the difference of Rs. 93. But there is no such pizza,
for which the difference of price for Double and Family size
is Rs. 93. So our assumption was wrong.
Next, let us assume that Isha Gupta ordered one Godfather’s
Choice in Double size. That means Isha must have ordered
the other pizza in Solo size. The price of Godfather’s Choice
in Double size is Rs. 20 more than Hawaiian Honeymoon in
Solo size. So, the other pizza must have created the
difference of Rs. 61. But there is no such pizza, for which
the difference of price for Solo and Double/Family size is
Rs. 61. So our assumption was again wrong.
So, Isha must have ordered one Godfather’s Choice in Family
size. That means Isha must have ordered the other pizza in
Solo size. The price of Godfather’s Choice in Family size is
Rs. 118 more than Hawaiian Honeymoon in Solo size. So,
the other pizza must have created the difference of Rs. 37.
Only for Verona’s Army, the difference of price for Solo and
Double size is Rs. 37.
Now, finally we can collate the derived information in the following table, which can help us to answer all the questions:
Name
Asha
Esha
Isha
Usha
Surname
Singh
Hazra
Gupta
Malhotra
Verona's Army
Raffaele's Bianco
Pizza 1
Pizza 2
Type
Size
Solo
Solo
Solo
Solo
Type
Sicilian Fugitive
Verona's Army
Godfather's Choice
Montreal Ham
Size
Family
Double
Family
Double
372
262
343
245
Total Bill (Rs.)
1. c
6. b
7. a
8. d
San Marino Imbottiti Hawaiian Honeymoon
2. a
3. b
30+25+40+5 for reminder mail, i.e. 100% space is
used
∴ free space is 0%
The deleted mails will go to trash and would still
occupy space. So all will bounce.
On day 3, 20 MB of space will be created in trash
folder as mails deleted on day 1 will be automatically
cleared.
So, none of the mail will bounce.
For questions 9 to 14:
The maximum possible lecturers were short-listed.
Since there are 5 age groups and 4 different
disciplines; and there can be at most two lecturers
of a particular age group representing a particular
discipline, the number of lecturers short-listed
= 5 × 4 × 2 = 40.
Number of them selected = 5 + 20 = 25.
Now,
(i) Tells that out of these 25, the number of lecturers
from various age groups is:
Page: 68
4. c
5. e
Young:
5
Middle-aged:
5
Senior:
4
Stalwarts:
4
Retired:
7
TOTAL
25
Now coming directly to (iii),
Phys ics
Che m is try M athe m atics Biology
Young
Young
Young
Young
Middle-aged Middle-aged Middle-aged
Middle-aged
Senior
Senior
Senior
Senior
Stalw art
Stalw art
Stalw art
Stalw art
Retired
Retired
Retired
Retired
MBA
Test Prep
Solution Book-2
Note that the column for Mathematics is complete.
And in Biology, count for Retired can be 2 at max.
And there is no provision for Middle Aged in Biology.
Each of them is having at least one lecturer.
So assigning one to each of them, the table looks
like:
Physics
Chem istry Mathem atics Biology
Young (1)
Young (1)
Young (1)
Middle-aged Middle-aged Middle-aged
(1)
(1)
(1)
Senior
Senior (1)
Senior (1)
Middleaged
Senior (1)
Stalw art (1) Stalw art (1) Stalw art (1)
Stalw art
Retired (1)
Retired (1)
Retired (1)
Retired (1)
Thus, out of the two balance retired lecturers, at
most one can go in Biology and in that case, the
remaining retired and middle aged ones must go into
Physics or Chemistry or both.
Young (1)
9. d
The balance lecturers are:
Young:
1
Middle-aged:
2
Senior:
1
Stalwarts:
1
Retired:
3
TOTAL
8
Case I: Middle-aged lecturer goes into Physics.
Now, out of the remaining two retired lecturers one
has to go into Physics and the remaining one can go
either in Chemistry or Biology.
Going back to (ii), total number of lecturers for
physics, chemistry and biology can be counted to
lie between 12 and 20. Moreover, the total is a
square.
⇒ Total in these three is 16 and there are 9 lecturers
from mathematics stream.
Proceeding, (iv) tells us that Rocky and Platy are
Young physics lecturers.
So the balance Young lecturer goes in Physics.
And since none of the cells can have more than 2
and total for Mathematics is 9, we get a better
version of the table as:
Case II: Retired lecturer goes into Physics.
Here also, the count for retired in Physics is full (2).
Middle Aged cannot go in Biology and if he goes into
Chemistry, the initial condition that "the number of
Physics lecturers is greater than the number of
Chemistry lecturers" can never get satisfied.
So the middle aged goes into Physics and we get
the same configuration as in case I
Physics (7)
Young (2)
Chem istry
(5 or 6)
Young (1)
Mathem atics Biology
(9)
(3 or 4)
Young (1)
Young (1)
Middle Aged
(2)
Senior
Middle Aged
(1)
Senior (1)
Middle Aged
(2)
Senior (2)
Middle
Aged
Senior (1)
Stalw art (1)
Stalw art (1)
Stalw art (2)
Stalw art
Retired (2)
Retired (1)
Retired (2)
Retired (1)
The one remaining retired lecturer goes into either
Chemistry or Biology.
Hence the number of retired Chemistry lecturers
cannot be determined.
Physics (5 or Chem istry Mathem atics Biology (3
m ore)
(5 or m ore) (9)
or m ore)
Young (2)
Young (1)
Young (1)
Young (1)
Middle Aged
(1)
Senior
Middle Aged Middle Aged
(1)
(2)
Senior (1)
Senior (2)
Middle
Aged
Senior (1)
Stalw art (1)
Stalw art (1) Stalw art (2)
Stalw art
Retired (1)
Retired (1)
Retired (1)
Retired (2)
10. b
It is already explained that the middle aged lecturer
and at least one retired lecturer goes into Physics
or Chemistry or both, the total is at least 12 and at
most 13. Only option (b) is correct.
11. d
Since Kandy is the lone retired Chemistry lecturer,
the balance 2 retired Chemistry lecturers go into
Physics and Biology (one each). So there are two
retired Biology lecturers for sure.
12. c
At least one of the retired lecturer must go into one
of Physics or Chemistry. Thus the count for both of
them cannot be one. Option (c) is not possible.
And the balance lecturers now:
Middle Aged:
1
Retired:
2
TOTAL
3
As of now, minimum lecturers for Physics as well
as that of Chemistry are 5. But since the number of
Physics lecturers is greater than the number of
Chemistry lecturers, at least one of the balance
three must go into Physics.
Explanations: Fundamentals of
Logical Reasoning & Data Interpretation
MBA
Test Prep
Page: 69
13. d
14. a
15. a
16. b
Laxman has scored 3,000 runs and has taken 60
catches. He has not taken any wickets. Even if we
assume that he has scored all runs in centuries,
then also his points would be:
Runs → 3,000, Catches → 180, Centuries → 30 ×
50 = 1,500
Total points 4,680, so grade E.
Zaheer has scored 1,000 runs and taken 50 catches.
He has taken 150 wickets (half of Kumble). 8 five
wicket hauls. If we assume that he has scored all
runs in centuries, then his maximum points would
be:
Runs → 1,000, Catches → 150, Centuries → 10 ×
50 = 500, Wickets → 3,000, Five wicket haul → 400
Total points, 5,050 , so at the most he could be in
grade D.
Ganguly has scored 10,000 runs and taken 50
wickets. If we assume that he has scored all runs
in centuries, then his points would be:
Runs → 10,000, Centuries → 100 × 50 = 5,000,
Wickets → 1,000
Total points 16,000, so grade A.
Sehwag could score maximum 7,999 runs and taken
50 catches. He has taken 40 wickets.
And taken 4 five wicket hauls. He has scored 10
centuries. His points would be:
Runs → 7,999, Catches → 150, Centuries → 500,
Wickets → 800, 5 wicket haul → 200
Total points 9,649, so he cannot be in grade B.
For questions 17 to 19:
18. d
If it breaks away at least five members, the total
strength gets reduced to 495, 60 abstain (40+20)
so for simple majority you need half of 435, i.e. 218,
Congress has backing of 220. CPM's support would
definitely be another way.
19. d
If members of BJP, Others, CPM and RJD support a
Congress led government then it would have a 75%
majority. Hence four parties.
20. b
Statement I is not sufficient as there are more than
one sequences satisfying the question condition.
Statement II is sufficient to answer the question.
Since the first term is 0, the 3rd term will be 6.
21. a
Statement I alone gives the relationship the three
shapes in terms of size and is enough to answer
the question.
Statement II gives the data to find the area of the
circle but not for the two others.
Thus, the answer is (a).
22. c
‘C’ is a set of integers, and ‘x’ an element in C. Then
from statement I, (x + 3) can also be an element in
‘C’ if we take x as negative integers and hence the
condition is fulfilled. Similarly, with the help of
statement II alone we can get the answer.
23. e
There can be many different relationships possible.
So statement I does not provide an answer.
Statement II gives us a very general piece of
information. The two statements combined together
are not sufficient.
24. d
From statement I, we get 15x + 20y = 130
The only possibility for x and y are (6, 2) and (2, 5).
From statement II, we get x = 2 and y = 5,
i.e. we have 5 tickets of $20 and 2 tickets of $15.
Thus, the answer is (d).
25. d
From statement I, we get the lowest = 20, highest =
10
From statement II, we get the two series are in AP.
20, 23, 26, 29, ..., 80
UPA (30 + 5 + 7)% = 42%
If Congress led ⇒ 42% + 2% (of CPI) = 44%
NDA = (20 + 2 + 2)% + 10% (“Others”) = 34%
17. d
(i) –(8% (of SP) + 4% (of BSP)) = –12%
(ii) 34% (of NDA) + 5% (of RJD) = 39%
(iii) 55 members ⇒ –11%
If all (i), (ii) and (iii) happen, then
NDA = 39% which is more than 50% of
(100 – 12 – 11) = 77%
Page: 70
∴ Average =
MBA
Test Prep
80 + 20
= 50
20
Solution Book-2
Explanations: Ratio & Proportion
Solutions (Non MCQ)
7.
Let father’s present age be F.
Son’s present age be S.
∴ (F – 5) = 5[S – 5] and (F + 2) = 3(S + 2) or,
F – 5S = –20 and F – 3S = 4
∴ 2S = 24; S = 12 and F = 40
∴ Ratio of ages of father and son = 10 : 3.
8.
Mohan’s expenditure = 4x
Mohan’s savings = x
∴ Income = 5x
New income = (5x)(1.2) = 6x; new saving = 1.12x
New expenditure = 4.88x
Percentage increase in expenditure
Level – 1
1.
2.
(54 – x) : (71 – x) : : (75 – x) : (99 – x)
Solving, x = 3
Ratio =
1
1
1
:
:
= 15 : 10 : 6
2
3
5
or first part =
15
× 465 = 225
31
4.88x − 4x
=
× 100 = 22%.
4x
10
× 465 = 150
Second part =
31
6
× 465 = 90
Third part =
31
Hence, the three parts are 225, 150, 90.
3.
Ratio of first, second and third is
2x : 2x – 10 : 5x
and 2x + 2x – 10 + 5x = 170
9x = 180
x = 20
So the three parts are 40, 30, 100 respectively.
4.
If numerator = x and Denominator = 3x
∴ Fraction ⇒
5.
6.
x+y
1
=
3x + 3y 3
9.
3 4 12
, i.e. 3 : 2 : 3
: :
1 2 4
∴ 3x + 2x + 3x = 1800
8x = 1800
⇒ x = 225
Value of 50-paisa coin = 2 × 225 = 450
and number of 50-paisa coins = 450 × 2 = 900
coin is
A
10.
The angles of the triangle are
If ∠A = 4x; ∠B = 5x; ∠C = 5x – 30
∠A + ∠B + ∠C = 180
⇒ 4x + 5x + 5x – 30 = 180
⇒ 14x = 210 ⇒ x = 15
So ∠A = 60°, ∠B = 75°, ∠C = 45°
11.
Let the initial number of employees = n.
15
x.
14
Ratio of total wages before and after the change
8 15
x = 21 : 20.
= (nx) : n
9 14
The ratio of volumes of two cylinders is x : y and
heights are p : q.
2
r 2 xq
V1 πr1 p x
d 2 xq
=
=
= 1 =
⇒ 12 =
2
V2 πr 2q y
py
py
d2
r
2
2
8
n
9
Let initial wages = x. New wages =
∴
C
B
Ratio of number of coins = 4 : 5 : 6
Suppose we have 4, 5 and 6 coins of Re. 1,
50-paisa and 25-paisa respectively. Then the total
value = Rs. 8.
Since the aggregate sum is Rs. 32 (4 times the sum
that we have arrived at), the number of coins of the
denomination Re. 1, Re. 0.5 and Re. 0.25 must be
16, 20 and 24 respectively.
New number of employees =
Ratio of values of 1 rupee, 50 paisa and 25-paisa
12.
The given expression can be written as
Q–p Q–q Q–r
=
=
=k
2
5
7
p = Q – 2k; q = Q – 5k; r = Q – 7k
So, Q =
Explanations:
Fundamentals of Ratio & Proportion
Q – 2k + Q – 5k – Q – 7k
2
MBA
Test Prep
Page: 71
2Q = 3Q –14k
Q = 14k
p = 12k;
q = 9 k;
r = 7k;
∴ p : q : r = 12 : 9 : 7
13.
Alternative method:
There is a loss of 4 kg on replacement of 12 kg
chair by 8 kg. This loss will be equally compensated
between all the four chairs. So, there is a decrease
of 1 kg in average.
Ram
Incomes
8x
11x
Expenditures
7y
10y
Savings
500
500
∴
8x – 7y = 500
⇒
19.
When the same number is added to all the
quantities, the average increases by the number
added.
Therefore, increase = Rs.1,000
20.
New average will also increase by 20% of the
present average.
Therefore, new average = 15,000 + 3,000
= Rs. 18,000
21.
Turnover from January to March = 12 × 3 = 36
Turnover from March to June = 14 × 4 = 56
Turnover from January to June = 36 + 56 – Turnover
of March.
Also average turnover from January to June is 12.
∴ 36 + 56 – Turnover in March = 12 × 6
So turnover in March = Rs. 20 crore.
22.
If John leaves the group, number of students in
the group will be 3.
So average height
Shyam
88x – 77y = 5500
11x – 10y = 500 ⇒ 88x – 80y = 4000
∴ 3y = 1500, y = 500
⇒ x = 500
∴ Incomes of Ram and Shyam are Rs. 4,000
Rs. 5,500 their expenditures are Rs. 3,500,
Rs. 5,000 respectively.
14.
If B joined for t months then profit ratio = (450 × 12)
: 300t = 2 : 1
∴
∴
15.
When A gets 60 points, B gets 40 points
When A gets 60 points, C gets 45 points
∴ When C gets 45 points, B gets 40 points
1
When C gets 60 points, B gets 53
points
3
∴
16.
17.
18.
450 × 12
= 2 ⇒ t = 9 months
300t
B joined after 3 months.
C gives B
6
2
points in 60
3
=
23.
Let Tom’s score be x.
72 × 4 + x
∴
= 70 ⇒ x = 62
5
24.
Time t = [20 men × 10 days × 5 hr ] = 25 days.
[15 men × 8 hr ]
3
20 × 10 × 5
× 1.25
15 × 0.75 × 8
= 13.88 days (approximately).
Multiplication by 1.25 is because of 25% extra work
load and 0.75 accounts for the efficiency.
168 × 4 − 170
= 167.37 cm (approximately)
3
Cost of 20 kg apples = 20 × 40 = Rs. 800
Cost of 10 kg apples = 10 × 50 = Rs. 500
Total cost of 30 kg = 800 + 500 = Rs. 1,300
∴ Average cost =
1300
= Rs. 43.33 per kilogram
30
Time t =
25.
Time taken to ride 20 km at 10 km/hr =
Time taken for another 20 km at 5 km/hr =
10 + 8 + 12 + 6
= 9 kg
4
Now a 12 kg chair is replaced by an 8 kg chair
=
Average speed =
40
= 6.67 km/hr
6
Average speed =
Total dis tan ce
Total time
∴ Average =
26.
=
Page: 72
20
= 4 hr
5
∴ Total time taken = 6 hr
Total distance covered = 40 km
Average weight of the chairs
10 + 8 + 8 + 6
= 8 kg
4
i.e. a decrease of 1 kg in average.
20
= 2 hr
10
10 + 20
= 10 km / hr.
2 +1
MBA
Test Prep
Solution Book-2
27.
Let total number of units = 100
Revenue when 50 units are sold at Rs. 20 per unit
= Rs. 1,000
Revenue when 30 units are sold at Rs. 40 per unit
= Rs. 1,200
Total price = Rs. 2,200
Average price =
28.
31.
2200
= Rs. 27.50
50 + 30
Averag e percentage
If 2 friends leave the group, then 3 are left.
New average =
To a mixture containing 10% water, pure water
(100%) is mixed to get the resultant solution
containing 20% water.
Applying the rule of alligation, average concentration
of constituents
10%
100%
Resu ltant average
(6.5 × 5 ) − (7 × 2 ) = 6.16 kg
20%
3
Q uantities
10
80
29.
Total score
Average =
Number of students
Ratio of the volumes = 8 : 1 or 40 : 5.
∴ To 40 L of solution, 5 L of water must be added.
Alternatively:
36 L of milk in the original mixture now becomes
80% of the mixture. Hence, the total volume of the
new solution = 45 L. So the extra 5 L must be the
water that was added.
30 × 70 + 20 × 60
=
= 66%
50
30.
Method 1:
7
.
10
Fraction of gold in first alloy =
Fraction of gold in second alloy =
7
.
20
32.
CPII 3
15 – 0
=
=
0 – (–10) CPI 2
Applying rule of alligation,
Averag e fractio nal
concen tration of
the con stituen ts
7
10
Using alligation,
=
7
20
CPII Cost price of second shirt
=
CPI
Cost price of first shirt
and CPI + CPII = 900
Resu ltant averag e
∴ CPI =
Q uantities
7
1
1
3
CPII = × 900 = 540
5
7
∴ x – 10 = 20 − x
⇒
2
× 900 = 360
5
7
7
+
7 3 21
20
10
x=
= =
2
2 20 40
∴ Ratio of gold to copper = 21 : 19.
Method 2:
The first alloy is a 70% gold alloy and the second is
a 35% gold alloy. If the two are mixed in equal
quantities, the average concentration of gold in the
70 + 35
= 52.5% .
2
Therefore, the ratio of gold to copper is
52.5 :47.5 = 21 : 19.
resulting alloy is
Explanations:
Fundamentals of Ratio & Proportion
33.
1
th .
6
Fraction of water in pure water (added) = 1.
Fraction of water in the adulterated milk =
Fraction of water in resultant mixture =
3
th .
8
∴ Apply rule of alligation.
Averag e fractio nal
constitu tion of the
constitu en ts
1
6
3
8
Resu ltant averag e
Q uantities
MBA
Test Prep
1
5
8
5
24
Page: 73
Ratio of the volumes of the solution to the water
added = 3 : 1 = 66 : 22.
∴ 22 kg of water should be added.
3
∴ A woman gets 5 of what a man gets
∴ 6m + 12w + 17b
Alternative method:
55 kg of milk is now
8
× 55 = 88 kg.
Hence, new weight =
5
Hence, the amount of water added = 88 – 66
= 22 kg.
34.
= 6m + 12 ×
5
th of the new mixture.
8
or m = Rs. 2.5. Similarly, b = Rs. 1, w = Rs. 1.5
m, b, w are shares of a man, a boy and a woman,
respectively.
37.
1
2
3
4
a = b = c = d
2
3
4
5
Then
10
7
7
is the ratio of
and
,
9
9
10
which are the average milk content in the initial and
the final mixtures.)
∴ a + 4 a + 3 a + 8 a = 7300
i.e. b =
Method 1:
Let CP of 1 L of milk = Rs. 100.
He mixed x litres of water.
SP of 1 L of mixture = 100.
∴ SP of (1 + x) L of mixture = 100(1 + x).
Percentage of water =
38.
Let the worth of the properties of A, B and C be
2x, 3x and 5x respectively. Total = 10x
1
1
(3x) = x and
(3x) = x to C.
3
3
A has 3x, B has x, C has 6x
∴
C sells
1
3
(6 x ) to A = x
4
2
3x 9
= x = Rs. 18,000
2
2
Hence, x = Rs. 4,000
∴ Total worth of properties = 10x = Rs. 40,000.
∴
39.
A has now 3 x +
Let A, B and C invested Rs. 5, Rs. 6 and
Rs. 8, respectively for a, b, and c months.
Then 5a : 6b : 8c is the ratio of profits which is given
to be 5 : 3 : 1.
∴ 5a : 6b : 8c = 5 : 3 : 1.
1
1
∴ a : b : c = 1 : 2 : 8 = 8 : 4 : 1.
2 men = 5 boys and 3 boys = 2 women
2
of what man gets.
5
1 woman gets
1× 3
or 1.5 times of what a boy
2
40.
Total amount deducted is 15 + 10 + 30 = 55.
Amount left = 535 – 55 = 480.
Now Rs. 480 is divided in the ratio 4 : 5 : 7.
So after deduction
share of A =
4
× 480 = 120 ,
16
share of B =
5
× 480 = 150 ,
16
gets.
Page: 74
5
B sells to A =
Level – 2
∴ 1 boy gets
2
⇒ a = 2400, b = 1800,
c = 1600 and d = 1500
0.25
× 100 = 20% .
1.25
Method 2:
If the total cost of 100 L of milk is Rs. 100, the
milkman is making a revenue of Rs. 125. Since he is
selling at the same cost price, he must actually be
selling 125 L. Hence, he is adding 25 L of water or
20% of the mixture is water.
5
3
2
a, c = a and d = a
8
4
3
3
SP − CP
100x
× 100 =
× 100 = 100x = 25
P% =
CP
100
∴ Volume of mixture = 1 + 0.25 = 1.25.
36.
Assume that the persons in 4 battalions are a, b, c
and d.
10
times the original
9
volume. Hence, if the original volume is 729 ml, the
new volume = 810 ml. The increase is due to the
addition of 81 ml of water.
The volume of the solution is
(Hint: the factor
35.
2
3
m + 17 m = 20m
5
5
MBA
Test Prep
Solution Book-2
7
× 480 = 210 .
16
∴ Initial share of A = 120 + 15 = Rs. 135.
Initial share of B = 150 + 10 = Rs. 160.
Initial share of C = 210 + 30 = Rs. 240.
share of C =
41.
Ratio of contributions expected from the three men
= (110 × 6) : (50 × 9) : (440 × 3) = 22 : 1 5 : 44.
15
47.
5
∴ Second man must pay 81 th or 27 th of the total.
42.
B should get =
44.
2 leaps of the dog = 5 leaps of the hare
or 1 leap of the dog = 2.5 leaps of the hare
∴ 6 leaps of the dog = 15 leaps of the hare
∴ Ratio of leaps of dog to hare = 15 : 4.
48.
49.
Total weight
Number of students
If water weights x kg/unit volume
weight of gold = 19x/unit volume
weight of copper = 9x/unit volume
∴ Weight of the mixture desired = 15x/unit volume
Apply rule of alligation:
Averag e of g old and cop per
In a race of 100 m, A travels 100 m while B travels
95 m. This means in a race of 500 m, when A travels
19x
9x
Q1
Q2
Resu ltant averag e
Q uantities
500
= 475 m.
200
Hence, when B travels 475 m, C travels 475 ×
19 x – 15 x
475
=
500
451.25.
Therefore, when A travels 500 m, C travels 451.25
m.
This means, A can give C a lead of 48.75 m.
Please recollect you would have solved this
problem in the chapter dealing with SI and CI. It is
repeated here to explain solving the problem using
alligation. On an average Shiv pays a interest rate
of 7.2%.
Explanations:
Fundamentals of Ratio & Proportion
Average =
62 × 5 − 70 + x
⇒ x = 60
5
Weight of new student x = 60 kg
500
500 m, B travels 95 ×
= 475 m.
100
Similarly, when B travels 200 m, C travels 190 m. So
46.
Total weight = 25 × 10 kg
2 cases of weight 30 kg each are replaced by
2 cases of weight 32 kg each.
∴ Total weight = (25 × 10) – (2 × 30) + (2 × 32)
= 254 kg
Number of students = 10
So 60 =
13x + 100% of 13x 3
=
17x + 2500
5
if B travels 500 m C travels 190 ×
2
× 10000 = 4000
5
Average is decreased by 2 kg.
∴ New average = 60 kg
26x
3
=
17x + 2500 5
130x = 51x + 7500
79x = 7500
x = 94.93
So the original prices are 13 × 94.93= Rs. 1234.08
and 17 × 94.93 = Rs. 1613.81.
45.
Thus amount lent at 6% =
254
= 25.4 kg
10
∴ The average will change by 0.4 kg.
Let the original prices by 13x and 17x. So ratio for
new prices =
8 − 7.2 0.8 2
=
=
7.2 − 6 1.2 3
∴ Average =
5100 × 3
(3200 × 4) + (5100 × 3) + (2700 × 5) 1248 = Rs. 459
43.
Using alligation, we have
15 x – 9 x
50.
=
Q2
Q1
⇒
4
6
=
2
3
Let the cost price of first variety be Rs. 2x per
kilogram.
So the cost price of second variety becomes Rs. x
per kilogram.
Selling price of mixture is Rs. 36 per kilogram after
making profit of 20%.
So the cost price of mixture =
MBA
Test Prep
36
= 30 per kilogram
1.2
Page: 75
Applying the principle of alligation,
2x – 30 7
=
30 – x
3
6x – 90 = 210 – 7x
13x = 300
x = 23.07
So the cost price of first variety = Rs. 46.14 and
cost price of the second variety = Rs. 23.07.
53.
Applying the rule of alligation,
Averag e percentage
12%
–8%
Resu ltant averag e
11%
Q uantities
1
19
51.
3
Ratio of copper in first alloy = .
9
The ratio of the items that he is selling at a profit of
12% and at a loss of 8% is 19 : 1.
5
Ratio of copper in second alloy = .
9
Hence, he sold
1
Ratio of copper in new alloy = .
2
and
Applying the principle of alligation,
54.
5
–
9
1
–
2
⇒
1
2 = QI
Quantity of first alloy
3 QII = Quantity of second alloy
9
Apply rule of alligation for fractional water content.
Q uantities
1
× 64 = 16 kg
4
3
× 64 = 48 kg .
4
1
5
8
then the original amount of the mixture =
5
.
13
Milk proportion in the total solution of second variety
(VI ) =
9
.
13
7
.
Milk proportion to total in the required mixture =
13
Applying the principle of alligation,
11
56
3
8
The ratio of the volumes of the alcohol mixture to
water = 21 : 11. If only 8 L of water was added,
Milk proportion in the total solution of first variety
(VII ) =
3
7
Resu ltant averag e
and quantity of second alloy =
52.
1
× 60 = 3 pens at a 8% loss.
20
Averag e fractio nal
constitu tion of the
constutu en ts
QI 1
=
QII 3
∴ Quantity of first alloy =
19
× 60 = 57 pens at a 12% profit
20
Alcohol content in that =
55.
21× 8
.
11
4 21× 8 96
×
=
.
7
11
11
Let the volume of the mixture be x. Then
7
5
x, amount of B =
x.
12
12
When 9 L of mixture is taken out,
amount of A =
7
21
L;
×9 =
12
4
9
7
–
13 13 = VI
7
5
VII
–
13 13
amount of A withdrawn =
VI 2 1
= =
VII 2 1
5
21
x+
.
12
4
(Since all the milk that was removed was substituted
by water.)
Page: 76
amount of A left in the mixture =
7
21
x−
.
12
4
∴ Final amount of B in the mixture =
MBA
Test Prep
Solution Book-2
x – 98 5
=
100 – x 3
8x = 794
x = Rs. 99.25 per litre
∴ Cost of first solution = 99.25 + 2 = Rs. 101.25 per
litre and cost of second solution = Rs. 99.25 per
litre.
21
7
x−
4
7
12
= , x = 36.
Now ratio of A and B =
21
9
5
x+
4
12
∴
Amount of A =
7
× 36 = 21 L.
12
Alternative method:
Use rule of alligation:
After 9 L is taken out, the ratio of A and B is 7 : 5. To
this is added pure B to obtain a solution containing A
and B in the ratio 7 : 9. Considering the rule of
alligation for fractional B content,
Averag e fractio nal
constitu tion of the
constitu en ts
5
12
58.
118
= 59 calories .
2
Use alligation.
=
104 – 59 Vm Volume of mango juice
=
=
59 – 54
V0
Volume of orange juice
1
9
16
Resu ltant averag e
7
16
Q uantities
45 Vm
=
5
V0
7
48
Vm 9
or = V = 1
0
∴ Ratio of the original solution left to that of the
water added = 3 : 1 = 27 : 9.
Hence, the initial volume = 36 L.
∴
56.
Amount of A initially =
7
× 36 = 21 L.
12
Let t be the time for which he travelled on foot.
∴ 8t + 16(7 – t) = 80 or t = 4 hr.
The distance travelled on foot = 32 km.
Alternatively:
The problem could also be done using the principle
of alligation. The average speeds on foot and on a
bicycle are given and the resulting average speed
80
i.e. 7 is also given. We can find the ratio of
times and hence, the time taken for each part of the
journey.
57.
If the cost of second liquid is Rs. x per litre, then
cost of first liquid is Rs. (x + 2) per litre.
Selling price of mixture is Rs. 120, after making a
profit of Rs. 20.
So the cost price of mixture =
Use Alligation
120
= 100 per litre .
1.2
312
= 104 calories .
3
9 L of mango juice contains = 54 calories and 9 L of
mixture (orange and mango) contains
9 L of orange juice contains =
So fraction of orange juice in 18 L mixture is
59.
1
.
10
Apply rule of alligation:
Averag e co sts of ho rses
–6%
Resu ltant averag e
Costs of horses
7.5%
0%
6
7.5
∴ The ratio of the costs of the horses is
7.5 : 6 or 5 : 4.
∴ 5 : 4 is the ratio of cost of each horse.
∴
5
× 1350 = 750 is the cost of one which was sold
9
at a loss of 6% and
4
× 1350 = 600 is the cost of the
9
other.
(x + 2) – 100 5
=
100 – x
3
Explanations:
Fundamentals of Ratio & Proportion
MBA
Test Prep
Page: 77
60.
= 40%. Now 30 L is withdrawn and replaced with
Total fractional amount of milk
Total fractional amount of water
3
× 100 .
water. Final concentration = 30% =
10
2 3 3
+ +
3
4 5 = 40 + 45 + 36 = 121
=
20 + 15 + 24 59
1 1 2
+ +
3 4 5
61.
Hence,
40 − 30 10 1
=
=
mixture. ∴
40
40 4
Ratio of quantities = 2 : 5.
Ratio of prices = 3 : 1.
∴ Ratio of the values of sugar to orange peels
= 6 : 5.
1
th of the mixture = 30 L
4
So the volume at the end of first state = 120 L.
Or
∴ 11 [5.2 − 0.8] = $2.4 worth is sugar in marmalade
6
62.
63.
Fresh grapes contain 90% water. So 20 kg fresh
grapes contain 18 kg water and 2 kg pulp. Now if 2
kg pulp contributes 80% of dried grapes, then the
total amount of dried grapes = 2.5 kg.
This is
n
27
1 V – 3
=
1728 1 V
3 V–3
=
12
V
Hence, the initial volume =
2
× 120 = 80 L.
3
Solutions Exercise 1 (Level – 1)
1. c
n
A 3 . Let A = 3x, B = 7x
=
B 7
A + B = 45;
3x + 7x = 45, x =
3
2. d
x
y
x 35 3 5
=
× =
y 27 7 9
4
3. a
4
4
16
160
2
=
= =
81
240
3
∴ Quantity of rum left =
Let the fraction be
x
1
7x 35
y 27
or,
=
=
3
1
3y 27
7 35
⇒V=4L
1 240 – 80
=
1 240
If a, b and c are in continued proportion, the mean
proportional is b.
Therefore, b2 = ac, b2 = 8 × 72, b =
16
× 240 = 47.40 L
81
If a, b, c and d are in proportion,
5. e
Third proportional, c =
6. b
Let the number added be x. Then,
At the first stage, the concentration of milk is 60%
3
= × 100 . Since the volume increases by 50%,
5
3
times of the original volume.
2
2
Hence, the concentration becomes × 60%
3
Page: 78
576 = 24
3 27
a c
,
=
= or
15
d
b d
or d = 135 is the fourth proportional.
4. b
Level – 3
the volume becomes
45
= 4.5
10
B = 7x = 31.5
Final ratio of rum to total = Initial ratio of rum to total
V – X
×
V
65.
3
times of the initial volume.
2
Final ratio of water to total = Initial ratio of water to
V – X
total x
V
As the process is repeated two more times.
∴ The number of times we do the same operation
will be 3.
64.
1
th of milk has been removed from the
4
b2
30 × 30
=
= 45
a
20
4+x 2
= , 12 + 3x = 18 + 2x
9+x 3
x=6
MBA
Test Prep
Solution Book-2
7. d
A’s share = P ×
x
2000
= 3610 ×
x+y+z
4750
16. b
= Rs. 1,520
Where P = Profit and x, y and z are respective
shares of A, B and C.
8. c
x 2
=
y 7
Let x’s share = 2a, y’s share = 7a
a + b + c 3k + 4k + 7k 14k
=
=
=2
c
7k
7k
17. b
x ' s share
2a 2
=
=
x ' s share − y ' s share 5a 5
9. d
10. a
B' s part =
=
11. d
or
3A = 2B = C
B
2
B + B + 2B
3
Option c and d:
4x = 84
x = 4; Possible.
18. b
19. b
x 0.07 1
=
=
y 0.35 5
B’s money =
5
× 800 = 1000
4
3
× 1000 = 1500
2
Therefore, total amount of money = Rs. 3,300
C’s money =
C’s share = 2 B’s share
2
B’ share
3
2
B' s share
2
3
=
⇒ required ratio is
(2 B' s share – B' s share) 3
0.35 of x = 0.07 of y
∴
× 3960
3
× 3960 = 3 × 360 = 1080
11
Option a:
Numbers = 9x, 3x
12x = 84
x = 7; possible.
Option b:
8x = 84 ⇒ No whole number x is possible.
68 – x 3
= , ⇒ 272 – 4x
49 – x 4
= 147 – 3x ⇒ x = 125
A B C
= =
2 3 6
a b c
= = =k
3 4 7
a = 3k, b = 4k, c = 7k
Let
A’s share =
12. c
13. d
5x − 9 23
Let the number be 5x and 3x, then
=
3x − 9 12
60x – 108 = 69x – 207; 9x = 99, x = 11
The first number is 11 × 5 = 55
15. a
x x 68
17x 68
or,
+
=
=
, x = 28
7 10 10
70 10
21. b
x’s share ⇒ 3x + 30
y’s share ⇒ 4x + 20
z’s share ⇒ 5x + 50
Sum is 9700
12x + 100 = 9700, 12x = 9600, x = 800
y’s share = 4x + 20 = 3200 + 20 = 3220
22. d
Let B get’s
A get’s
C get’s
x 3 y 2 4
= & = =
y 4 z 3 6
⇒ x:y:z =3:4:6
In a ratio A:B, A is called the antecedent and B is
called the consequent.
Let antecedent = 4x & consequent be 9x.
4x = 36 ⇒ 9x = 81.
M 3
in 100 L mixture
=
W 1
Milk = 75 L, Water = 25 L
After adding 200 L of water, water = 225 L and
milk = 75 L
Ratio =
Let the two parts be x, (68 – x)
x
1
68 x
=
(68 − x ) = −
7 10
10 10
4x = 3y = 2z
⇒
14. d
20. c
Rs. x
Rs. x + 70
Rs. x – 80
x + x + 70 + x – 80 = 530
3x – 10 = 530, 3x = 540
x = 180
A ' s share 180 + 70 250 5
=
=
=
C' s share 180 − 80 100 2
75
=1:3
225
Explanations:
Fundamentals of Ratio & Proportion
MBA
Test Prep
Page: 79
23. b
A=
1
A 1
= ,
B, B = 2C or
2
B 2
Water available = 30 L – 6 L = 24 litres.
Now, 16 litres of milk are added,
⇒ Milk = 40l + 16 L = 56 L
Water = 24 litres
⇒ Milk : Water ratio is 56 : 24 = 7:3.
B 2
=
C 1
A:B:C=1:2:1
Shares of A, B, C are x, 2x, x
29. b
B
2x
2x 2
=
=
=
A + B x + 2x 3x 3
24. c
500
= 25 kg
20
Copper = 13 × 25 = 325 kg
x=
Initially, Number of boys = 250
Number of girls = 250
New Batch:
1
× 250 = 50 girls left & 25 boys joined
5
the class
30. d
Let present age of son = x and
present age of father = y
(y – 4) = 6(x – 4)
… (i)
… (ii)
(y + 12) = 2(x + 12)
from (i) and (ii),
y – 6x + 20 = 0
y – 2x – 12 = 0
4x = 32, x = 8 years, y = 28 years
Ratio of their present ages =
27. d
5
× 80 = 50 liters
8
Water = (80 – 50) = 30 liters.
When we take a 16 liter sample of this mixture.
5
Milk taken out = × 16 = 10 litres
8
Water taken out = (16 – 20) = 6 litres.
⇒ Milk available = 50 L – 10 L = 40 litres
Page: 80
16
.
56
1. b
B 7
=
G 5
Let number of boys = 7x
Let number of girls = 5x
7x + 5x = 72,
x=6
Number of boys = 42
Number of girls = 30
12 more girls should be admitted to make the ratio
equal.
2. d
Let A’s income = Rs. 3x
B’s income = Rs. 2x
A’s expenses = Rs. 5y
B’s expenses = Rs. 3y
A’s savings = 3x – 5y = 3000
B’s savings = 2x – 3y = 3000
Solving (i) and (ii), x = 6000
B’s income = 2x = Rs. 12,000
C1 5 C2 5
= ,
=
I1 8 I2
3
In 80 liters of mixture, milk =
2
7
Solutions Exercise 2 (Level 1)
28
=7:2
8
5 5
+
Copper 13 8 40 + 65 105
=
=
=
8 3 64 + 39 103
Iron
+
13 8
28. c
Number is
Let Father’s age be 7x, son’s age = 2x
After 15 years,
7x + 15 2
= , 7x + 15 = 4x + 30, x = 5
2x + 15 1
Present age of father = 35
Present age of son = 10
Father’s age when son was born = 35 – 10
= 25
26. d
Simplest form =
Numerator = 2x
Denominator = 7x
7x – 2x = 40, x = 8
Boys 275 11
=
=
Girls 200 8
25. a
Let copper = 13x
Zinc = 7x
In 500 kg, 13x + 7x = 500
3. a
Let the number to be added be, x.
Then,
4. a
… (i)
… (ii)
7+x
1
= ,x=5
19 + x 2
Take 7 L of the solution of each mixture,
(LCM of 5 + 2, 6 + 1, 4 + 3)
∴ net amount of water in the resulting mixture,
5
6
4
× 7 + ×7 + × 7 = 15 L
7
7
7
The net amount of alcohol is,
2
1
3
×7+ ×7+ ×7 =6L
7
7
7
Ratio is, 15 : 6 = 5 : 2
MBA
Test Prep
Solution Book-2
5. c
If the ratio is a : b, then a + b should divide 12
completely. Only 3 : 2 does not divide, the rest
are all possible ratios.
6. d
If the ship weighs 5 kg, weight above water
= 3 kg.
Weight submerged = 2 kg.
Therefore, desired ratio = 2 : 3.
∴ Cost of glass lenses = 3 (52 − 40) = $4 .
1
(Compare with the above algebraic method.)
7. c
5% of x = 30. Therefore, x = 600. Therefore,
number of entrants who lost = 570.
8. c
If the ratio is a : b, then a + b should divide 8
completely. Only 36 does not fulfil this condition.
9. e
Considering the ratio,
B’s + C’s shares = 20x and A’s share = 5x.
Hence, the second sentence gives no additional
information.
Since the total amount is not given, we cannot say
anything about A’s share.
10. d
LCM(15, 25) = 75 min is the time when both of
them would meet at the starting point for the
first time. In this time A takes a lead of two rounds
over B. If they have to meet for the first time,
then A has to take a lead of one round, which
happens after 37.5 min.
11. a
13.a
Ratio of savings of A and B is 5 : 3. So it is
2
obvious that A’s savings is 66 % more than that
3
of B.
14.b
Here Sujith earns more but spends less so
definitely his savings will be more than that of
Pratima.
15.d
1 : 5 and 3 : 5 solutions
Since equal quantities of both are mixed, we can
take LCM of (1 + 5 = 6) and (3 + 5 = 8), i.e. 24.
Assuming that 24 L of both solutions are mixed
together in the resultant solution.
Milk = 4 + 9 = 13; Water = 20 + 15 = 35
Hence, the ratio of water and milk = 35 : 13.
Alternative method:
Amount
1
1
of milk in 1 : 5 solution =
.
6
6
Amount
3
3
of milk in 3 : 5 solution =
.
8
8
Let the incomes are 3x and 2x and the
expenditures are 5y and 3y.
Therefore, 3x – 5y = 1000 and 2x – 3y = 1000.
y = 1000, x = 2000.
Therefore, incomes are Rs. 6000 and Rs. 4000.
Alternative method:
In first step see which option gives a ratio 3 : 2 in
incomes. Unfortunately, all three are in the ratio 3 :
2.
Now subtract Rs. 1,000 (savings) from each option
to
get the ratio of expenditures.
We see that only option (a) gives a ratio 5 : 3.
12.b
Plastic lenses are four times costlier than glass
lenses.
∴ If cost of glass lenses = x, cost of plastic
lenses = 4x.
Cost of age examination and frames = y.
∴ y + x = 40 and y + 4x = 52
⇒ 3x = 12 ⇒ x = $4
Alternative method:
The price differential is due to plastic lenses
which cost four times as much as the glass
lenses.
∴ Price differential = Cost of plastic lenses over
and above glass lenses = 3 (Cost of glass
lenses).
Explanations:
Fundamentals of Ratio & Proportion
1
6 =1
3
–x 1
8
x–
⇒ 2x =
1 3 4 + 9 13
+ =
=
6 8
24
24
13
48
∴ Water : Milk = (48 – 13) : 13
= 35 : 13
⇒x=
MBA
Test Prep
Page: 81
16. c
Mileage when petrol is adulterated
=
3
of (16) = 12 km/L
4
4
Hence, the cost increases to a factor of the
3
original cost of maintenance. So it increases by
1
× 100 = 33.33% .
3
Questions 20 and 21: Half of the volumes of the containers
A and B are emptied into the third container. Hence, the
volumes emptied are 108 cm3 and 32 cm3 respectively.
20. c
21. e
Hence, the ratio of water to milk is 108 : 32 or 216 :
64.
140
× 100 = 14%
1000
22. a
Alternative method:
We can safely assume that the car runs for the
same
number of kilometres.
3
of its original value,
Now mileage is reduced to
4
which is, say, x.
∴ Distance = Mileage × Number of litres
⇒ x × y1 =
Elder son
Younger son now
20x
5x
4x
After 10x years,
30x
15x
14x
So 30x = 2 × 14x + 3
⇒ x = 1.5 ∴ 20x = 30 years.
23. a
3
y
4
x × y2 ⇒ 2 =
= 1.33
4
y1 3
The final amount of milk after two operations
=
Therefore, we see that the consumption of fuel
24. c
increases by 33.3% .
Father
4 4 3
× × × 50 = 19.2 L
5 5 5
2xy
The harmonic mean of 2 numbers x and y is (x + y )
and the geometric mean is xy .
17.a
5
5
If the number is x, x −
x = 125 or x = 360.
8
18
∴
18. e
19. a
If A’s age is 5x, B’s age = 7x.
Case I: 7x – (5x + 6) = 2
So x = 4.
Case II: (5x + 6) – 7x = 2
So x = 2
Hence, sum of their present ages = 12 × 4 = 48
years or 12 × 2 = 24 years.
⇒
or
This can be solved using alligation.
2xy
x+y
xy
=
12
,
13
2 xy 12
xy
6
=
=
or
x + y 13
x + y 13
x
+
xy
y
13
x
or
=
+
6
y
xy
y 13
=
x
6
From here, you can put options in the above equation
and check. Only option (c) satisfies.
25. c
Let X ml be added to the solution.
⇒
[(15% of
[400
400) + X]
+ X]
=
32
100
⇒ X = 100 ml
So the amounts invested are in the ratio 2.5 : 1.5
= 5 : 3.
Since the amounts invested at the two rates are in
the ratio 5 : 3, and Rs. 680 is invested at 6%,
amount invested at 10% =
Page: 82
680
× 3 = Rs. 408.
5
Alternative Method:
15% of 400 ml = 60 ml
Rest is water = 400 – 60 = 340 ml
Since, water is constant ⇒ 340 ml will be 68% (in
the new mixture)
⇒ 340 is 68%
340
× 100 = 500 ml ; Extra alcohol
68
required 500 – 400 = 100 ml
⇒ 100 is
MBA
Test Prep
Solution Book-2
Solutions Exercise 3 (Level 2)
1. d
2. a
3. e
The question uses the same concept as in Q 17.
Number of mugs = 12
The sum of broken and unbroken mugs must be
equal to 12.
Sum of ratio = 3, 12, 5, 6, (5 does not divide 12)
∴ 3 : 2 cannot be the ratio.
A’s effective investment = 16000 × 6 =
B’s effective investment = 12000 × 8 =
C’s effective investment = 1000 × 12 =
Profit sharing ratio = 96 : 96 : 12 or 8 :
Rs. 96,000
Rs. 96,000
Rs. 12,000
8:1
Bs’ share =
5. d
For every 100 paise that A gets, B gets 65 paise
and C gets 35 paise.
35
C' s share = 560 = sum ×
200
∴ Sum = Rs. 3,200
9. b
Let the amount of second and third quality of sugar
in the mixture be ‘x’ and ‘y’ kg respectively.
∴ 15 × 4 + 18x + 22y = 22 ×
10. a
r=
11. e
∴ using the rule of alligation,
11
9
1 0.5
0 .5
Let the quantity of second and third variety be x
and y kg respectively.
100
∴ 20 × 1 + 30x + 24y = 30 ×
[1 + x + y ]
120
⇒ 5x − y = 5 …(i)
All the four options (a), (b), (c) and (d) satisfy
equation (i).
12. e
Let’s take
12 litre of 5 : 1 sample
24 litre of 2 : 1 sample and
36 litre of 3 : 1 sample
Amount of milk in the final mixture
5
2
3
× 12 + × 24 + × 36 = 53 L
6
3
4
Amount of water in the final mixture = 72 – 53
= 19 L
⇒ milk : water = 53 : 19.
=
84
× 26500 = 10,500
212
2100
× 100 = 10.5% per annum
20,000
100
= Rs. 16 / kg
50
Let the quantity of second and third variety be ‘a’
and ‘b’ kg.
∴ 15 × 6 + 18a + 20b = 16[6 + a + b]
Mean rice of the mixture is 8 ×
⇒ a + 2b = 3
a = 1 and b = 1 is the only integral solution.
Let A’s, B’s and C’s investments be 8x, 7x, 5x
A’s effective investment = (8x) × 5 + (4x) × 7
= 40x + 28x = 68 x
B’s effective investment = 7x × 12 = 84x
C’s effective investment = 5x × 12 = 60x
The net rate of interest paid is,
13. e
Using alligation we find the amount of copper
(average amount of copper in the final mixture is
x)
Thus ratio of loan from first to second bank is
0.5 : 1.5 = 1:3
Explanations:
Fundamentals of Ratio & Proportion
7
10
2
5
x
2
1 .5
100
[4 + x + y ]
110
⇒ y − x = 10
L:R:O=5:7:3
Let Labour cost be 5x, Raw Material Cost
= 7x
Overheads = 3x, Total cost = 15x
Profits = 20% of 15x = 3x
B’s share =
7. a
Let the required quantity be x kg.
∴ 20 × 2 + 24x + 30 × 3 = 25 (3 + x + 2)
⇒x=5
3
× 5200 = 2600
6
Material cos t 7x 7
=
=
Pr ofit
3x 3
6. c
8.b
Ratio of their profits = Ratio of their investments
= 4000 × 12 : 8000 × 9 : 12000 × 2
48 : 72 : 24
2 : 3 : 1
4. c
1
∴ Loan from 1st bank = × 20,000 = 5,000.
4
and Loan from 2nd bank = 15000.
1
7
– x
2 ⇒x=1
⇒ 10
=
2
x – 2 1
5
1
of the new alloy the final ratio
2
of copper to nickel will be 1 : 1.
Since, copper is
MBA
Test Prep
Page: 83
14. b
The ratio between the distance and the speed is
the same. Since the ratios between the two are the
same, the time has to be constant and hence the
ratio needs to be 1 : 1 : 1 or mathematically,
Time =
Dis tan ce
Speed
Hence, Ratio of time =
15. e
20. c
Ratio of dis tan ce
= 1: 1: 1 .
Ratio of speed
Selling at Rs. 11 per kilogram, profit percentage
= 25%.
Suppose the person buys A apples and M
mangoes and cost price of an apple is Rs. x.
Therefore, cost price of a mango will be 2x.
Total cost price = Ax + 2Mx.
Now selling price of an apple is 2x.
∴ SP of a mango will be 6x.
Total SP = 2Ax + 6Mx.
Now we have
2Ax + 6Mx =
21. a
11
= Rs. 8.8
1.25
Applying the rule of alligation,
Therefore, CP =
5
M 1
(Ax + 2Mx) or
= .
2
A 2
Let us assume that out of 100, he obtained 72
marks. If he had attempted four more questions,
he would have made one more mistake. Hence,
three of them are correct. The three correct
questions secure him 12 marks more. This
12
= 4 marks and
3
hence, number of questions is 25.
means, each question carries
Alternative method:
We can clearly see that three correct answers
increase the score by 12%. [Three correct questions
as four new questions and one mistake]
100% score corresponds to
3
× 100% = 25 questions.
12%
The required ratio is 2 : 3.
16. c
of the examination =
17. b
18. d
22. a
Let the maximum marks in the quizzes be 1 and
that in the examination is 3. Hence, the weightage
3 ×1
1
= .
6 × 1+ 3 × 1 3
Alternative method:
In these type of questions it is always easier to
work
with man-hours (or man-days, etc). 50 men working
10 hr a day for 8 days is 50 × 10 × 8 = 4,000 manhours.
Now 35 (boys and girls) working 15 hr a day for
If he spends ‘t’ minutes on the second half, then
he spends
2
t on the first half.
3
Therefore,
2
t + t = 90 or t = 54.
3
1
(say) x days is × 35 × 15 × x man-hours
2
If there were 2, 5, 1 coins of the denominations
Rs. 1.50, one-rupee and 25-paisa respectively,
then value of the coins would be Rs. 8.25. Hence,
if the total value has to be Rs. 330, then the
number of coins would have to be multiplied by
a factor of 40. There would be in all 80 of Rs.
1.50 coins, 200 one-rupee coins and 40 25-paisa
coins.
⇒
Let the value of the turban be Rs. x.
Then
Page: 84
100 + x 12
=
⇒ or x = Rs. 40.
65 + x
9
1
× 35 × 15 × x = 4,000
2
⇒x=
23.c
19. c
20 boys = 10 men. 15 girls = 7.5 men.
Therefore, total number of men = 17.5
Now 50 × 10 × 8 = (17.5 × 15)x, where x is the
number of days.
Therefore, x = 15.24 (approximately).
8000
= 15.24 days.
35 × 15
Suppose numbers of students who pass and
fail are 3x and x respectively.
∴ Total number of students appearing
= 3x + x = 4x.
MBA
Test Prep
Solution Book-2
In second case,
total number of students who appear = 4x + 8;
total number of students who pass = 3x – 6.
∴ Total number of students who fail
= (4x + 8) – (3x – 6) = x + 14.
3x − 6
= 1 ⇒ x = 10.
x + 14
∴ Total number of students is 40.
30
× 100 = 30%
100
of water is added to pure milk.
Therefore, 30 L is added or
28. e
Now we have
24.c
25. c
Ratio of white to yellow balls = 6 : 5.
Difference in the number of white and yellow
balls
= 6x – 5x = x = 45.
Therefore, number of white balls now available
= 45 × 6.
Number of white balls ordered
= (45 × 6) – 45 = 225.
29.c
Ratio of white mice to total mice
30. c
=
1
1
2
:
1
1
8
= 1: 4 =
1
1
3
:
1
1
: 1.
4
1
9
= 3 : 9 = 1: 3 =
⇒
235 + x
+ 10
8
7
1
x = (235 ) + 10 ⇒ x = 45
8
8
∴ Total collection = 235 + 45 = Rs. 280.
1 1
: = 3:4 .
4 3
27. a
4
1
− 140 × = 80 − 35 = 45 L.
7
4
Six people contribute a total of Rs. 180.
Let the seventh person contributes Rs. x.
Eighth person contributes Rs. 55.
Total contributions of these eight persons
= 235 + x.
Now x =
1
: 1.
3
∴ Ratio of white mice to gray mice =
26. e
4
1
v − x = v ⇒ 9v = 28x
7
4
Since by adding 35 L the level rises from a
quarter to a half, the volume of the vessel = 35 ×
4 = 140 L.
Therefore, x = 140 ×
Ratio of gray mice to total mice
=
No price values (either of profit amount or of
cost price) is given. Obviously, SP cannot be
determined.
When 5 L brandy is transferred, volume of
brandy left in vessel I = 15 L.
Volume of brandy in vessel II = 5 L.
Volume of water in vessel II = 20 L.
Now, 4 L of mixture containing brandy and water
in the ratio 1 : 4 is poured into pure brandy of
volume 15 L.
4 L of mixture contains 3.2 L water and
0.8 L brandy.
∴ Volume of brandy left in vessel II
= 5 – 0.8 = 4.2 L.
Ratio of water in vessel I to brandy in vessel II
= 3.2 : 4.2 = 16 : 21.
Solutions Exercise 4 (Level 2)
1. c
Percentage of whisky in mixture = 75%.
Percentage of whisky in water = 0%.
In 100 L pure milk, let him add ‘a’ litres water.
Let CP of 100 L pure milk = Rs. 100.
CP of (100 + x) L adulterated milk = 100.
10000
CP of 100 L adulterated milk =
.
100 + x
SP of 100 L adulterated milk = 100.
(
(
100
Profit percentage =
100
1 − 100
+x
10000
100 + x
)
) × 100
So mixture and water should be mixed in the
1
ratio 2 : 1 or we need to replace
3
of the mixture
with water.
or x = 30.
Explanations:
Fundamentals of Ratio & Proportion
MBA
Test Prep
Page: 85
2. a
Let x be the distance travelled by bus.
Quantity of milk after third draw
x 1000 − x
∴ 80 + 36 = 16
⇒ x = 770 km (approximately)
3. d
9
9
9
×
×
× 80 = 58.32
10 10 10
∴ Ratio = 100 : 81.
=
Applying the rule of alligation:
Alternative method:
We know that
New amount of liquid X
=
Original amount of liquid X
n
Amount of liquid taken out and replaced
1 −
Total amount of liquid
where n = Number of times the process is repeated
2
8
9
= 1 –
= 10
80
∴
4. e
∴ Ratio = 102 : 92 = 100 : 81
Ratio of volumes of two solutions = 1 : 2.
Tin content in the alloy = 100 – 77.78% = 22.22%.
Using the principle of alligation,
2
6. b
Initial respective amount of milk and water is 28
L and 7 L.
7 L of water added means now we have 14 L of
water. To keep the ratio M : W same, amount of
milk must be doubled, i.e. 28 L of milk must be
added.
Alternative method:
In order to keep the solution of the same ratio, we
need to add milk in the same ratio as already there in
the solution.
∴ We need to add
7. a
4
× 7 = 28 L of milk.
1
Rule of alligation for fractional content of water:
50
× 18 : 18 = 32.4 : 18 .
Ratio of volumes =
27
.
78
∴ Total tin content in the new alloy
= (22.22% of 32.4) + 18 = 25.2 kg.
Total copper = (77.78% of 32.4) = 25.2 kg.
Hence, none of these is the correct option.
By the way, is there any need to do any
calculations? In the question it is given that the
new alloy has 50%, i.e. equal amount of copper
and tin.
5.b
Quantity of milk after first draw
90
× 80 = 72 kg .
100
Quantity of milk after second draw
=
=
Page: 86
or 11 : 1 or 33 : 3
∴ 3 L of water must be added to 33 L of solution
of milk.
90 90
× 80 = 64 kg .
100 100
MBA
Test Prep
Solution Book-2
8.b
9.b
In fact, she takes out 5% of the existing volume
of wine in every operation.
After the first operation concentration of wine
is 95%.
After the second operation concentration of
wine is 95 × 0.95 = 90.25%.
After the third operation concentration of wine
is90.25 × (0.95) = 85.7375%.
Obviously, after the fourth operation the
concentration of wine will be less than 85%.
13. b
450
= 3.75
120
Using the alligation method,
Hence, CP =
The ratio of incomes of Rashmi and Divya is
3 : 5. Ratio of their expenditures is 2 : 3, i.e.
3 : 4.5. Had the ratio of expenditures been 3 : 5,
ratio of savings also would have been 3 : 5; but
since ratio of their expenditures is 3 : 4.5 only,
obviously savings of Divya will be something
0 .25
5
of savings of Rashmi and thus
3
Divya will save more.
more than
10.c
Income
Rohit
Sunil
5x
7x
Expenditure
5y
7y
Saving
5x – 5y
7x – 7y
Ratio of savings is
11.b
SP = 4.5 and profit = 20%.
i.e. 1 : 1
Hence, in a 40 kg mixture, there are 20 kg of the first
variety and 20 kg of the second.
14. a
Here we calculate the number of birds in terms of
number of trees. We know that there are 630 birds.
Let the number of trees be x.
∴ Number of sparrows =
5( x − y )
=5:7 .
7( x − y )
Other birds =
n(n + 1)(2n + 1)
.
6
Square of sum of first n natural numbers is
Thus, x +
n2 (n + 1)2
= 17 : 325
4
x 3
× ×4 .
2 4
x 3 x 21
x = 630 ⇒ x = 240
+
=
8
2
8
Hence, other birds are there in
3x
= 90 trees.
8
On solving, we get n = 25.
6a2 = 2(lb + bh + lh)
l : b : h = 1: 2 : 4. Therefore, b = 2l; h = 4l
Hence, 6a2 = 2[2l2 + 8l2 + 4l2]
6a2 = 28l2
If IBM initially quoted Rs. 7x lakh, SGI quoted 4x lakh.
IBM’s final quote = (4x – 1) lakh
14 2
14 2 3
a = l ∴ a3 = l
3
3
Now
= 17 : 325 .
4
15.a
1
12.c
x
×2 .
2
x 1
Also number of pigeons = × × 1 .
2 4
Sum of squares of first n natural number is
n(n +1) ( 2n +1)
6
n 2 (n +1)2
0 .25
4x − 1 3
=
or x = 1
4x
4
IBM’s bid winning price = Rs. 3 lakh.
So IBM wins the bid at 4x – 1 = Rs. 3 lakh.
Note: Whoever quotes the minimum price, wins the
bid.
Explanations:
Fundamentals of Ratio & Proportion
3
Ratio of the volumes of the cube and cuboid
3
14 2
= a3 : lbh = : 8
3
MBA
Test Prep
Page: 87
16. a
If three kinds of birds are taken to be 3x, 7x and 5x
respectively, then 7x – 3x = 63y (where y is any
positive integer). As the number is a multiple of both
9 and 7, it has to be a multiple of 63.
20. d
A, B and C are to share profit in the ratio 4 : 3 : 7.
But, since A is to receive 10% of profit, only 90% is
to be shared.
A’s total share = 10% of profit +
63 y
4
Minimum value of y for which x is a natural number
is 4.
⇒ x = 63
Hence, the number of birds = 15x = 945.
⇒x=
4
× 90% of profit
14
= 6000
or (10 +
180
)% of profit = 6000
7
250
% P = 6000 ⇒ P = Rs. 16,800
7
B and C’s share combined
or
17. e
18. d
Price = k × Weight, where k is any constant.
Original weight of the stone = 28x.
Original price = 28kx.
New price = k(15x + 13x) = 28kx.
Hence, there is no profit, no loss.
Concept: Please note that since we have a linear
function, there is neither profit nor loss.
10
10 90
of 90% profit =
×
× 16800
14
14 100
= 10,800
=
21. a
In a mixture of 1,000 ml, milk : water = 3 : 1.
Hence, milk = 750 ml, water 250 ml.
A 250 ml of 3 : 2 solution contains 150 ml milk and
100 ml water.
Total milk = 900 ml, total water = 350 ml.
After using 250 ml to make curd milk used
250
× 900 = 180 ml.
1250
Pure milk left = 900 – 180 = 720 ml.
30
5
=
or x = 2
(10 + x ) 2
=
19. d
Let income and expenditure of the first person be
5x and 9y respectively.
Then income and expenditure of the second person
are 4x and 7y respectively.
5x – 9y = 4x – 7y = 500
or x = 2y.
Therefore, y = 500, and x = 1000.
Incomes are Rs. 5,000 and Rs. 4,000 respectively.
OR
By checking the choices you will find that only for
choice (d) is the ratio of incomes 5 : 4.
Alternative method:
In the given choices, the choices (a) and (c) are not
in
the ratio 5 : 4. Now we can subtract Rs. 500 from
incomes and check the ratio of expenditures in
choice (b). Ratio of expenditures
= (3750 – 500) : (3000 – 500)
= 3250 : 2500 = 13 : 10
Hence, the solution is (d).
Please note that we need not calculate the ratio of
expenditures in (d).
In the original mixture,
3
spirit =
× 40 = 30 L,
4
1
water =
× 40 = 10 L.
4
To get the ratio 5 : 2, let’s assume that we add x
litres of water.
Hence, 2 L of water has to be added.
22.d
i.e. 2 : 1
Hence x litres must be 12 L, i.e. (6 × 2).
23. c
This is a classic case for approximations. We see
that the ratio 1075 : 1000, when converted to
percentage terms, is approximately 52% : 48%. Thus,
we can calculate the literacy percentage as
24
8
× 52 +
× 48 ≈ 16.3%
100
100
Now as 100 – 16.3 = 83.7%, only choices (c) and
(e) are left. We calculate the number of literate
people as
16.3
× 311250 = 50,700 .
100
Page: 88
MBA
Test Prep
Solution Book-2
Thus, assuming we have x students initially.
x(26 – 27) + 3(29 – 27) = 0
We could have approximated the ratio of men to
women as 50% : 50% also.
24. b
Let x be the number of children employed.
Then numbers of men and women are 8x and 5x
respectively.
Let y be the common factor for wages.
Then wages of men, women and children are Rs.
5y, 2y and 3y respectively.
Hence, total wages of different categories will be
40xy, 10xy and 3xy respectively.
Total wages = (40 + 10 + 3)xy = 53xy = 318.
So xy = 6.
Thus, they will be getting 40 × 6, 10 × 6 and 3 × 6
respectively.
Alternative method:
⇒ –x+6=0
x=6
Please note that always subtract the average from
the number and not vice versa.
28. a
Assuming that there is 1 L of spirit and 3 L of water,
the total volume of the mixture is 4 L.
After increasing its volume by 25%, volume of the
mixture is 4 + 25% of 4 = 5 L.
So the ratio of volumes of spirit to water = 2 : 3.
29.c
A ratio
Sum of all three wages have to be equal to Rs. 318.
Only (b) displays this property.
25. a
26. a
Speed in upstream
= Speed of the boat in still water – Speed of the river
= b – r.
Speed in downstream
= Speed of the boat in still water + Speed of the river
= b + r.
Therefore, b + r = 3(b – r)
or 2b = 4r
or b = 2r
Since r = 4 km/hr, b = 8 km/hr.
Alternative method:
1 1 1
: : = 0.5 : 0.33 : 0.25
2 3 4
Thus, C, who was getting the least, finally got the
maximum amount.
This eliminates (a) and (b).
Thus, gain of
Milk in final/Milk in original = [(a – b)/a]n,
where a = Quantity of mixture,
b = Amount taken out during each operation,
n = Number of times the operation is repeated.
3
4
0.25
–
C = Rs.117
4 + 3 + 2 0.25 + 1/ 3 + 0.5
= Rs. 25
Hence, the answer is (c).
3
40 − 10
3
Milk in final/Milk in original =
= 4 .
40
1
1 1
:
:
is equivalent to 6 : 4 : 3.
2
3 4
So in this case, A, B and C would have got Rs. 54,
Rs. 36 and Rs. 27 respectively.
But actually the money was divided in the ratio
2 : 3 : 4 and shares of A, B and C in this case was
26, 39 and 52 respectively.
Hence, the answer is (c).
30. d
27
= 16.875 L.
Milk in final solution = 40 ×
64
Water = 23.125 L.
Required ratio = 16.875 : 23.125 = 27 : 37.
27. b
This question can either be done by substituting
choices, or by using weighted averages.
If x be the number of members in club initially, then
sum of ages of (x + 3) members now = 26x + 87
= 27(x + 3)
Hence, x = 6.
So number of members now = 9.
Alternative method:
This method is called sum of deviations, as follows.
“The sum of all weighted deviations from the average
is zero.”
Explanations:
Fundamentals of Ratio & Proportion
i.e. 7 : 1
Hence, milk and water are in the ratio 7 : 1 since
there is 28 L milk. Hence, water should be 4 L.
MBA
Test Prep
Page: 89
Solutions Exercise 5 (Level 2)
1. a
If A, B and C represent the number of three
denominations in increasing order of their values,
then A = 7B, C = 4B.
Since C = 12, B = 3 and A = 21, the total money in the
purse = 21 × 1 + 3 × 10 + 12 × 20 = Rs. 291.
2.c
If there are x number of one-rupee coins in the bag,
then there are 8x and 16x, 50-paisa and 25-paise
coins respectively.
Therefore, money in the bag = x + 4x + 4x = 9x =
495.
So x = 55.
Hence, there are 8 × 55 = 440 50-paisa coins in the
bag.
3. a
A
B
1
B
+
= (2C – A) =
2
3
2
2
∴ B = 3A, C = 2A
So amount spent
3A
= (A + 3A + 2A) –
= 4.5A
2
Percentage of amount spent = 75%.
∴
10.e
We cannot determine the final ratio unless we know
the volume in 3 jars.
11. b
Mixing two lumps, we have 18 g from each metal.
So price of the second metal
(87 + 78.60 − 6.7 × 18)
= Rs. 2.50 per gram
18
∴ 6.7x + 2.5(18 – x) = 87
∴ x = 10 g and 18 – x = 8 g
=
There are 880 25-paisa coins.
Value of the money = 0.25 × 880 = Rs. 220.
4.c
2
12.c
V1
V2
V3
Initial volume:
0L
40 L milk
80 L W water
After operation 1:
0
20 L M + 20 L W 60 L w + 20 L milk
After operation 2:
0
15 L M + 25 L W 55 L w + 25 L milk
5.a
Check from the above data.
6.a
Suppose he invested Rs. x and Rs. y in the
matches of South Africa and India
respectively. From the first match he will get 3
x, but from the second he will get nothing.
Now 3x = (x + y) + (x + y)
⇒ 1−
8.c
When 25 L is withdrawn, half the solution is
removed. Hence, 9.6 L of milk is left.
14. c
Let Tina’s share be T, Issan’s be I, Abhishek’s be A
and Fatima’s be F.
I 80A
=F−4
Given that T + 3 = I + =
3 100
You get
T=F–7
... (i)
x
150
5
or y =
.
100
1
Suppose he invested 3X on the first match
and 4X on the second match.
He will get back 3X × 3 = 9X .
So gain = 9X – 7X = 2X = 200000
or X = 100000
:. Total investment = 7X = 7 × 100000 = 700000
9. c
8x
× 100 = 266.66%.
3x
A +B+C
⇒ Originally B had an amount
C=
3
= 2C – A
Now A′ =
Page: 90
A
1
, B′ = B , C′ = 0
2
3
3
(F – 4)
4
... (ii)
5
(F – 4)
... (iii)
4
Also given that T + I + F + A = 80
... (iv)
Substituting the values from (i), (ii) and (iii) in (iv),
we get
A=
3F 5F
F + 4 + 4 + F – 7 – 3 – 5 = 80 or F = 23.75
Investment = x + 2x = 3x
He gets back 3x + 8x = 11x. So gain = 8x.
Gain percentage =
x
2
= ⇒ x = 18
54
3
13. a
I=
7.a
2
24
4
x
x
54 1 −
=
= 24 ⇒ 1 − 54 =
54
9
54
15. a
Let the price of one tea = x
⇒ the price of other tea=
x
2
Price of 1 kg = 2x + 3 × x = 7x
5 5 2 10
But CP =
MBA
Test Prep
17.50 × 100
= 14
125
Solution Book-2
(Only the first option satisfies the first condition.)
22. d
10 kg copper is from alloy II.
Hence, 28 kg copper is from alloy I.
In alloy I, tin : copper = 3 : 4
Since, 4 units = 28 kg
Alloy = 7 units = 49 kg
23 c
Use alligations.
7x
⇒
= 14 ⇒ x = 20
10
So the price of the tea’s are 20 and 10.
16. c
1
of the whole gets thrice
4
of what the others get on an average, each one will
get
If the person who gets
So lution 1
Averag e percentage
of acid in so lutions
1 1
1
× =
of the whole.
3 4 12
Therefore, if there are x persons other than the
person who gets one-fourth of the whole, then
=
90%
75%
Resu ltant average
1
x
+
=1
4
12
So x = 9
Hence, total number of people = 10
17. c
So lution 2
Ratio of q uantities
of acid in so lutions
∴
6p + 8q = 7.5(p + q); 6q + 8p = n(p + q)
14(p + q) = (7.5 + n) p + q ⇒ n = 6.5
X
Y
90 − 78 Y
=
78 − 75 X
12 Y
X 1
= ⇒ =
3
X
Y 4
So ratio of quantities of solutions = 1 : 4
Total quantity = 30 L
∴ Quantity of 90% solution of concentrated acid
⇒
18. b
Amount of 16% iodine in the mixture
16
gm
100
Now the amount of iodine becomes 20% of the
mixture.
Amount
16
100
evaporated = 735 − 735 ×
× 20 = 147 g
100
= 735 ×
19. b
Whether you add them in the ratio 2 : 3 or 1 : 1, we
get the same concentration.
Hence, both of them must have a 35% concentration.
So (b) is the answer.
20. c
Only 5 women were college graduates while 20
women were employed. Therefore, maximum
number of ‘employed graduate women’ can be 5.
Since number of ‘unemployed men’ was (55 – 15) =
40. Answer cannot be greater than
21. b
1
.
8
Only possibility is if 650 is divided by a multiple of
13, because the other jars have a capacity that is
not a multiple of 13. 403 is the only choice possible.
Alternative method:
As the beakers filled by same fractions,
a
b
c
a+b+c
=
=
=
250 650 200 250 + 650 + 200
3 a + b + c = 682 ∴
= 30 ×
24. d
1
=6L
5
s−a s−b s−c
=
=
=k
1
7
4
Simplifying this, we get
a : b : c = 11 : 5 : 8
Solutions Exercise 6 (Level 3)
1.c
After first
amount of
amount of
amount of
amount of
transfer,
wine in first vessel = 9 L,
water in first vessel = 3 L,
wine in second vessel = 3 L,
water in second vessel = 1 L.
3 L drawn from first vessel contains
of wine and
3
9
×3 =
L
4
4
1
3
×3 =
L of water.
4
4
3 L drawn from second vessel contains
wine and
b
682
=
650 1100
9
L of
4
3
L of water.
4
b = 403 ml
Explanations:
Fundamentals of Ratio & Proportion
MBA
Test Prep
Page: 91
∴
Amount of wine in first vessel after second
3. c
9 9
+ =9.
4 4
Amount of water in first vessel after second
transfer = 9 −
3 3
+ =3.
4 4
for second
From (ii), a =
transfer = 3 −
Similarly,
5a + b > 51
3a – b = 21
8a > 72 or a > 9.
… (i)
… (ii)
21 + b
3
21 + b
>9⇒b>6
3
Combining a > 9, b > 6.
⇒
vessel,
wine
=
9 9
+ =3
4 4
3 3
Water = 1 − + = 1
4 4
∴ Ratio is same in both the vessels.
3−
Alternative method:
Ratio of wine to water in first vessel after first
transfer
is (12 – 3) : 3 = 3 : 1.
Ratio of wine to water in second vessel after first
transfer is 3 : (4 – 3) = 3 : 1.
Since the ratio of wine to water in both the vessels
is equal. Any amount of such exchange any number
of times will not alter the ratio of wine to water.
2.e
4. b
Average age = 13. Number of students = 50
W ∝ H and H ∝ A ⇒ W ∝ A
∴ W = K1 H and H = K2A ⇒ W = K1K2A
For A = 11, H = 165 and W = 33
33
= 3 ∴ W = 3A
11
Since W ∝ A, the averages should also be in direct
proportion.
∴ Wavg = K1K2 A avg = 3 × 13 = 39 kg
K1K2 =
Consider two cases:
i.
Princy
Kunjumol
Income
100
200
Expenditure
1
5
Saving
100 – 1 = 99
200 – 5 = 195
Thus, Kunjumol is saving more.
ii.
Princy
Kunjumol
Income
100
200
Expenditure
39
195
Saving
100 – 39 = 61
200 – 195 =5
Thus, Princy is saving more.
Hence, their savings depends upon their
respective expenditures.
Page: 92
MBA
Test Prep
Solution Book-2
Explanations: Time, Speed & Distance
Solutions (Non MCQ)
6.
2.
Speed of second train =
Distance to be covered = 100 m.
Speed of the person = 36 km/hr = 10 m/s.
40 100
=
3
3
Time taken by the two trains to completely pass
100 m
= 10 s.
Therefore, time taken =
10 m / s
= 20 +
2 × 75 × 50
= 60 km / hr
Average speed =
[75 + 50]
each other =
7.
3.
One revolution means circumference of the wheel
= 250 cm = 2.5 m.
In 10 revolutions 25 m is covered by the wheel.
Speed of the wheel =
25
25 18
×
m/s =
km/hr =
4
4
5
22.5 km/hr
4.
Method 1:
The man takes 2 hr to travel 14 km, where he stops
for 10 min. Therefore, time between the start of
journey and start after first stoppage
= 2 hr 10 min.
This pattern of his travel repeats. Now, 98 km
= 14 × 7. But when he completes 98 km, the case of
his stopping for 10 min does not arise. Therefore,
total time taken
= 6 (2 hrs 10 min.) + 2 hr = 15 hr
Method 2:
If the man had travelled non-stop he would have
covered the distance in 14 hr. But since he stops 6
times on the way he would take an extra 6 × 10 = 60
min to cover the same distance. Hence, he would
reach the destination after 15 hr.
5.
Total relative distance travelled by the car
= [40 + 60] m = 100 m.
The relative speed =
200
m / s = 20 m/s.
10
40
200
m/s =
m/s.
3
15
Relative speed of the two trains (when they
move in opposite directions)
Level – I
1.
Speed of first train =
100 m
= 5 m / s.
20 sec
The relative speed is also equal to
(30 – v) km/hr
5
= (30 – v)× m/s, where ‘v’ is the speed of the
18
200 + 200
= 12 s
100 / 3
As distance covered by faster one is 4 times
the distance covered by slower one while their
speed ratio is 3 : 2. So faster one is moving
downstream and slower one is moving
upstream.
6 +u 4
=
So
4 – u 1
They will meet first time in (6 + u) t + (4 – u)
t = 100;
t = 10 s.
For second time meeting, faster is coming back
to its starting point after reaching starting point
of slower one.
20
80
S low e r
B
Fa ster
A
20 m will cover by A to reach slower’s starting
20
= 2.5 s to reach B. In
8
this time slower one has cover 2 × 2.5 = 5 m.
Now, A has to cover (20+ 5 = 25 m) with
respect to B to catch B. Now A is also moving
upstream so his speed is now 6 – 2 = 4 m/s. So
relative speed of A with respect to B is
4 – 2 = 2 m/s.
point. So he will take
25
= 12.5 s .
2
So in all = 10 + 2.5 + 12.5 = 25 s.
They will meet second time after 25 s.
So A will take
bus.
Therefore, (30 – v) ×
5
= 5 or, v = 12 km/hr.
18
Explanations: Fundamentals of
Time, Speed & Distance
MBA
Test Prep
Page: 93
8.
Speed downstream = v + u (where, v = Speed of
the boat in still water, and u = Speed of the stream).
Speed upstream = v – u.
Distance downstream = Distance upstream
(v + u) 45 = (v – u)75 or, u : v = 1 : 4.
9.
a.
Ratio of distances = Ratio of speeds
= 20 : 23 :27.
b.
Let us calculate the time of meeting of A and B
for the first time. A will meet B after
13.
At 10 a.m. a train from station B is just coming to
station A. That train must have started at
5 a.m. from station B. The train that starts at 10
a.m.. will reach at 3 p.m. at station B.
So from 5 am to 3 p.m. every train started from
station B will cross the train. So total number of
train = 11.
14.
From A a train is starting at 10 a.m. So from B every
1
train will start at an (integer + ) hr. interval i.e. 5 :
2
30, 6 : 30, 7: 30 .... and so on.
200
200
=
s.
23 − 20
3
This means that A and B meet every
Similarly, B meets C every
So the 10 a.m train from A will meet first train which
is suppose to reach at 10 : 30 starting from B at
5 : 30.
Also, 10 a.m. train from A will reach station B at
3 p.m. So the last train it will cross is the one started
at 2 : 30 p.m. from station B. So from
5 : 30 to 2 : 30, we have a total of 10 trains.
200
s.
3
200
200
=
27 − 23
4
= 50 s.
Therefore, LCM of these time periods will give
us the time when all the three meet the first time
15.
So at 10 a.m. a train is just reaching station A that
started from B at 7 a.m. So, the train that left at
10 a.m. train will reach at 3 p.m. Thus, from 7 a.m. to
3 p.m. each train will cross the train from B that
started at 10 a.m.
So total number of trains = 8 + 1 = 9 trains.
16.
In this case we cannot find the exact number of
trains crossing because we do not know the
starting time of any train started from station B.
May be possible it will reach station A at 10 a.m. or
at 11 a.m. If it reaches at 10 a.m. to station A, then
it starts at 5 a.m. and 10 a.m. train will reach at 3
p.m. So total number of trains crossing.
= 5 a.m., 7 a.m., 9 a.m., 11 a.m., 1 p.m., 3 p.m. = 6
Or if train started from B reach A at 11 am should
start from Bat 6 a.m. So number of crossing trains:
= 6 a.m., 8 a.m., 10 a.m., 12 a.m., 2 a.m. = 5 trains.
17.
They will finish 100% of the work in
200
, 50
which is LCM
3
= 200 s or (3 min. 20 s).
10.
A’s speed is 2 m/s with respect to B. (i.e. relative
speed)
So to catch B he has to overcome 250 m because B
is 250 m ahead of A. So A will take =
250
= 125 s .
2
x
11.
B
P
Q
A
12.
Let the train increase its speed after ‘t’ hours.
Total time taken by the train = 1 hr 15 min = 1.25 hr.
Therefore, 40t + 50(1.25 – t) = 55 km.
Or t = 45 min.
Page: 94
= 12 days.
Therefore, 75% of the work will be done in 75% of
12 = 9 days.
–x
500
At the first time they will meet at point P. After that
they meet for the second time at point Q. Up to point
P distance travelled by A and B is add up to 250 m.
After that when they meet at Q together they have
travelled one circle. So total distance travelled by
them is 750 m [250 + 500]. So time taken by them to
cover this much distance is (5 + 3)t = 750
750
= 93.75
t=
8
20 × 30
20 + 30
18.
Method 1:
Let a, b and c be the times taken by A, B and C to do
the work alone.
Also let x =
1
1
1
,y=
,z=
.
a
b
c
1
1
1
, y+z=
,z+x=
.
20
15
18
From the last two equations, y – x
Therefore, x + y =
=
1
1
1
–
=
.
15 18
90
MBA
Test Prep
Solution Book-2
Therefore, 2x =
=
1
1
7
–
=
, which means x
20
90
180
21.
Ratio of distance = Ratio of speed = 32 : 41.
Therefore, distance travelled by both the trains
= 32x and 41x.
According to given condition, 41x – 32x = 45 km
Therefore, x = 5. Hence, distance between
A and B = 41x + 32x = 365 km
22.
Let the trains meet at some place M between
A and B. Time taken from A to M = time taken from B
to M.
Therefore, ratio of speed = ratio of distance
= 4 : 3.
23.
A takes a lead of (3.5 km/hr) (2.5 hr) = 8.75 km over
B.
Hence, when B starts he has to travel a relative
distance of 8.75 km and at a relative speed of
(4.5 – 3.5) km/hr = 1 km/hr, which means B will take
8.75 hr to catch up and then overtake A.
24.
In one day, the time gap between the two faulty
watches is 6 + 4 = 10 min.
The two watches will show the same time next
when the faster watch takes a lead of exactly
12 hr over the slower. This would happen after
7
360
or a =
days.
360
7
Method 2:
Assume that the number of units of work is 180
units.
A and B put in 9 units per day B and C put in
12 units per day; and A and C put in 10 units per day.
(9 + 10 + 12)
2
units per day, i.e. 15.5 units per day. So A alone puts
So A, B and C working together put in
in 15.5 – 12 =
takes 180 ×
7
2
units per day. So he
2 360
days to complete the job
=
7
7
working alone.
Level – II
19.
1× 12 × 60
= 72 days. In 72 days the faster watch
10
would gain
Let the distance travelled by the man be x km.
Therefore, time taken at a speed of 10 km/hr
=
72 × 4
= 4.8 hr = 4 hr and 48 min.
60
So the time according to the faster watch would
be 7:48 p.m. and the slower would be 12 hr behind
but would also show the same time.
x
hr.
10
Time taken at a speed of 20 km/hr =
x
hr.
20
x
x
45
−
=
10 20 60
Solving, we get x = 15 km.
Therefore,
Hence,
25.
On that particular day it completes
x
x 15 15
–
–
=
= 0.225 hr = 13.5 min
25 40 25 40
20.
But in half time due to engine problem it takes 2
6
60
When he moves at 12 km/hr, time of travel = t −
9
60
Difference in time taken in the two cases
6
9
1
3
1
+
= hr
= t +
=
− t −
60 60 10 20 4
If x km is the distance between school and house,
difference in time taken
=
1
the journey in
2
same time (12.5 hr).
Let us say ‘t’ hr is the usual time. When Sanju
moves at 10 km/hr, time of travel = t +
Normal day it travels a distance = 25 (50 + u)
1
hr
2
extra.
So it take 15 hr to complete the remaining half
journey.
1
× 25(50 + u)
2
30(40+ u) = 25 × 50 + 25 u
30u – 25u = 1250 – 1200
5 u = 50; u = 10 km/hr
So, 15(40 + u) =
x
x
x
−
=
10 12 60
Therefore,
x
1
= hr .
60
4
or x = 15 km.
Explanations: Fundamentals of
Time, Speed & Distance
MBA
Test Prep
Page: 95
26.
Ratio of speed of Bhim and Arjun = 7 : 4.
(Since the ratio of the times = 4 : 7)
a. If the length of circular track = 28m, the speeds
of Bhim and Arjun are 7 and 4 m/min.
The time when they are together the first time
will be when Bhim (the faster one) has taken
one round more than Arjun (the slower one).
Therefore, if time when they meet is ‘t’, then 7t
28
– 4t = 28, which means t =
min.
3
b. They will meet at the starting place the first time
at a time which is the LCM of the times each
one of them takes to reach the starting place.
Therefore, LCM of 4 and 7 is 28 min.
29.
c.
30.
Diametrically opposite point is at a circular
distance of 14 m. Bhim reaches this point in
Method 2:
(49 + 25)
x=
×145 = 429.2 km
25
As A starts puffing, the length of burnt cigarette in
→ B →
one cycle is A
gap
gap
14
14
= 2 min and Arjun reaches this point in
=
7
4
3.5 min. Bhim reaches this point in the 2nd
min, 2 + 4 = 6th min, 6 + 4 = 10th min ... so on.
Arjun reaches after 3.5 min, 10.5 min, 17.5
min. so on.
The time after the start when Bhim reaches the
point is a natural number, whereas the time
when Arjun reaches this point will always be
a non-natural number.
So they will never meet.
Method 1:
Let the total distance between A and B be x.
Since B had to travel 145 km to reach A, this means
they met at a distance of 145 km from A.
Hence, distance between the meeting place and B
= (x – 145) km.
Ratio of speed = Ratios of distance
25 : 49 : : 145 : (x – 145).
Therefore, x = 429.2 km.
=2×3+3+3×3+3
= 6 + 3 + 9 + 3 = 21 mm.
63
= 3 ,
As there are three such cycles Q
21
time taken in one cycle = 2 + 3 + 3 + 3 = 11s.
So total time = 3 × 11 = 33 s.
1
31.
5
G oing o ut
C o m ing in
So the hour hand and minute hand move 360° in this
time (together).
1°
per minute and
2
speed of minute hand = 6° per minute. Let t is the
required time in minutes.
As the speed of the hour hand =
When Bhim gives Arjun a lead of 4 min, in 4 min
Arjun travels 16 m. Bhim now starts running towards
Arjun at a relative speed of 3 m/min. To catch up
So 6t +
with a distance of 16 m, Bhim takes
Page: 96
4
5
the time when they would meet for the first time.
28.
3
4
28
min to take a lead of one round. This is
take
3
16
min.
3
Let x be the distance between A and B.
The train which leaves A travels a total distance of
x + (x – 200) = 2x – 200.
The train which leaves B travels a total distance of
x + 200.
The ratio of distances travelled by the trains
= Ratio of the speeds.
Therefore, (2x – 200) : (x + 200) : : 40 : 60
Or x = 250 km.
2
3
Alternatively for (a) and (b):
If the time when they would meet for the first time at
the starting point = LCM(4, 7) = 28 min, in this time
Bhim does 7 rounds and Arjun completes 4 rounds.
So Bhim takes a lead of 3 rounds. Hence, he would
27.
1
2
1
t = 360°
2
13
t = 360°
2
32.
t=
720
min.
13
Let us say total work is of 20 × 20 units.
Each woman does 1 unit per day.
Each man does 2 units per day while each boy
1
unit per day.
2
So on first day 20 units are completed.
On second day 21 units, on third day 20.5 units and
on fourth day 21.5 units are completed.
does
MBA
Test Prep
Solution Book-2
33.
34.
Combining two terms, we get 41 + 42 + 43 + ... + 48
+ 49 = 405 > 400.
So there are 9 such pairs. So on the 18th day work
will be complete.
Let the inlet pipe take ‘t’ minutes to fill the tank alone.
Let the total work be of 4,00 units.
So a woman does 1 unit per day,
while a man does 2 units per day.
So work done on first day = 20 units
and work done on second day = 21 units.
Thus, the pattern goes like 20 + 21 + 22 + ... + 30
= 275
275 + 31 + 32 + 33 = 371 up to 13th day, 371 units
has been completed. On 14th day job will be
complete.
Method 2:
The net inflow is 50 l per minute. Hence, the actual
inflow = 80 l per min (50 + 30). So the inlet pipe will
Then
2000
= 25 min to fill the tank independently.
take
80
39.
C can start the work on first day but there is
nothing to demolish then. So this day is wasted.
The work effectively starts from second day
onwards.
Let the total work be of LCM(10, 15, 20) = 60 units.
Now A can do 6 units in a day.
B can do 4 units in a day.
C can demolish 3 units in a day.
So in 3 days (starting from A) 6 + 4 – 3 = 7 units will
be complete. So in 8 × 3 days, 8 × 7 = 56 units of
work will be done by A, B and C. The rest 4 units
36.
Let the total job be of LCM(10,15) = 30 units.
A + B can do 3 units per day.
B + C can do 2 units per day.
A + B + C can do 5 units a day.
A + 2B + C can do 5 units.
Here it can be clearly observed that B’s share in the
total work is null. Hence, he cannot complete the job
by himself.
So,
1 1 2 1
+ +
=
t 2t 3t 3
1
1 2 1
× 1 + + =
t
2 3 3
1 13 1
× =
t 6 3
t = 6.5 hr.
Level – III
Let the leakage take ‘t’ minutes to empty the tank
completely. When all the three pipes (two inlets and
1
1 1 1
– =
+
10 20 t 25
or, t = 100/11 = 9.09 minutes (approximately).
1
m/s per second.
2
Thus, reduction in the speed of the slower truck =
Method 1:
The outlet pipe will discharge the water in
1
m/s per second.
2
As A alone can do the job in 10 days and A and B
together also taking 10 days to complete the job. So
100% of A = 60% of A + B’s efficiency. So B will take
one leakage) are open, then
38.
t + 2t 3
= t hr .
2
2
36 km/hr ⇔ 10 m/s
18 km/hr ⇔ 5 m/s
It is mentioned that the reduction in speed of the
faster truck = 1 m/s per second. Note that the brakes
are applied successively.
Thus, the faster truck travels 9 m in the 1st second,
8 m in the 2nd second, 7 m in the 3rd second and so
on.
Final speed of the faster truck = 0 m/s
⇒ Truck stops at the 10th second.
Since both the trucks stop simultaneously, speed of
slower at 10th second = 0 m/s
If the reduction in speed of the slower truck
= ‘a’ m/s per second, (5 – 10a) = 0
10
= 25 days .
0.4
37.
A → First pipe
B → Second pipe
C → Third pipe
A can fill in t hr.
B can fill in 2t hr
C can fill in
2
will be done by A in day . So total number of days
3
2
2
= 1 + 8 × 3 + = 25 days.
3
3
35.
1
3
1
=
–
. Therefore, t = 25 min.
t 200 40
40.
⇒a=
2000 200
=
minutes alone.
30
3
Explanations: Fundamentals of
Time, Speed & Distance
MBA
Test Prep
Page: 97
Evaporation which is similar to a leakage
Alternative Method:
Ratio of speeds = 18 : 36 = 1 : 2
Since both the trucks are equally efficient, the
reduction in the speed of the slower truck must be
1
m/s per second
2
Now, distance travelled by the first truck in 10
seconds = (9 + 8 + 7 + 6 + … + 1 + 0) = 45 m
Distance travelled by the second truck in 10 seconds
= (4.5 + 4 + 3.5 + 3 + … + 0.5 + 0)
= 22.5 m
Thus, net separation between the trucks when they
notice each other = 45 + 22.5 = 67.5 m
41.
42.
45.
The rate of burning of first candle = 1 cm/hr.
The rate of burning of second candle = 1.5 cm/hr.
They will be of the same length when the longer
candle melts by a relative length of (10 – 8)
= 2 cm.
Therefore, time after which they will be of the same
After 4 hr, the candle of length 8 cm is now
8 – 4 = 4 cm long which would be of the same
length as the length of the other candle.
Rate of flow for the first pipe = λ(1)2 = λ.
Rate of flow for the second pipe = λ(2)2 = 4λ.
Rate of flow for the third pipe = λ(3)2 = 9λ.
Since rates are inversely proportional to time taken
by each to fill the tank, time taken by the 3 pipes are
9
min and 1 min respectively. Therefore,
4
when all the three pipes are opened, the fraction of
the tank filled in 1 min
9 min,
3t = 6
Hence, t = 2
Distance travelled by bus is equal to area under
1
1
× Vm1 × 2 + Vm1 × 2 + × Vm1 × 2
2
2
= 4Vm1 [Vm1 = Bus (maximum speed)]
Distance travelled by jeep.
the curve, i.e.
The thief travels for 2 hr (7.00 p.m. to 9.00 p.m.)
and takes a lead of (4.5 km/hr)(2hr) = 9 km.
The policeman is required to cover up a distance of
9 km at a relative speed of
1.5 km/hr. Therefore, the policeman will take
6 hr after 9 p.m. to catch up with the thief, i.e at
3 a.m.
2 cm
length = [1.5 cm / hr – 1 cm / hr ] = 4 hr.
43.
2.5
× 40 = 1 l / hr.
100
So total leakage is 3 l/hr.
So added water in an hour is 4 – 3 = 1 l
So it will take 40 hr.
=
1
1
V × 2 + Vm2 × 2 + × Vm2 × 2
2 m2
2
= 4Vm2 [ Vm2 = jeep’s maximum speed]
So 4Vm1 + 4Vm2 = 280
Vm1 + Vm2 = 70
Vm1 – Vm2 = 10
Vm1 = 40 km/hr
Vm2 = 30 km/hr
46.
Average speed by bus
=
Dis tance travelled by bus
Time taken by bus
=
4Vm1
6
2
80
km / hr
× 40 =
3
3
Average speed by jeep
=
=
Dis tance travelled by Jeep
Time taken by jeep
=
4Vm2
6
=
2
× 30 = 20 km / hr
3
1 4 1 14
+ + =
9 9 1 9
The time taken by all the pipes to fill the tank is
=
9
min. (λ is the constant of proportionality)
14
44.
Tap 1 can fill the tank in 10 hr.
So tap 1 gives
40
= 4 lit / hr .
10
Similarly, tap 2 leaks =
Page: 98
40
= 2 l / hr .
20
MBA
Test Prep
Solution Book-2
Solutions Exercise 1 (Level 1)
1. a
5. e
A takes 7 × 9 = 63 hr.
∴ in 1 hour A does =
1
of work
63
Assume that units to fill up
= LCM of (20, 15, 12) = 60.
Units filled up by A, B, C in 1 min
= 3, 4, 5 respectively.
Time taken by (A + B + C) to fill up =
B takes 6 × 7 = 42 hr.
∴ in 1 hr B does =
A + B in 1 hr =
A + B in
1
1
+
63 42
1
=3
0.33
1
part
80
∴10 days’ work =
10
1
=
part
80
8
Work to be done by B = (1 –
= 6, 5 and ( −3) units respectively.
∴ A + B = 11 units, C = –3 units
∴ Per hr intake = 11 – 3 = 8 units
Time to fill up = 60/8 = 7.5 hr.
7. d
1
7
)=
8
8
Number of days required by B =
3. b
8. a
4. b
60 60
+
=9
12 15
3 min total intake in units = 9 × 3 = 27
Units to be filled = 60 – 27 = 33
Time taken by B = 33/4 = 8.25 (8 min 15 sec)
336
= 48 days
7
9. c
Pipes A + B both are there,
1 1 1
+ =
, say A takes x hr to fill then
A B 12
B will take x – 10 hr to fill the tank.
∴
80× 48
= 30 days
128
Let us assume that work done by A and B in 1 day
are 2 and 1 unit respectively.
Work done in 1 day by (A + B) = 3 units
∴ Work done in 14 days by (A + B)= 14 × 3 = 42
42
= 21 days
2
Assume total wages = 21 × 28
A’s wages for 1 day = 21 × 28/21 = 28
B’s wages for 1 day = 28 × 21/28 = 21
A + B’s wages of 1 day = 49.
A + B’s wages will last for 21 × 28/49 = 12 days
Explanations: Fundamentals of
Time, Speed & Distance
Say, capacity of tank = LCM (12, 15)
= 60 units
1 min total intake in units =
1
1
128
+
=
80 48 80× 48
∴ A finishes the work in =
90
= 10 units per hr.
9
90
= 9 units per hr.
10
∴ Resultant outflow = 10 – 9 = 1 unit per hr.
Total outflow = 90 units, Min. = 90/1 = 90 hr.
7
7
7
, x=
=
8
42×8
336
∴ Number of days =
Capacity in units = LCM (9 ,10) = 90.
Inflow, when there is a leak =
Let ‘B’ do ‘x’ part of the work in 1 day
A+B=
60 60 -60
,
,
10 12 20
Inflow, when there is no leak
In 1 day, A does =
x × 42 =
Assume that capacity of the tank
= LCM (10, 12, 20) = 60 units.
Rate of A, B, C per hr =
1 42
42
1
+
hr =
= 0.33
×
5
63 42 5
∴ Number of days =
2. e
6. c
1
of work
42
60
= 5 min.
12
1
1
1
x –10 + x
+
=
⇒
x x – 10 12
x ( x –10 )
=
1
2x − 10
1
⇒
=
12
x ( x − 10 ) 12
24x – 120 = x2 – 10x
x2 – 34x + 120 = 0
x2 – 30x – 4x + 120 = 0
x(x – 30) – 4 (x – 30) = 0
(x – 4) (x – 30) = 0
x = 4, 30
∴ 30 hr ( Q x cannot be 4)
MBA
Test Prep
Page: 99
10. d
(a) Distance = Speed × Time = 48 × 10
= 480 km
(b) To cover the same distance in 8 hr.
17. d
the time =
d 480
= 60 km/hr
Speed = =
t
8
∴Speed must be increased by 60 – 48
= 12 km/hr
11. a
Let his usual time be t hr
and his usual speed be s km/hr.
Distance, d = st =
18. b
=
SA 3
=
SB 4
19. b
110
=11 : 9
90
To cross the platform the train has to cover
165 + 110 = 275 m.
5 110
=
m/sec
132 km/hr = 132 ×
18
3
t=
13. a
Thief starts at 2.30 p.m.
Policeman starts at 3.00 p.m.
In 30 min (= 1/2 hour), the thief covers
1
60 × = 30 km
2
At this instance, the distance between them = 30
km
difference of their speeds = 75 – 60
= 15 km/hr
∴ Policeman will overtake the thief in
14. d
15. b
Since ratio of speeds of A : B = 3 : 4. So, ratio of
time taken would be 4 : 3. If A takes 30 min more,
then 4x – 3x = 30 min
⇒ x = 30 min.
A takes 4x min, or 4 × 30 = 120 min = 2 hr.
Average speed =
=
16. d
=
2xy
2 × 24 × 36
=
x+y
24 + 36
144
= 28.8 km/hr
5
Average speed =
Total distance
Total time
2500 + 1200 + 500
4200
=
= 420 kmph
2500 1200 500
10
+
+
500
400 250
Page: 100
20. a
275
× 3 = 7.5 sec
110
t = 10 sec,
5
= 25 m/s
18
Length = s × t = 25 × 10 = 250 m
Speed = 90 km/hr = 90 ×
21. d
30
hr
15
= 2 hr, i.e. at 5.00 p.m.
27
min = 13.5 min.
2
Speed of A = 3x km/min
Speed of B = 4x km/min
distance is same.
d = SA × t A = S B × tB
3x × tA = 4x × 36
tA = 48 min
3
s × (t + 2.5)
4
By the time the trains crossed each other, one of
them had covered 110 km and the other had
covered 90 km.
Ratio of speeds = Ratios of the distances covered
54
= 27 min.
2
So, time taken be A =
Or, 4t = 3t + 7.5
t = 7.5 hr
12. c
A = 2B, B = 2C. C takes 54 min.
So, B would cover the same distance in half of
Total length to be covered = Length of train +
Length of platform = 900 + 300 = 1200 m
Total time = 60 + 12 = 72 sec
Speed =
22. b
1200 50
50 18
m/s =
×
=
= 60 km/hr
72
3
3 5
A + B = 72 days, B + C = 120 days,
A + C = 90 days.
Let total units of work = 360
… ( = LCM (72, 120, 90 ))
∴ (A + B) do units of work in 1 day =
(B + C) do units of work in 1 day =
360
= 5.
72
360
= 3.
120
360
= 4.
90
∴2(A + B + C) units in 1 day = 5 + 3 + 4 = 12.
(A + C) do units of work in 1 day =
12
=6
2
∴ A units in 1 day = 6 – 3 = 3
A + B + C units in 1 day =
A will finish 360 units in
MBA
Test Prep
360
= 120 days.
3
Solution Book-2
Solutions Exercise 2 (Level 1)
4t –
1. d
Let total units of work = LCM of (12, 16) = 48.
Work done by:
(A + B) in 1 day = 48/12 = 4 units.
(B + C) in 1 day = 48/16 = 3 units.
(A + B) worked for 5 days ⇒ 4 × 5 = 20 units
done.
(B + C) worked for 2 days ⇒ 3 × 2 = 6 units done.
Work remains = 48 – 26 = 22
C finishes in (13 – 2) days = 11 days.
∴ Units of work in 1 day by C =
2. c
1
1
1
40 + 30
8. c
At 5 km/hr t 2 =
1/ 3d 1
d hrs
=
5
15
Let his normal speed be s km/hr.
Let his normal time be t hr
1
d = st ⇒ 4 t − = 3 t + 1
6
6
Explanations: Fundamentals of
Time, Speed & Distance
Length of train= 110 m,
Speed of train = 58 ×
5
m/s = 16.11 m/s
18
Speed of man = 4 km/hr = 4 ×
days
Time =
10. a
110 m
110
=
= 7.33 sec
16.11 − 1.11 15
108 + 112
220 × 18
=
5
5
(50 + x )
(50 + x ) ×
18
250 + 5x = 660,
x = 82 km/hr
11. b
5x = 410
Length of the faster train
= (36 + 45) ×
12. d
5
= 1.11 m/s
18
Length of first train = 108 m. Its speed
= 50 km/hr
Length of second train = 112 m.
Let its speed = x
time to cross = 6 sec
⇒ 6=
Total time = 60 + 24 = 84/60 hr
2 / 3d 1
= d hr
4
6
d=s×t
1
3
journey ( = 40 km ) in of time ( = 6 hrs )
2
5
Total distance = 80 km
Total time = 10 hr
80 = 40 + S2 × 4, S2 = 10 km/hr
Let units of work = 18 units.
In one hour, man completes work = 9 units
(= 6 units + 3 units)
∴ Time taken = 2 hr.
d d 84
=
Total time = +
6 15 60
Or, d = 6 km
6. d
44
25
= 5t +
60
60
t = 19/60 hr = 19 min
4t +
9. d
1
1200
days = 17
days.
7
70
At 4 km/hr t1 =
Let normal speed be s km/hr.
Let normal time be t hr.
11
5
d = st ⇒ 40 t +
= 50 t +
60
60
10 men in 20 days or 20 women in 15 days.
∴ 5 men in 40 days and 10 women in 30 days.
=
5. c
48
= 24
2
1
= 1.5 days
2
⇒ 5 men + 10 women in
4. b
7. d
Let total work = 12 × 8 = 96.
In 6 days, 12 men complete work = 12 × 6
= 72
∴ Work left after 6 days = 96 – 72 = 24
Now, number of men = 12 + 4 = 16
∴ Time = 24/16 = 1
3. b
7 1
Distance = 4 × − = 4 km
6 6
22
=2
11
∴ Number of days required for C =
4
3
7
= 3t +
, t=
hr
6
6
6
5
5
× 8 m = 81 ×
× 8 = 180 m
18
18
Let x be the rowing speed of the man in still waters
Speed of the river = 2 km/hr = ‘y’
Speed upstream = x – y km/hr
Speed downstream = x + y km/hr
(x – y) 2t = (x + y) t
2 (x – 2) = x + 2
x = 6 km/hr
MBA
Test Prep
Page: 101
13. b
Let length of train be l m and its speed be s m/s.
It crosses a pole in 15 sec
17. a
… (i)
∴ l = 15 s
It crosses a platform 100-m long in 25 sec
100 + l
= 25
s
Solving (i) and (ii)
100 + 15s = 25 s
∴
∴ s=
14. a
15. c
… (ii)
or
1
1
1
1
+
+
=
22 33 44 x
or x =
100
=10 m/s, ⇒ l = 150 m.
10
18. b
The time taken to cover 240 km without any stops =
6 hr. Since he stops every 80 km, he would stop
twice before reaching the destination.
Hence, total time taken = 6 hr 40 min.
11× 12
132
2
days
=
= 10
6 + 4 + 3 13
13
Here the total distance = 180 km.
Total time =
30 60 90 2 2 6 38
hr .
+
+
= + + =
45 90 75 3 3 5 15
Average speed = 180 ÷
38
1
= 71 km / hr.
15
19
Let ‘l’ = Length of train in metres. Then speed of the
train =
19. d
l
l + 100
m/s =
m/s .
15
30
∴ l = 100 m
16. c
1
1
1
1
+
+
=
22 33 44 x
when x = Number of days in which they together
can finish the work.
1 1 1 1
+ + =
a b c 15
Alternative method:
The ratio of speeds of Sujit and Rishi = 100 : 95
= 20 : 19.
Similarly, the ratio of speeds of Rishi and Parveen =
20 : 19.
∴ The ratio of speeds of Sujit and Parveen
= 202 : 102.
⇒ When Sujit goes 100 m, Parveen goes
… (i)
5 25 25
+
+
=1
a
b
c
1 1 1
1
+ + =
… (ii)
5a b c 25
Subtracting (ii) from (i),
4
2
=
⇒ a = 30 days
5a 75
⇒
Alternative method:
Let the total work be of 15 units. They complete this
work in 15 days. It means one day work is 1 unit. In
5 days, they will complete 5 units of work.
Remaining 10 unit of work, B and C complete in 20
days. It means one day work of (B + C)
=
361
× 100 = 90.25 m .
400
∴ The lead that can be given is 100 – 90.25
= 9.75 m.
20. c
1
unit
2
1 1
=
unit.
2 2
So total days taken by A alone to complete
15 units of work = 30 days.
Suppose Pallavi takes t seconds to finish the race. It
means Anuva and Richa would be taking (t + 50)s
and (t + 90)s respectively to finish the race. When
Pallavi runs 1,000 m, Richa will run 550 m
⇒ Richa runs 1,000 m in (t + 90) s.
t + 90 1000
=
of t = 110 s
t
550
= 110 + 90 = 200 s.
So we have
⇒ A’s one day work is 1 −
Page: 102
When Sujit runs 100 m, Rishi runs 95 m.
When Rishi runs 100 m, Praveen runs 95 m.
∴ When Rishi runs 95 m, Praveen runs 90.25 m.
When Sujit runs 100 m, Praveen runs 90.25 m and is
beaten by 9.75 m.
21. a
Take (+) for the person walking down and
(–) for walking up. Therefore, his relative position
from the starting point = +4 – 3 + 6 – 2 – 9 + 2 = –
2.
Hence, the answer is 2 steps above the starting
step.
MBA
Test Prep
Solution Book-2
22. a
Since they move at the same time on different days,
the situation is equivalent to the that of two persons
where one moves uphill and the other downhill,
both starting at the same time, viz. 7 a.m.
Ratio of speeds of uphill to downhill = Ratio of
distances = 10 : 15 = 2 : 3.
Therefore,
1
1
+
work is done by (A + B) in 1 day
25 20
3
100 3
× = 6.66
work is done by them in =
5
9
5
days
Number of days = 10 + 6.66 = 16.66
⇒
3
of 25 = 15 km from the top of the hill
5
2
of 25 = 10 km from the bottom of
5
the hill is the point where they will meet.
6. b
they will meet or
Solutions Exercise 3 (Level 2)
1. a
9
× 60 min
Time taken to cover 9 km =
54
= 10 min
7. d
Let speed of boat be x km/hr.
Let speed of stream be y km/hr.
13
=x–y
5
Rowing upstream =
… (i)
28
=x+y
5
Solving for y, y = 1.5 km/hr
Rowing downstream =
Let speed of boat in still waters be ’x’ = 6 km/hr
Let speed of stream be ‘y’.
x – y = 4.5 km/hr
⇒ y = 1.5 km/hr
∴ Rate along the stream = 1.5 + 6 = 7.5 km/hr
3. c
Let the total work be (LCM of 45,40) = 360 units.
In one day,
A does
360
= 8 units and
45
8. d
4. d
Let r lt/hr be the rate at which the hole empties the
tank and let the tank has a capacity of V liters.
The inlet pipe fills the tank at 4 liters/min or at 240
liters/hour, we have two equations,
r×6=V
…(i)
& V + 8 × 240 = r × 8 …(ii)
V = 5760 liters
10. b
Length of bridge = 1 km
Length of train = 0.5 km
5. d
Time to clear the bridge = 2 min =
Speed =
12
= 7.5 days
1.60
In 10 day’s A completes work =
1
2
×10 = of the
25
5
11. a
2
hr
60
1 + 0.5 1.5
=
× 60 = 45 kmph
2 / 60
2
Speed of stream = 1 km/hr
Let speed of boat in still waters = x km/hr
Total time = 12 hr
35
35
+
x −1 x +1
Now put the value of x and check the options, only
(a) satisfies.
12 =
total work.
Remaining work =
60
= 13.33
4.5
9. e
A takes 12 days,
B is 60% more efficient
⇒ B will take time =
A:B=1:2
C : (A + B) = 1.5 : 3
If C does 1.5 units of work per day,
A does 1 and B does 2.
Total units done by C = 40 × 1.5 = 60 units
In a day work done by (A + B + C) = 4.5
⇒ Number of days taken by (A + B + C) to complete
work =
360
= 9 units
40
Both A and B together do 8 + 9 = 17 units of work.
In last 23 days, B does 9 × 23 = 207 units of work.
So work done by both A and B together
= 360 – 207 units = 153 units.
153
= 9 days.
17
1333.33
= 0.66
2000
⇒ Total number of hr = 4.66
B does
Number of days =
A in 1 hr = 2000
B in 1 hr = 1333.33
⇒ (A + B) in 2 hr = 3333.33
⇒ (A + B) in 4 hr = 6666.66
In 5th hr work to be done = 8000 – 6666.66
= 1333.33.
Number of hr required by A =
… (ii)
2. d
Due to stoppage, it covers 9 km less.
3
of the total work.
5
Explanations: Fundamentals of
Time, Speed & Distance
MBA
Test Prep
Page: 103
12. a
Ratio of distances travelled by the trains when they
meet = 9 : 10.
Suppose first train travels 9x. Then the second
would travel 10x km.
∴ Total distance = 10x + 9x = 19x km.
We have 10x – 9x = 120 or x = 120 km.
∴ 19x = 19 × 120 = 2,280 km.
Alternative method:
It is given that speed is 45 and 50 km/hr respectively.
So in 1 hr, the faster train moves 5 km extra than the
slower one. It is given that the faster one moves
120 km extra than the slower one.
Alternative method:
Solve using the relative velocity concept, V is the
speed of train.
10(V + 330) = 330 × 12
V =
v
17. c
37
= 7.4 hr, which is same as
5
the time taken at the original rate.
18. e
1
= 30 km,
2
which is the relative distance to be covered by the
owner
moving
at
relative
speed
of
(70 – 60) km/hr = 10 km/hr.
∴ Time taken by the owner to catch up with the
Let ‘w’ is the time of walking and ‘r’ is the time of
riding
∴ w + r = 7 hr.
Riding both ways he gains 2 hr. Therefore, he takes
5 hr to ride both ways or 2.5 hr to ride one way.
∴ r = 2.5 hr ⇒ w = 4.5 hr.
Time taken to walk both ways = 2 × 4.5 hr = 9 hr.
19. b
From 7 a.m. to 1 p.m., i.e. in 6 hr, the first train
travels a distance of 60 × 6 = 360 km.
∴ Distance between the trains at 1 p.m.
= 1200 – 360 = 840 km.
∴ Time taken to cover this relative distance
In half an hour, the thief travels 60 ×
Suppose the distance between the college and
residence is d. Then
2
d
d 2
40 min = hr ⇒ d = 20 km
−
=
3
10 15 3
15. d
=
20. c
Average speed
=
16. e
When he travels the entire journey at 5 km/hr, total
distance travelled = 35 + 2 = 37 km.
∴ Time of travel =
30 km
= 3 hr which is at 5 p.m.
thief =
10 km / hr
14. c
3 30
120
= 24 hr.
5
There relative speed is 95 km/hr.
So distance travelled by them = 95 × 24
= 2,280 km.
So they travel for
13. e
330
= 66 m / s
5
(50)(1) + (48)(2) + (52)(3)
1
= 50 km / hr.
6
3
If the person had not moved towards the source of
the gunfire, he would have heard the second shot
12 min after the first shot. Since the person is
actually moving towards the source, the shot is
now heard after 10 min. It means the sound would
have taken 2 min more to reach the initial position of
the person; but this very distance was travelled by
the train in 10 min. It means it is the speed of the
1
train in 10 min. It means speed of the train is
of
5
the speed of the sound, i.e.
= 66 m/s =
Page: 104
330
m/s
5
66 × 3600
= 237.6 km / hr
1000
21. d
840
= 6 hr, i.e. 7 p.m.
140
In 6 hr, the second train travels 6 × 80 = 480 km.
∴ They meet at 480 km from Q.
3
4
Speed in upstream = 1 × 60 km/hr.
11
4
3
4
Speed in downstream = 1 × 60 km/hr.
7
2
Speed of man in still water
1
(Speed in downstream + Speed in upstream)
2
=
1
1 3
1
+
× 60
2 4
11 1 7 1
4
2
=
MBA
Test Prep
= 5 km/hr.
Solution Book-2
22. b
Let x be the distance per litre of petrol and v
= Speed at which it is driven.
⇒ v′ = 30 km/hr
Alternative method 2:
Let the total work be 180 units. 10 men complete
this in 18 days. So one day work of 10 men
= 10 units.
(All men are equally efficient, so one man’s work in
one day is 1 unit.)
In 6 days, 10 men will complete 60 units work.
Remaining 120 unit is to be done by (10 + 5) men in
Let a = Speed in upstream and
b = Speed in downstream
120
days = 8 days.
15
Then x ∝
1
v2
or xv2 = k = Constant. 25(36)2
= 36 ( v′ )2 ( v′ is the required speed)
23. b
40 90
60 60
+
= 5 and
+
=5
a
b
a
b
∴ a = 20 km/hr and b = 30 km/hr.
∴ Speed of the water (current)
=
Solutions Exercise 4 (Level 2)
1. c
1
(b − a) = 5 km/hr
2
3 men = 4 women. 4 women complete the job in 12
days. Hence, 5 women and 3 men, i.e.
9 women can do the job in
Q {b = VW + VB and a = VB − VW }
2. c
24. a
25. c
Let the job be completed in x days. Then the young
man worked for 2 days.
His father worked for x – 1 days and the grandfather
worked for x days.
∴
2 x −1 x
+
+
=1
10
15
20
or
3
12 + 4x − 4 + 3 x
= 1 or x = 7 days
7
60
To complete the given work, 10 × 18
= 180 man-days will be required.
In 6 days, amount of work done
= 60 man-days.
Number of men now available = 10 + 5 = 15.
∴ Number of days in which the job can be done =
6
× 500 = Rs. 375 .
8
Earning of 4 women = Rs. 375.
Earning of 3 men = Rs. 375.
Earning of 1 man = Rs. 125.
Alternative method:
5 3 4
The man earns 50 × = Rs. 125 .
4 2 3
3. b
4. a
3
of people becomes times the earlier number,
2
2
the job would be done in times the earlier
3
number of days, i.e. 8 days.
Earning of 1 girl = Rs. 50.
Earning of 10 girls = Rs. 500.
Earning of 8 boys = Rs. 500.
Earning of 6 boys = Rs.
120
= 8 days more.
15
Alternative method 1:
After 6 days, 12 days of the job is left. If the number
12 × 4
1
= 5 days .
9
3
Amount of work to be done = 10n,
where n = Number of workers originally available.
Now 10n = 12(n – 5) ⇒ 2n = 60.
Therefore, n = 30.
1
1
of a man =
of a woman = 1 child.
3
2
⇒ 2 men = 3 women = 6 children.
20m + 30w + 36c = 60c + 60c + 36c = 156c
78
.
156
Now 15m + 21w + 30c = 45c + 42c + 30c
= 117c.
If 156 children get Rs. 78, 1 child gets
78
×117 per day.
∴ 117 children should get Rs.
156
∴ For (18 × 7) days, they should get
78
×117 = Rs. 7371.
= Rs. (18 × 7)
156
Explanations: Fundamentals of
Time, Speed & Distance
MBA
Test Prep
Page: 105
Alternative method:
Let a man, a woman, a child put in 3, 2, 1 units per
day.
Then they are paid Rs. 78 for putting in
(20 × 3 + 30 × 2 + 36 × 1) units, i.e. 156 units.
Hence, if (15 × 3 + 21 × 2 + 30 × 1), i.e. 117 units is
6. a
12n = 18(n – 5) ⇒ n = 15
7. c
4
of work is done by them in 7 days.
5
Hence, they take
117
per day or Rs.
put in, then the payment = Rs.
2
7,371 in 18 weeks.
5. b
1 1 1
1
+ + =
a b c 18
job.
∴
… (i)
… (ii)
1 1 3
+ =
a c b
… (iii)
2
10
1.25 1
=
−1=
=
a 8.75
8.75 7
a = 14 days
Alternative method 1:
A and B work together for 8 days.
1 1 2
− =
or c = 54 days
18 c c
WA + B in 7 days is 4 W [W is total work]
5
1 1 3
− =
or b = 72 days
18 b b
∴ a = 43.2 days
4
8
32
× W =
W
⇒ WA + B in 8 days is
5
7
35
(i) and (iii) ⇒
Now A leaves and B does
Alternative method 1:
Let the number of units of work = 18.
⇒ B will do W in 2 ×
1
C puts in of that in 18 days = 6 units.
3
1
3
5
1
–
=
=
10
70 70 14
So A will do the work in 14 days.
18
× 18 = 43.2 days.
In order to finish it he takes
7.5
Alternative method 2:
Let the total work be 35 units (LCM of 7 and 5).
4
of work = 28 units, is done by A and B in
6
7 days.
∴ 1 day’s work of A + B = 4 units.
B completes (7 – 4) units in 2 days.
(Out of a total of 10 days, 7 + 1 = 8 days’ work = 8
× 4 = 32 units)
∴ 1 day’s work of B = 1.5 units.
∴ A’s 1 day’s work = 2.5 units.
Total time taken by A alone to complete 35 units of
Alternative method 2:
Let the total work be 18 × 12 = 216 units.
Ratio of (A and B) : C = 2 : 1.
1
× 12 = 4 units .
3
Ratio of (A and C) : B = 3 : 1.
∴ C does
1
× 12 = 3 units .
4
∴ A does = 12 – (4 + 3) = 5 units.
∴ D does =
216
= 43.2 days.
5
35 70
=
days.
3
3
=
Hence, A does 7.5 units in 18 days.
Total work will be done in
3
W in 2 days.
35
WA = WA + B – WB
1
B puts in of that in 18 days = 4.5 units.
4
Page: 106
1 1
1
8 10
=1
+ =
and +
a b
a b 8.75
⇒
1 1 2
+ =
a b c
(i) and (ii) ⇒
7
= 8.75 days to complete the
4
5
work =
8. d
35
= 14 days .
2 .5
In 30 days, only half the work could be done.
Therefore, number of men must be doubled to get
the work done in half the time (i.e. half of 30 = 15
days, because only 15 days are remaining).
MBA
Test Prep
Solution Book-2
9. c
Alternative method:
In 4 hr, lead taken by Shivku is (13 – 8) × 4 = 20 km.
Now speed of Sujeet = 16 km/hr and of Shivku = 12
km/hr.
Sujeet has to cover 20 km with relative speed 4 km/
hr in 5 hr. So total time = 4 + 5 = 9 hr.
15m = 24w = 36b
x men must be associated.
36
x + 18 + 6 boys.
∴ (x)m + 12w + 6b =
15
36 x
30 × 6
12 × 8
Therefore,
= (36 )
+ 24
15
21
1
4
10. a
M aya nk
A
Suppose the work is of 150 units. So in a day A and
B together do 15 units; B and C together do 10 units;
and A and C together do 6 units.
We have A + B = 15, B + C = 10 and C + A = 6.
⇒ A = 5.5, B = 9.5 and C = 0.5
In first 4 days, A, B and C together will do
4(5.5 + 9.5 + 0.5) = 62 units.
In next 5 days B and C will do 5(9.5 + 0.5) = 50 units.
So C alone has to do 150 – (62 + 50) = 38 units, that
Let X be the point where they meet on the way.
38
= 76 days .
0.5
Let us assume that the distance between the two
places be 60 km. This means that to go 120 km at
120
12 km/hr, she can use
= 10 hr.
12
Also, as she goes at 6 km/hr, the time taken for this
60
= 10 hr.
6
Thus, there is no time to come back.
⇒
Alternative method:
If distance from Kashipur to Bareily is D, Nupur takes
D
hr to travel distance D.
6
Since average speed of journey is 12, total time
2D D
= .
12
6
Hence, there is no time to return.
X
AX 6
=
[As their speeds are in this ratio]
XB 5
t m dm / Sm dm SG
Now t = d / S = d × S
G
G
G
G
m
dm AX 6
t m 6 6 36
⇒ Since d = XB = 5 ⇒ t = 5 × 5 = 25
G
G
tm =
2.5 × 36
= 3.6 hr = 3 hr 36 min.
25
Alternative method 1:
Conventional method of solving
D
13. a
Suppose the total time was x hr.
Since distances travelled by both would be same,
we have (8 × 4) + (x – 4)16 = (13 × 4) + (x – 4)12
or x = 9
Explanations: Fundamentals of
Time, Speed & Distance
B
G
M
Let Golu and Mayank met at point C which is
x kilometres from A, and A and B are D kilometres
apart.
VG
6
The ratio of speeds = V = 5 .
M
They take same time to reach point C.
⇒
x D – x
=
6
5
taken is
Suppose they meet after (100 + x) m. Shivku starts
running when Pawan has already run 100 m. Now
if Pawan runs x km, Shivku will run 2x km.
So we have 100 + x = 2x or x = 100.
Thus, they will meet at 100 + 100 = 200 m from the
starting point.
If the length of the track is 150 m, they will meet at a
distance of 50 m from the starting point.
C
x
A
journey is
12. c
B
⇒ x = 8 men
he will do in
11. e
G ollu
14. d
⇒x =
6
D
11
Now Golu covers
So VG =
5D
5
11 ×
2
5
5
D in
hr.
11
2
=
2D
km/hr
11
5
5
2
5D
D=
⇒ VM =
× VG =
×
6
6
11
33
Mayank covers 5D in 33 hr.
So he will cover
MBA
Test Prep
6
33
6
D in
= 3.6 hr
×
11
5
11
= 3 hr 36 min
Page: 107
Alternative method 2:
G
15. a
3
t
t − t = = 45 min
2
2
or t = 90 min.
Had the accident occurred 60 km away, the train
would have been delayed by 1 hr only.
Normal time
M
2
VG
=
VM
tM
t
6
⇒ = M
tG
5
tG
∴ tM =
36 5
× = 3 hr 36 min
25 2
2
of the speed, the delay will be 15 min only
3
(1 hr – 45 min stoppage)
∴ Normal time to cover the remaining distance
⇒ t = 2 × 15 = 30 min.
It means the train covers 60 km in 60 min
(90 – 30 min).
Speed of the train = 60 km/hr.
Total journey = 30 km + 90 km = 120 km.
Due to
Suppose the winning point is x metres far if the
girls finish the race at the same time. Then Pallavi
will have to cover x metres when Richa covers (x
– 350) m.
Ratio of their speeds is 20 : 13.
20
x
=
13 x − 350
or x = 1,000 m = 1 km.
So we have
18. a
200
used 8 gallons
for the journey.
25
Alternative method:
Pallavi has to cover 350 m with a relative speed of
7x.
350
.
7x
In this time, the actual distance covered by Pallavi
19. c
Required time will be
is
16. a
17. b
350
× 20x = 1,000 m, i.e. length of the track.
7x
It happens when both of them meet for first time at
the starting point. Ratios of speeds
= 4 : 12. Hence, ratios of distances covered
= 1 : 3. Hence, when A makes one round, B makes
3 rounds, so the answer is 1.
∴ x = 16
90
4
min .
= 16
5.5
11
[Converting minutes to hours]
Solving for x, we get x = 120 km and v = 60 km/hr.
Alternative method:
Overall, the train was late by 1 hr 30 min.
Out of this for 45 min, the train was stopped. So
due to
4
min
11
Alternative method:
At 3 o’ clock, the angle between the hour hand and
the minute hand is 90°. The minute hand will cover
the distance with a relative speed of (6 – 0.5) = 5.5°
per minute, i.e.
v
x−
1 3
2 − x = 11
+ +
2v
2 4
v
2
3
v
v
x − − 60
+ 60
3
x
2
2
+
+
− =1
Also
2v
v
4
v
3
We know that in 60 min, the minute hand gains 55
min over the hour hand.
At 3 o’ clock, the minute hand and the hour hand are
15 min apart.
To cross the latter, the minute hand must gain 15
min.
Since 55 min is gained by the minute hand in 60 min,
15 min is gained by the minute hand in
60
4
× 15 = 16
min.
55
11
Let length of the journey = x km and
speed of the train = v km/hr.
Then
Vibhor had lost 4 gallons.
Since he travelled at 50 mph for 4 hr = 200 miles, he
20. b
Let’s assume the distance to be 24 km.
(LCM of 6, 4, 3)
Then time taken to cover that distance will be in the
24 24 24
:
:
= 4 : 6 : 8 = 2 : 3 : 4.
6 4 3
Hence, the ratio of their speed will also be 2 : 3 : 4.
ratio
2
of the speed, the train delayed by 45 min
3
only.
Page: 108
MBA
Test Prep
Solution Book-2
Alternative method:
Since the distance travelled by the three is same,
the ratio of speeds will be inverse of the ratio of
time taken.
1 1 1
: :
Hence, the ratio of speeds =
6 4 3
= 4 : 6 : 8 = 2 : 3 : 4.
21. c
23. a
Fraction of the tank filled in 2 hr =
1 1
9
+ =
.
4 5 20
So in 4 hr, fraction of the tank filled =
The remaining
Ratio of speeds = 3 : 4.
Distance remaining constant, the ratio of time taken
= 4 : 3.
A takes 0.5 hr more than B.
Hence, time taken by A = 4 × 0.5 = 2 hr
18
.
20
2
1
or
of the tank will be filled by
20
10
pipe A.
Pipe A fills
1
of the tank in 1 hr.
4
1
1
of the tank in 4 ×
= 0.4
10
10
= 0.4 hr = 24 min.
Hence, the total time taken to fill the tank
= 4 hr 24 min.
So it will fill
22. a
2 1
= .
6 3
Work done by B and C in 2 days
Work done by A in 2 days =
1 1
5 + 4 9
= 2 +
= 2 40 = 20 .
8 10
Remaining work
= 1−
24. c
1 9
60 − 20 − 27 13
.
−
=
=
3 20
60
60
Hence,
13
part of the job will be done by C in
60
1000
= 200 s .
25 − 20
In 200 s, A will have made
13
60 = 13
1
= 2 days.
1
6
6
10
200 ÷
Total time taken to complete the job
=2+2+2
Alternative method:
Let the total work be LCM (6, 8, 10)
= 120 units.
A’s 1 day’s work = 20 units.
B’s 1 day’s work = 15 units.
C’s 1 day’s work = 12 units.
In 2 days A will work = 40 units.
(B + C) will work in 2 days = 54 units.
Work remaining after 4 days = 26 units.
∴ Time taken by C =
26 13
=
.
12
6
Total time taken = 4 +
13
days .
6
1000 200 × 25
=
= 5 rounds.
25
1000
Alternative method:
From the given data, in 1,000 m race A can give B a
start of 200 m. It means when A has covered 1,000
m, B can cover 800 m. The ratio of their speeds is 5
: 4. From this ratio of speed, it can be easily found
that they will meet only at the starting point. So A
has covered 5 rounds.
1
1
= 6 days.
6
6
Answer is 6
A can give B a lead of 200 m for every 1,000 m, i.e.
every one kilometre. This means that when A covers
1,000 m, B covers 800 m.
Suppose the speed of A is 25 m/s and of B is 20 m/
s.
A will meet B for the first time after
25. a
When B covers 800 m A has covered 1,000 m.
So when B has covered 3 rounds = 3,000 m, A will
cover 3,750 m.
But to meet at some point, the overall lead should be
of one round = 1,000 m.
So there must be an initial lead of 250 m.
1
days .
6
Explanations: Fundamentals of
Time, Speed & Distance
MBA
Test Prep
Page: 109
Alternative method:
We see that the car reduces its travelling time by 20
min. So it reduces its one-way travelling time by
Solutions Exercise 5 (Level 2)
1. d
The hands of the clock are 35 min (= 35 × 6
= 210º) apart at 7 o’ clock.
For the hands to be together, the minute hand has
to gain 210° min over the hour hand.
Speed of the minute hand relative to the hours
hand
= 6º – 1/2º = 5.5º per minute.
Hence, the hands will meet after
20
= 10 min.
2
Thus, it would have met the children
10 min earlier, i.e. 3.50 p.m. Now this means that the
children walked from 3 p.m. to 3.50 p.m., i.e. 50 min.
4. d
210 420 382
=
=
min.
5.5
11
11
OR
Between x and (x + 1) o’ clock, the two hands will
12
= 11.11 hr.
27
Hence, the total time = 600 + 11.11 = 611.11 hr.
25 ×
12
be together at 5x × min past x.
11
12
2
min past 7.
i.e. 5 × 7 × min past 7 = 38
11
11
5. d
Alternative method:
Since the hour hand is between 7 and 8, the minute
hand has to show the time between 35 min and 40
min. Hence, there is only one choice in between
them.
2. a
If original speed and time are s and t, and new
speed and time are t1 and t1 respectively, then
s1 =
3. c
6. b
7
10t
s ⇒ t1 =
−t =1
10
7
or t =
7
hr = 2 hr 20 min.
3
Let X minutes be the time required for going to
school from home. Usually the car leaves home by
(4 p.m. – X) min and the children are back home by
(4 p.m. + X) min.
On Saturday, the children reached home 20 min
earlier, i.e. by (4 p.m. + X – 20) min.
⇒ The total time for which the car was in use
= 4 p.m. + X – 20 – [4 p.m. – X] = (2X – 20) min.
Hence, the car was used for (X – 10) min to go
and come back from school. If the children were
walking for Y minutes, then
⇒ 3 p.m. + Y + (X – 10) = 4 p.m. + (X – 20) min
⇒ Y – 10 = 4 p.m. – 3 p.m. – 20.
Hence, Y = 60 – 20 + 10 = 50 min.
In 24 hr, i.e. 1 day, the snail goes up by only
11 inch. height of the pole = 12 × 25 = 300 inches.
In 25 days, i.e. 600 hr it will go up (effectively) by
25 × 11 = 275 inches.
Now it can creep up 27 inches in 12 hr.
So to climb up 25 inches it will take
Two of the dog’s leaps cover 4 m,
three of the fox’s leaps cover 3 m,
i.e. each time the dog runs 4 m.
The distance between them is reduced by
4 m – 3 m = 1 m.
The initial distance between them is 30 m.
Hence, the dog will catch up with the fox when it
covers 4 × 30 = 120 m.
148
= 8 hr, i.e. it reaches 8 hr
18.5
after 7 a.m., i.e. 3 p.m.
So the second train reaches at 3.15 p.m.
Ratio of speeds of two trains is 8 : 5.
Hence, the ratio of time taken (distance being
constant) is 5 : 8.
Therefore, time taken by the second train
The first train takes
5
= × 8 = 5 hr, i.e. it starts 5 hr before 3.15 p.m.
8
⇒ 10.15 a.m.
7. c
Since Ajay is faster than Mallu and they start
together, to meet at 10 m from B, Ajay would have
covered a distance from A to B and would meet
Mallu on his way back to A.
Mallu would be on his way from A to B.
So Ajay covers 200 + 10 = 210 m in 10 s.
Hence, Ajay’s speed = 21 m/s.
So he will take
190
s to cover the remaining 190
21
m.
The time required for Ajay to reach A will be
10 +
Page: 110
190 400
s.
=
21
21
MBA
Test Prep
Solution Book-2
8. c
In the first 15 min, the thief will cover
Alternative method:
1
× 60 = 15 km.
4
Hence, to cover this 15 km, a policeman will take =
384
= 16 days for New York-Delhi travel.
24
Since the speed gets added up, when two people
come from opposite sides, anybody going from New
York to Delhi will meet two people going from Delhi
It takes
Dis tance
15
15
=
=
= 3 hr.
Relative speed 65 − 60 5
to New York every
9. a
When Zigma-S leaves Jaipur, the thief has covered
a distance of 15 km.
Distance between them = 300 – 15 = 285 km.
Relative speed of Zigma-S with respect to the
thief
= 60 + 60 = 120 km.
So time taken to catch the thief
285
= 2.375 hr = 2 hr 22.5 min.
120
According to question 8, Sigma-Z takes 3 hr to
catch the thief. Hence, Sigma-Z takes 37.5 min
more than Zigma-S in catching the thief.
∴ A person will take
11. a
13. b
16
hr to complete the race.
7x
Since they move in the same direction, relative speed
= 4x.
4
1
= .
4x
x
Therefore, number of times they meet in the entire
Time when they first meet =
1
of the
8
16
x
7 = 16
race =
= 2.3.
1
7
x
work.
Similarly, the second photocopier will complete
1
12
of the work in 1 hr.
If both work together, then fraction of the work
completed in 1 hr
1 1
3+2
5
.
+
=
=
8 12
24
24
Hence, both the photocopiers
Therefore, they meet twice before the race finishes.
14. d
working
384
= 16 days.
24
Thus, the person starting from Delhi (at 3 p.m.) is
bound to meet the 17 persons already enrouted,
and 16 more (who will have started after he starts)
during his journey.
Therefore, the total number of people he meets =
17 + 16 = 33.
4
2
=
.
10 x 5 x
16
x
7 = 80
Therefore, they meet
= 5.7 times,
2
14
x
5
24
= 4.8 hr.
simultaneously will take
5
i.e. 5 times.
Duration of the journey =
Explanations: Fundamentals of
Time, Speed & Distance
When they move in opposite directions, the time
when they meet the first time =
=
12. d
Let the speeds of two persons are 3x and 7x.
Y takes
Let the total distance covered by them when they
first meet is d kilometres.
When they meet for the second time, the total
distance covered by them is 3d.
Hence, they will meet after 12 × 3 = 36 min.
In 1 hr the first photocopier will complete
384
= 32 intervals to complete
12
his journey.
⇒ He will meet (32 + 1) = 33 people during the
journey (As he will meet the first person as soon as
he leaves New York and last person as soon as he
reaches Delhi.)
=
10. d
24
= 12 hr.
2
15. c
1
of the
6
tank. Hence, if the capacity of the tank = V m3, 60
In 1 hr, pipe A fills 60 m3, and pipe B fills
v v
=
(Portion of the tank filled in 1 hr when both
6 4
A and B are open)
720 + 2V = 3V ⇒ V = 720 m3
+
MBA
Test Prep
Page: 111
16. d
720
= 12 hr.
60
Therefore, 1 hr net task of all three pipes, when
opened simultaneously, is
Pipe A can fill the tank in
and
d
=3
u−v
or
u−v 2
=
u+v 3
Each maid takes 7 hr to clean a floor. Hence,
3 maids would also take 7 hr.
Here the concept of chain rule can also be used,
but the mop and the person should be taken as one
entity.
Since no work can be done if there are only mops
or only persons without mops.
24. c
Area per revolution = (2πR) × L = 1.32 sq. m
⇒ Total area = 1.32 × 400 = 528 sq. m
⇒ Total cost = Rs. 5280
25. d
In a mile race Akshay can be given a start of 128m
by of Bhairav. Or in 100m race Bhairav can give a
start off 8 m to Akshay.
So, for 100 m race:
u 5
= .
v 1
Thus, we see that we can get the ratio of the speeds
and not the exact speeds.
Let the job consists of 120 units [i.e. LCM
(20, 24, 15)]
∴ A + B do 6 units per day.
B + C do 5 units per day.
A + C do 8 units per day.
A + B + C do 9.5 units per day.
So B + C in 8 days do 40 units. A in 4 days
does 18 units.
Hence,
19. e
10
190
min
× 19 =
=
4
4
Hence, total distance covered
180 190 1
×
×
⇒ 7.5 km
19
4
60
PQ = 3.75 km
=
Let the speed of the boat in still water is u
Let the speed of the flowing river is v
when boat goes upstream (against the flow of river)
speed of the boat = u – v
for downstream (with the flow of the river)
=u+v
4 : 1 = (u + v) : (u – v) ⇒ u + v = 4u – 4v
⇒ 3u = 5v
Page: 112
Bhairav
100
Chinmay
96
So Chinmay is faster than Akshay.
Now Chinmay can give a start off 4 m in 96 m race
to Akshay.
So in 2,400 m race start given to Akshay
2 × 10 × 9 180
=
km / hr
19
19
180
: 12 ⇒ 15 : 19
Ratio of times of B and A =
19
Hence, if A takes 10 min more than Bibek
20. a
Akshay
92
58 29
=
part of the job is completed.
120 60
Average speed of A =
Assume speed of the river and the boat and make
an equation by equating their time.
Let M be the speed of the motor boat and R be the
speed of the raft/river. Then
23. c
or
18. c
2. 4
2. 4
+
=1
5−x 5+x
40
28
4
M 17
+
=
⇒
=
3(M + R) 3(M − R) R
R
3
Let the speed of the boat in still water be u km/hr
and that of the river be v km/hr.
If the distance from A to B is d, then we have
d
=2
u+v
Time taken =
⇒ x = 1 km/hr
22. c
1 1 1 2+4−3 1
+ − =
= .
12 6 8
24
8
Hence, the tank will be filled in 8 hr.
17. e
21. a
=
4
1
× 2400 = 100 m =
mile
96
16
Solutions Exercise 6 (Level 3)
1. a
Let the speed of the goods train be X m/s, and that
of the passenger train be Y m/s. In 28 s the goods
train covered 28X (m), and the passenger train,
28Y (m). Therefore, 28X + 28Y = 700.
The goods train passes the signal lights in
and the passenger train in
490
s
X
210
s.
Y
490
210
–
= 35 .
X
Y
Solving the two equations, we get X = 36 km/hr and
Y = 54 km/hr.
Therefore,
MBA
Test Prep
Solution Book-2
Short cut:
Going from the given answer choices, we see that
option (a): 36 km/hr = 10 m/s and 54 km/hr = 15 m/s.
So the goods train takes 49 s
490
10 and the
210
.
passenger train takes 14 s
15
Hence, the goods train takes 35 s longer than the
passenger train.
2. c
16
+ 11 hr or
The two trains would meet after =
3
Let’s assume that A takes x days to finish the job.
Then B will take 3x days.
3x – x = 60 ⇒ x = 30 days
1
.
Work done by A in one day =
30
1
.
Work done by B in one day =
90
at 4 : 20 p.m.
6. d
As in question 5,
speeds of the first train before and after the
accident would be 1 km/hr and 0.5 km/hr
respectively.
Since they meet 7 hr after they leave, the distance
covered by the first train till 4 p.m.
= 12 – 7 = 5 km.
Average speed of the first train till 4 p.m. =
Using alligations,
1 2
1
+
of the
Both combined would do
=
30 90 45
job in one day. Hence, they together will do the job
45
days = 22.5 days.
in
2
3. a
Since the two trains have the same speed, they
would take same time to cover the same distance
(between A and B). Hence, they would meet exactly
6 hr after they start, i.e. at 3 p.m.
4. b
If the distance between A and B is assumed to be
12 km, then relative speed of the two trains = 1 km/
hr.
Therefore, by 1 p.m., they would have both covered
4 km each. Distance left between them = 4 km.
Speed of the first train = 0.5 km/hr; speed of the
second train = 1 km/hr.
(Relative speed = 1 + 0.5 = 1.5 kmhr)
Hence, time taken to meet, after the accident
1
2
5
7
Hence, the time for which the train travels at two
speeds is in the ratio 3 : 4, i.e. after 3 hr of departure,
the accident takes place.
Hence, the accident takes place at 12 noon.
Alternative method:
Assume that the total distance d = 12 km.
Speed of the train from A to B and B to A is 1 km/hr.
4
2
= 2 hr = 2 hr 40 min.
1.5
3
So they meet at 3.40 p.m.
d
B
A
Now if the accident occurred after t hours from
start, the distance covered by the first train in t
hours = t km.
The train is late by 10 hr due to reduction in speed
after the accident as the speed is half. So time
taken now will be twice that of the original. Hence,
the accident must have occurred 10 hr before the
actual arrival time at the destination, i.e. at 11 o’
clock. Now till 11 o’ clock, if we assume the speed
of each train to be 1 km/hr and total distance = 12
km, in 2 hr they together will cover 4 km. The
remaining distance of 8 km will be covered with
1
km/hr.
relative speed of 1 km/hr and
2
8 16
∴ Time = =
hrs
3
3
2
1
2
(Because after the accident the speed is half)
(9 to 4 → 7 hr)
Distance covered by the second train in 7 hr
= 7 km.
To meet at 4 a.m., the total distance = 12 km.
=
5. c
5
km/hr.
7
Explanations: Fundamentals of
Time, Speed & Distance
And distance covered in (7 – t) hr = (7 − t ) ×
∴ 7 + t + (7 − t ) ×
1
= 12
2
⇒ t = 3 hr after start, i.e. at 12 o’ clock.
MBA
Test Prep
Page: 113
7. b
Let the length of the wall be 30 m. Bamdev can
demolish 1 unit per day and his son can demolish
0.5 unit per day. Manas can construct 3 units per
day. If they work simultaneously with Manas starting
they would do effectively 1.5 units per day. So in 36
days, 18 × 1.5 units is completed, i.e. 27 units is
completed. Hence, on the 37th day, Manas would
complete the job. Manas would have worked for 19
days and constructed 19 × 3 m.
Hence, the fraction of the job completed
9. d
=
57 19 38
=
=
=
.
30 10 20
10. a
1
work in 1 day.
10
Also Bamdev and his son can demolish
∴ He does
23 min + 55 s
~ 51 s.
28
Let us calculate the tonne-hours done.
3 tonnes × 30 trucks × 8 hr = 720 tonne-hours.
5 tonnes × 9 trucks × 6 hr = 270 tonne-hours.
∴ Total tonne-hours = 720 + 270 = 990 tonne-hours
∴ Hours required for a single three-tonne truck
990
= 330 hr
3
Hours required for five-tonne truck
1
1
1
+
=
of the wall in next day.
30 60 20
Thus, in 2 days Manas effectively constructs
=
1
1
1
–
=
of the wall.
10 20 20
=
20 × 9
9
×2
of the wall in
10
10
= 36 days and on 37th day, he finishes the
construction. Since he has worked for 19 days, he
∴ He constructs
constructed
11. b
19 38
=
portion of the wall.
10 20
If Manas started when there were 24 units to be
completed, then working at 1.5 units per 2 days, 21
units
would
be
completed
in
Let the rate of growth of the grass be x straws per
day and the initial amount of grass be y straws.
Assuming that one cow grazes on one straw of
grass in a day, (40 cows) × (40 days)
= 1600 = y + 40x.
(30 cows) × (60 days) = 1800 = y + 60x.
⇒ y = 1200 and x = 10.
So for 20 cows,
(20 cows) × (n days) = y + nx = 1200 + 10n
Alternative method:
Let G be the amount of grass in the field and g be
the growth rate of grass per day.
⇒ G + 40g = 40 × 40
and G + 60g = 60 × 30
Alternative method 1:
⇒ g = 10, G = 1200
2
of the
10
wall, i.e. 8 days of work is already done.
∴ So 37 – 8 = 29 days are required.
This is identical to question 22, only that
Alternative method 2:
From choices it is evident that there is only one odd
choice. We know that the answer has to be odd
since
on the last day Manas has to complete the job.
990
= 198 hr.
5
⇒ n = 120 days.
21
1. 5
= 14 pairs of days or 28 days.
Hence, on 29th day Manas would complete the job.
Page: 114
4 min 18 s + 7 min 13 s + 12 min 24 s
6 + 9 + 13
=
Alternative method:
Manas can construct a wall in 10 days.
8. b
In his first shift, he uses the container six times.
(45 glasses in 5 rounds and the remaining 6 glasses
in the 6th round)
Similarly, he prepared 73 glasses in 9 rounds, and
112 glasses in 13 rounds.
Hence, the required average
Total time
=
Total number of rounds
⇒ 1200 + N × 10 = 20 N
⇒ N = 120 days
12. d
t1
t2
Let t1 be the time at which B switches the speed
and t1 + t 2 be the total time between start and finish.
Let x be the speed of B initially. So A’s speed
= 1.2x and B’s final speed = 1.44x.
MBA
Test Prep
Solution Book-2
13. d
Now lag of B in time t1
= (1.2 x – x )t1 = 0.2 × t1
20
= 4 days.
5
So time taken by 4 men to complete the job
= 4 × 2 = 8 days.
And time taken by 4 women = 8 × 2 = 16 days.
Therefore, 12 men + 10 children + 8 women
… (i)
Also gain of B in time t 2
… (ii)
= (1.44 x – 1.2 x )t 2 = 0.24 × t 2
Since both reach at the same time,
∴ Lag = lead
⇒
t1 0.24 6
=
=
t 2 0.20 5
complete =
1760 × 6
= 960 m.
11
14. a
x
20 × 20
= 44% more
100
VB + t1 + 1.44 VB t 2 = 1760
6
t2
5
Now distance covered by A in t1 time = 1.2 VBt1
⇒ t1 =
5
1.2 VB t1 + t1 = 1760
6
1.2 VB t1 =
1760 × 6
= 960 m .
11
= 9 ⇒ x = 9y – 5
⇒ 90y – 9x = 100
Solving (i) and (ii) x = 50 m
Since race ended in dry heat, equating time of A and
B
∴ 1.2 VB ( t1 + t 2 ) = 1760
5
m/s )
9
... (i)
x
Similarly, for second 9 y − 10 = 10
9
1.44 u
Alternative method 2:
Assume that A runs 20% faster than B for t1 time
and for t2 time B runs 20% faster than A.
Speed of B initially is VB for t1 time.
So VA = 1.2 VB for (t1 + t2) time
VB’ = 1.44 VB for t2 time
(As 2 km/hr =
9
x1
x1 1760 − x1 1760
+
=
⇒ x1 = 797.819 m .
u
1.44u
1.2u
5 9y − 5
=
9
9
9y − 5
= 1.44 u
u
x = Length of train, y = Speed of train (m/s)
Relative speed with respect to first one
= y−
Alternative method 1:
Let the speed of B → u.
Speed of A → 1.2u
New speed of
B = 20 + 20 +
3
1
1 (3 + 4 + 1) 8
+ + =
= 1 of the
=
8
2 8
8
8
work.
Hence, time taken = 1 day.
1760 × t1
∴ A covers
of the distance
( t1 + t2 )
=
The time taken by 5 children to complete the job =
y=
... (ii)
55
m/s
9
15. b
When Ram completes second round, they pat each
other once. In the 20th round, Ram will finish the
race and the total number of pats = 20 – 1 = 19
This solution is only valid when greater ones speed
is less than twice the first one speed.
16. c
They reverse directions on reaching A.
Ram with 12 km/hr speed will reach A first and
immediately reverse directions.
Suraj will reverse direction only after he reaches A.
Time taken by Ram to complete the circle
= 1.2 km/12 km/hr = 0.1 hr = 6 min
Time taken by Suraj to complete the circle
= 1.2 km/8 km/hr = 0.15 hr = 9 min
In those extra (9 – 6) = 3 min, Ram would have
already reversed direction and done another half
circle in the same direction as Suraj.
1.2
The effective distance between them =
2
= 0.6 km
Time taken to meet after Suraj reverses direction
0 .6
= 0.03 hr = 1.8 min
(12 + 8)
Therefore, time taken to meet second time
= 9 + 1.8 min = 10.8 min
=
Explanations: Fundamentals of
Time, Speed & Distance
MBA
Test Prep
Page: 115
17. b
Time taken will be double the time taken to meet the
first time = 3.6 × 2 = 7.2 min
Since each time the effective distance between
them will be 1,200 m.
18. c
The time taken by P and Q to meet after their start is
2. e
For every day's work, he can afford to miss
3 days. Hence, to break even it has to be 7 days out
of 28 days.
3. e
In 1 day A and B together produce
1
1
3
+
=
units
15 12 20
∴ In 20 days,
400
400
=
= 4 hr
40 + 60 100
Hence, they must have started at 11 a.m.
A and B together will produce
Alternative method:
which will fetch them
Rs. 270.
3
Q covers
of the distance, i.e. 240 km.
5
It takes 4 hr to cover this distance.
19. e
R would take
3
× 20 = 3 units ,
20
Now,
60 × 4
⇒ 12 hr to meet Q.
(80 − 60)
Efficiency of B 15 5
=
=
Efficiency of A 12 4
⇒ Share of B is
5
× 270 = Rs. 150
9
By that time, Q would have already reached Guntur.
4. a
20. a
Normally P and Q meet after 4 hr of journey.
Now they are meeting after 5 hr.
Hence, the distance travelled by
Q = 400 – (40 × 5) = 200 km
If the accident occurred after x hours, then
60(x) + 20(5 – x) = 200
or 40x = 100 or x = 2
1. a
2 1
=
of the remaining is to be scarified by
6 3
each.
∴
1
hr
2
Solutions Exercise 7 (Level 3)
5. b
Let ship sends the radiowave from point A and
receives it at point B
A
5 00 m
M
12 bottles have to be shared by has 10 people, i.e.
a man consumes 1.2 bottle and 50% has been
consumed
⇒ 6 bottles or (0.6 bottle by each).
2 bottles out of 6 have to be scarified to give them
their original share.
V = Speed, N = Wagons
∴ The quantity by which speed is diminished = 40
–V
⇒ 40 – V = K N ; V = 28 and N = 16
⇒ K=3
B
∴ 40 – V = 3 N
When V ≥ 10, the least value of V = 10 km/hr
⇒ 40 – 10 = 3 N ⇒ N = 100
O
O cea n b e d
0.5
1
time taken =
=
= 1 min
30 60
Distance travelled by radiowave in 1 min.
= 200 × 60 = 12 km
6. d
1 28 x
1
x
+
=
12 16 192
Left capacity = 1 –
28x
192
∴ OA = OB = 6km
In ∆AOM
OM2 = OA2 – AM2
1
= 62 −
2
⇒ OM =
Page: 116
2
This is filled by P in 5 min and fills
⇒
1
in 1 min
12
192 − 28 x
5
⇒ x = 4 min
=
192
12
143
km
2
MBA
Test Prep
Solution Book-2
L
7. e
Shallow end
12. b
S1
S2
10.5m O
2
× 33
5
D eep end
O 18.5m
Trained labourer Days
First they meet at point O,
S2 L − 18.5
for which S = 18.5
1
y
...(i)
as both stay at the ends for equal time, so time
taken from the starting up to the second meeting(at
O’) will be same.
S
L + (L − 10.5)
⇒ 2 =
...(ii)
S1
L + 10.5
L − 18.5 2L − 10.5
=
18.5
L + 10.5
apply componendo & dividendo-
Thus, y =
13. d
Hours
Work
15
12
1
11
9
1.5
2
15
3
12
× 33 ×
×
×
= 36
5
11
2
9
6 33 7
8 36
+
=
+
= 2 and
u
v
4
u
v
Solving these, we get u = 16 and v = 24.
Hence, rate of stream is 4 km/hr.
from (i) and (ii)
14. b
L
3L
=
⇒ L = 45m
L − 37 L − 21
8. b
The two trains start at the same time.
If they meet after x hr, then 21x – 16x = 60
⇒ x = 12
⇒ Distance = (16 × 12 + 21 × 12) = 444 km
9. c
Ratio of time taken is 3 : 2.
If difference is 1 min, A takes 3 min.
If difference is 10 min, A takes 30 min.
⇒ At double speed A takes 15 min.
10. c
3 men = 4 boys
Hence, 27 boys can reap a field in 15 days.
3
of the
So 20 boys + 16 boys = 36 boys will take
4
time, i.e. 11.25 days.
LCM of 224, i.e. (25 × 7) and 364, i.e. (22 × 7 × 13) =
25 × 7 × 13
25 × 7 divides this 13 times.
So A does 13 rounds before meeting again.
22 × 7 × 13 divides this 8 times.
So B does 8 rounds before meeting again.
Thus, the difference in the number of laps is
(13 – 8) = 5
15. b
11. c
If A takes x seconds and B takes y seconds to
run 1 km, then
960
960x
+ 30 = y
y and
1000
1000
⇒ y = 150 s and x = 125 s
X's rate
=
Y's rate
or
45
=
Y ' s rate
150
× 5000 = 750 s
1000
16. d
If the cross-sectional circumference of pipe A is 1
unit, that of B would be 2 units and that of C would
be 3 units.
So areas would be in the ratio 1 : 4 : 9 or the areas
of B and C together is 13 times that of A.
So it takes
1
3
× 16 min or 1 min to fill the
13
13
original tank.
But now we have to fill the second tank. Thus, time
required will be
Explanations: Fundamentals of
Time, Speed & Distance
200 5
=
288 6
⇒ Y’s rate = 54 km/hr
x + 19 =
⇒ answer =
Time taken by Y to reach Jodhpur
Time taken by X to reach Jaipur
MBA
Test Prep
16 × 2 32
=
min .
13
13
Page: 117
17. c
18. c
Alternative method:
Let x be the initial speed. Equating time
Let him walk at 4 km/hr for t hr and at 3 km for h hr
Now 4 × t + 3 × h = 36
Also 4 × h + 3 × t = 34
Solving, we get h = 4, t = 6
Therefore, h + t = 6 + 4 = 10 hr
420
60
9
480
6
–
+
+
=
x
x + 20
60
x
60
420
60
1 480
+
+
=
x
x + 20
4
x
1 man = 2.5 boys; 1 woman = 1.5 boys; convert all
work in terms of boys.
Thus, group 1 = 43 boys, group 2
= 107.5 boys.
Hence, group 2 would take 25 ×
⇒
43
×3
107.5
60
1 60
+
=
x + 20
4
x
Check from options, x = 60.
= 30 days
20. e
19. d
It is given that
7
of the distance = 420 km or total
8
8
= 480 km.
7
It is given that he stopped for 9 min and when he
increased his speed by 20 km/hr, he reached his
destination by 6 min earlier. Hence, there is a
difference of 9 + 6 = 15 min. Now take options into
work.
Taking option (d), i.e. 60 km/hr,
distance = 420 ×
Average speed =
=
Total dis tance
Total time taken
480
480
480
=
=
7 hr + 9 min + 45 min
7 hr 54 min 7.9
= 60.75 km/hr.
420
60
+
= 7 hr 45 min.
60
60 + 20
If he travelled the entire 480 km at the rate 60 km/
hr, then time taken = 8 hr.
Since there is a difference of 15 min, (d) is the
answer.
time taken =
Page: 118
MBA
Test Prep
Solution Book-2
Explanations: Fundamentals of Grammar
Unit – 1
Chapter 1
6.
who baked the winning pie modifies woman
7.
when I was unable to answer modifies time
8.
for whom you are looking modifies one
9.
who are willing to serve others modifies those
10.
to whom much is given modifies one
Exercise 1
1.
time - subject,
may be - verb
Exercise 4
2.
mail - subject,
was - verb
1.
How the prisoner escaped = subject
3.
letter- subject,
has been - verb
2.
that the robbery was an inside job = predicate
nominative
4.
men - subject,
were - verb
3.
how he could just disappear = direct object
5.
tracks - subject, were – verb
4.
that he had escaped = appositive
5.
whoever finds the stolen diamonds = indirect object
6.
where he might be = object of the preposition
7.
That we were ready to go = subject
8.
whoever wants to go = indirect object
Exercise 2
1.
compound verb
2.
compound sentence
3.
compound sentence
9.
That you are losing ground = subject
4.
compound subject
10.
Whoever injured the handicapped woman = subject
5.
compound verb
Exercise 5
6.
compound sentence
1.
that we have had = adjective clause modifying the
predicate nominative year
7.
compound object of the preposition
2.
8.
compound sentence
until we received word of his rescue = adverb clause
modifying the verb waited
9.
compound verb
3.
whom I saw on Mount Kilimanjaroo = adjective clause
modifying the subject hiker
10.
compound sentence
4.
that he will win the lottery = noun clause used as the
direct object
5.
Who lost this wallet = noun clause used as the subject
Exercise 3
1.
who listens to his men modifies leader
2.
which I loved dearly modifies dog
3.
who takes responsibility well modifies person
4.
who purchased tickets modifies individuals
5.
that you bought for me modifies shirt
Exercise 6
Explanations:
Fundamentals of Grammar
1.
If the manager is unable to help = adverb clause
modifying the verb try
2.
whom you should write the letter = noun clause used
as the object of the preposition
MBA
Test Prep
Page: 119
3.
whose neck was broken = adjective clause modifying
the subject man
5.
She said that she did not know where her shoes
were.
4.
that the ozone levels were dangerous = noun clause
used as the direct object
6.
She asked, “Why are you studying English?”
7.
She said, “May I open a new browser?”
5.
when the governor changed his mind = adverb clause
modifying the verb objected
8.
She said she had to have a computer to teach
English online.
6.
that Sarika will not return = adverb clause modifying
the predicate adjective unfortunate
7.
Why you don’t do your work = noun clause used as
the subject.
8.
where your Aunt is buried = noun clause used as the
predicate nominative.
9.
that the island is under water = noun clause used as
the appositive.
10.
whoever told the truth = noun clause used as the
indirect object.
Unit – 2
Chapter 5
Test 1
1.
She wants an orange from that tree. (there can be
many oranges on the tree and the noun is starting
with a vowel sound).
2.
The edifice on the corner is huge. (some specific
building, which is huge).
3.
The Chinese that Mrs. Sarla speaks is very easy to
learn. (that specific Chinese, which Mrs Sarla speaks).
4.
I borrowed a pencil from your sister’s pile of pens
and pencils. (She borrowed one of the many pencils
that her sister has).
5.
One of the officers said, “ The chairman is late today.”
(The officers are talking about someone specific, their
chairman)
6
Ankur likes to play volleyball. (No article is required
here.)
7.
I bought an umbrella recently. (the modified noun is
beginning with a vowel; u)
8.
Kirti is learning to play the violin at her school. (Definite
article comes before a)
Please give me the book that is on the counter. (The
speaker is asking for a particular book, which is lying
on the table.)
Chapter 2
Exercise 1
1.
Mistakes are made by us.
2.
Extensive research is done to determine by which
gene autism is caused in the body.
3.
People all over the world speak English.
4.
The dog bit Latika.
5.
Characters are revealed by manners.
6.
Why did Sheena say such a thing?
7.
He was laughed at by everyone.
8.
I will be obliged to go by the circumstances.
9.
Chapter 3
Exercise 1
10.
We lived on Pali Street when I first came to your town.
(No article is required before names of cities, states
, countries or streets)
Julia Roberts asked him about his plans for that
day.
11.
Delhi is the capital of India. (No article is required
before names of cities, states , countries or streets)
3.
Richa said that all of us would go there in the
morning tomorrow.
12.
My colleague’s family speaks Polish. (No article is
required before names of languages.)
4.
She said, “How long have you worked here?”
1.
Vandana says that her husband loves her a lot.
2.
Page: 120
MBA
Test Prep
Solution Book-2
13.
The apples in my basket are red. (Those particular
apples, which are in my basket)
14.
Our friends have a cat and a dog. (our friend has
both, one dog and one cat)
15.
Go to a university near your house. (The noun is
starting with a vowel but is pronounced with a ‘u’
sound, therefore, a instead of an is used.)
16.
a lot
17.
several of
18.
a lot of
19.
some
20.
only, a few
21.
a majority of, enough
22.
many
23.
much of
Test 3
1. d
A is incorrect use an 'a' before store, B is incorrect
use 'the' before commodities, E is incorrect, use 'an'
before enormous.
2. c
Options A, C & D are incorrect, all need the article 'a'
before 'far' in A, 'spirit' in C and 'watch tower' in D.
3. d
B is incorrect, needs 'the' before hero, E is incorrect,
needs 'A' before crush.
4. b
Option A is incorrect because it should be 'an' epigram
not 'the' as it embodies the generic, C is incorrect
since it uses an before earliest which should be 'the'
instead.
5. d
A is incorrect, use 'the' before 'West', C is incorrect,
use 'the' before the word 'roots'.
6. a
Option B is incorrect, use 'a ' instead of 'the', E is
incorrect, use 'The' instead of 'A' to describe the
definite.
7. b
Ans. b: A is incorrect as the article 'a' should be used
before certain, C is incorrect 'the' should be used to
emphasize 'nutritional'.
Test 2
1. a
In 'A' it should be 'a tedious loser' and in 'C' it should
be 'a similar level…'.
8. d
A is incorrect; use 'the' before Olympics, B is incorrect,
use the article 'the' before 'big contract'.
2. c
In 'E' it should be 'in …the sunshine".
9. b
3. c
In 'B' 'enough' is not required. The correct expression
is 'large enough' not 'larger enough'. In 'D' 'various' is
inappropriate. It should be replaced by 'several'.
A is incorrect, use definite article 'the' before
'dangers', D is ambiguous, needs a determiner before
'people' use 'some' to make the sentence correct.
4. c
'A' should have 'a friend'. 'C' should have 'gardening
is a joy....'and 'D' should have 'pick a better.."
5. a
In 'B' it should be 'ever more....' and not 'even..'. In 'C'
it should be 'many' not 'more'.
6. e
'E' should have '..the fleece'.
7. a
In A it should be 'a prudent board..' because it can be
any board.
8. e
In E it should b 'a mistake'.
10. e C is incorrect the definite article (the) should precede
'two' and E is incorrect as 'the' should precede the
word 'pleasure'.
Unit – 3
Chapter 6
Exercise 1
1.
petrol - mass; class - collective
2.
group - collective; bus - count
3.
orchestra - collective; arena and evening - count
4.
water and oil - mass; beach - count
9. c Money is quantified in terms of ‘little’.
10. d ‘People’ are numbered. ‘Few’ is used for numbers.
Explanations:
Fundamentals of Grammar
MBA
Test Prep
Page: 121
Exercise 2
7. a
It should be ‘Sumeet, Rekha and I’ because in such
cases the sequence should be third person, second
person and finally first person.
8. e
In A the right word is ‘baggage’; in C the right word is
‘equipment’ and in E the right word is ‘data’.
9. b
In B the correct word is ‘strength’ and not ‘strengths’,
and in E it will not be ‘shore’ but ‘shores’.
1.
Objective case
2.
Objective case
3.
Objective case
4.
Subjective case
5.
Subjective case
10. e ‘Suspicion’ will not take ‘s’.
6.
Objective case
Test 2
7.
Possessive case
1. c
8.
Possessive case
9.
Objective case
10.
Possessive case
C is the correct option. In A it should be ‘Soviet dictator’
(singular). In B the correct word should be ‘they’ since
it refers to ‘people’ and ‘people’ is plural. In D ‘often’
should come after ‘Stalin’ and in E it should not be
‘Soviet’s’; rather it should be ‘Soviet’ because it is
telling about the nationality of the concerned person
and thus qualifying the noun.
2. c
C is the correct option because ‘our house’ is being
compared to ‘that of’ our neighbours. B & D suggest
that ‘our houses or house’ is bigger than ‘our
neighbours’.
3. a
B is incorrect because the ‘apostrophe’ should come
with the ‘novelist’ and not ‘Charles Dickens’. In case
of two nouns coming together the apostrophe is
always placed with the second noun. D is incorrect
because ‘respect’ is an abstract noun and it can never
be in plural. In E ‘emotions’ will be followed by ‘has’
and not ‘have’ because it is referring to the ‘portrayal’
and not to ‘emotions’.
4. e
In A ‘boomers’ should have an apostrophe. In B ‘civil
right’ is a wrong expression because it’s always
‘rights’ in the given context. In C ‘credits’ is incorrect
because here ‘credit’ means ‘trustworthiness;
credibility’ and it cannot be used as plural. In D ‘too
much’ is not appropriate in the given context because
it changes the positive tone of the sentence into a
negative one.
5. b
In ‘A’ ‘coverages’ is incorrect. The right word is
‘coverage’. Certain words remain same in both singular
and plural states. For example, ‘aircraft’, information’,
‘sheep’ etc. ‘C’ has improper structure. In D it should
be ‘coverage makes..’ because ‘coverage’ is a
collective noun here. In E it should be ‘days’ since the
context is referring to a general kind of atmosphere
for which the author yearns therefore ‘days’ will be
more appropriate.
Exercise 3
1.
People and wall are concrete nouns. Home run is a
compound noun.
2.
Son is a concrete noun while post office and Udaipur
Lake city are compound nouns.
3.
Mumbai Island is compound; success is abstract;
Singapore is concrete.
4.
Respect, honesty and policy are abstract nouns.
Compound nouns can also be concrete or abstract.
Test 1
1. e
All the parts of the sentence are grammatically
correct.
2. d
In D it should be ‘another’ and not other since it is
talking about one more.
3. c
In C ‘barren century’ should be preceded by ‘some’ or
‘a’.
4. b
In B the right way of writing will be ‘each called a
scissor’.
5. a
In A the correct way of writing is ‘information does
not..’ since ‘information’ is always treated as a
singular.
6. c
In A the correct word is ‘damage’ and not damages
and in E the right word is ‘illness’ since it is ‘a chronic’.
Page: 122
MBA
Test Prep
Solution Book-2
6. c
A is incorrect because ‘hairs’ is incorrect. B is incorrect
because it says ‘not the…’. A process can be any
process therefore it will be ‘an ..extremely…’. D is
incorrect because ‘growing’ will not take ‘of’ here. E
is incorrect because ‘are’ and ‘processes’ are
incorrect out here, since the noun (gerund) referred
to, in this is ‘growing’.
7. a
‘Tourists’ and not tourism should be used.
8. c
The climate of Delhi can be compared to the climate of
Jaipur and not to the city of Jaipur.
9. a
‘alike’ means resembling each other. Similar would
mean not exactly the same.
Chapter 7
Exercise 1
1.
She, him, his
2.
I, you, your
3.
He, himself, our
4.
It, me, you
5.
They, her, me
Exercise 2
10. c When the sentence is positive, the tag has to be
negative.
1.
‘He’ is the antecedent for ‘his’.
Test 3
2.
‘Madonna’ is the antecedent for ‘her’.
1. c
3.
‘Rabbit’ is the antecedent for ‘its’.
4.
‘They’ is the antecedent for themselves, and you is
the antecedent for ‘your’.
5.
‘Teacher’ is the antecedent for she, and us is the
antecedent for ‘our’.
In A ‘politician’ should be in plural because the sentence
talks about ‘other’ which is an undefined number. In
case it were a singular then it should have been ‘..the
another’. In B ‘a time’ is incorrect; it should have been
‘some..’
2. c
In C ‘colonial’ is an incorrect word to be used because
it is an adjective and not a noun.
3. d
In A instead of ‘enough’ it should be ‘some’. In D an
article ‘the’ should precede ‘Atlantic’.
Exercise 3
1.
his
In A it should be ‘a degree of showmanship’ because
degree has to be somewhat quantified and
showmanship does not require a ‘the’ before it.
2.
yours
3.
theirs
In A it should be ‘a rebuke’. In D it should be ‘indiscretion’
and not ‘indiscrete’, because the previous word is a
noun and the latter is an adjective.
4.
his, hers
In A it should be ‘surge’ and not ‘surging’ since the
context requires a noun. In E logging will not take an
‘s’.
1.
myself
2.
herself
7. c
In C prediction should be in plural form and in D it will
be ‘effects’ and not ‘effect’.
3.
themselves
4.
himself
8. b
In B ‘experience’ is the right word and not ‘experiences’.
9. e
In E, it will be ‘imminence’ and not ‘imminent..’ since in
the context the emphasis is on the very fact of
‘imminence’ which is a noun. If we put ‘imminent’ the
meaning will be entirely out of context.
4. a
5. d
6. a
10. e In D it will be ‘tank bombardment’; since ‘bombarding’
is a verb.
Explanations:
Fundamentals of Grammar
Exercise 4
Exercise 5
1.
ourselves
2.
himself
3.
herself
4.
themselves
MBA
Test Prep
Page: 123
Exercise 6
4.
what - interrogative,
this - demonstrative,
me - personal
I - personal, that - relative, you - personal
1.
‘Which’ is the relative pronoun.
2.
‘That’ is the relative pronoun.
5.
3.
‘Who’ is the relative pronoun.
Exercise 11
4.
‘Whom’ is the relative pronoun.
1.
5.
‘Whose’ is the relative pronoun.
he - personal, himself - personal, intensive, my personal, possessive. He is the antecedent for
himself. (something is a noun)
Exercise 7
2.
Which - interrogative, this - demonstrative
1.
that
3.
These - demonstrative, mine - personal, possessive,
Whose - interrogative, these - demonstrative
2.
those
4.
3.
this
this - demonstrative, that - relative, I - personal, you personal
4.
these
5.
everyone, some, many, no one, none - all are indefinite
6.
he - personal, himself - personal, reflexive, his personal. He is the antecedent for himself and his.
7.
I - personal, myself - personal, intensive, him personal, himself - personal, reflexive, everybody indefinite. I is the antecedent for myself, and him is
the antecedent for himself.
8.
neither - indefinite, them - personal, anyone - indefinite,
who - relative, us - personal
9.
who - interrogative, that - relative, that - demonstrative
Exercise 8
1.
everybody, someone
2.
both, everything
3.
anyone
4.
no one, others
5.
somebody, one, neither
Exercise 9
1.
what
2.
who
3.
which
4.
whose
5.
whom
Test 1
1.
Here the subjective case of pronoun will be used.
2.
‘She’ is also a subject in this case along with ‘Payal’.
3.
In this sentence ‘They’ is the subject and the action is
directed towards ‘him’ and ‘me’ therefore these two
words are in objective cases.
4.
The ‘call’ is directed towards ‘him and her’ therefore
these two are the objects of the call in the sentence
thus in objective case.
5.
Same explanation as in ‘4’.
6.
Here both ‘I’ and ‘he’ are the subjective cases of
pronouns.
7.
‘Rohit’ and ‘she’ both are subjects in this sentence.
8.
In this sentence ‘we’ will be used because the action
of ‘decide’ is being done by them in the sentence;
thus the subjective case.
Exercise 10
1.
2.
3.
whom - interrogative,
you - personal,
that - demonstrative
neither - indefinite, my - personal,
me - personal
you - personal, someone - indefinite,
who - relative, others - indefinite
Page: 124
MBA
Test Prep
Solution Book-2
9.
Here ‘I’ will be used because ‘you’ is the subject
and the condition referred to is ‘if you were ..’.
10.
‘One’ is the pronoun antecedent therefore it should
be followed by ‘oneself’.
3. b
A is incorrect, use Everybody / Everyone instead of
‘Everyman’, B is incorrect use ‘who’ instead of ‘whom’
– whom is used as the object of a verb or preposition.
D is incorrect- use ‘they’ instead of them.
4. e
B is incorrect, use ‘what’ instead of which, ‘what’
here is used relatively to indicate that which; D is
incorrect, use ‘This’ instead of the plural ‘these’, ‘this’
is used to indicate a person, thing, idea, state, event.
5. b
B is incorrect, should be the possessive pronoun ‘its’,
C is incorrect , the pronoun should agree in number
with the noun that is the object of the preposition, use
‘is’ instead of ‘are’.
6. e
D is incorrect, use singular pronoun instead of ‘their’,
E is incorrect use ‘your’ instead of the possessive
yours.
7. c
E is incorrect, should be ‘her speech’- with compound
antecedents joined by or, nor, either...or, neither...nor,
use pronouns that agree with the nearest antecedent.
8. a
A is incorrect- When a pronoun is used along with a
noun, choose the pronoun case that matches the
noun’s function, should be ‘us’, B ‘They’- When a
pronoun is part of a compound element, choose the
pronoun case that would be correct if the pronoun
were not part of a compound element, C- Should be
‘he’ instead of himself - when a personal pronoun is
used in a comparison, choose the correct pronoun
case by carrying the sentence out to its logical
conclusion.
9. e
B is incorrect uses the contraction of it is should be
the possessive ‘its’, D is incorrect should use ‘which’
instead of ‘who’.
Test 2
1.
2.
‘whom‘ is the correct pronoun.
In this sentence a relative pronoun that is in the
objective case because the woman is the object of
the action and ‘we’ is the subject.
Similarly in the second sentence the subject is ‘you’
and the object has to be ‘whom’ and not ‘who’ because
the latter is a subjective case of pronoun.
3.
Same explanation as ‘2’.
4.
‘Whomever’ is the objective case.
5.
Same explanation as ‘4’
6.
Same explanation as ‘4’.
7.
Same explanation as ‘4’.
8.
Politicians are the subjects out here, therefore the
relative pronoun should be in the objective case.
9.
In this sentence the action is ‘write’, therefore the
doer has to be in the subjective case. And in this case
‘whoever’ is the doer.
10.
Similar explanation as ‘9’.
11.
In this sentence a particular and definite set of book is
being referred to, therefore ‘that’ has been used.
12.
The sentence is referring to one of the books,
therefore ‘which’ is being used.
13.
Similar explanation.
14.
Here the reference is to a definite law therefore
‘that’.
Chapter 8
15.
The law referred to is one of the laws thus ‘which’.
Exercise 1
10. e B is incorrect use the reflexive form ‘herself’ instead
of her, C is incorrect use ‘they’ instead of which.
Unit – 4
Test 3
1.
are going,
1. e
C is incorrect, uses the pronoun ‘myself’ , ‘me’ is the
correct pronoun to be used here, E is incorrect use ‘I’
instead of ‘me’.
2.
have been resting
3.
must be
B is incorrect; use singular verb with ‘each one’ should
be works, E is incorrect there should be a pronoun
before ‘young ones’, use ‘their’ to make it correct.
4.
will be finished.
2. e
Explanations:
Fundamentals of Grammar
MBA
Test Prep
Page: 125
Exercise 2
Exercise 6
1.
has
1.
have been driving
2.
do
2.
was parked
3.
was
3.
can be.
4.
are being
Exercise 7
‘Do’ is action verb, and was, has and are being are
state of being verbs.
1.
have helped
2.
will be
3.
had choked.
Exercise 3
1.
plays - action
2.
will return - action
3.
is - linking or state of being
Exercise 8
4.
have been - state of being
1.
is reading
5.
should have been playing - action
2.
can be altered
6.
go - action
3.
is being planned
Exercise 4
4.
should be
1.
Can understand
5.
‘ve (have) run.
2.
must have told
3.
shall go
4.
was howling
The first and third sentences are action verbs, and
the second sentence a state of being verb.
There are twenty-three helping verbs viz. is, am, are,
was, were, be, being, been, have, has, had, do, does,
did, shall, will, should, would, may, might, must, can,
and could.
Exercise 9
The use of helping verbs causes certain changes in
verb phrases. One change is the use of contractions.
1.
will be finished
Exercise 5
2.
should have been worked
1.
‘ve (have) done
3.
would give. Not and n’t are not part of the verb phrase.
2.
are going,
3.
‘s (is) staying.
Have and is are in contracted form. Are is connected
to the contracted form of not. All three-verb phrases
are action verbs.
Verb phrases can have one, two or three helping
verbs in them
Verb phrases with two or more helping verbs always
keep a definite order. Most helping verbs can combine
with other helping verbs but will not combine with all
of them. e.g., is being said, has been said, will be said,
could have been said, may have said, had been said.
We can change the form of a verb: A verb can have
an s added to it as in eat, eats or run, runs. Other
changes could be eating, ate, or eaten for the verb
eat. Run could be changed to running, or ran. Some
Irregular verbs have several confusing changes.
Page: 126
MBA
Test Prep
Solution Book-2
Exercise 10
Exercise 14
1.
am coming
1.
farming = predicate nominative
2.
came
2.
traveling = appositive
3.
comes
3.
writing = subject
4.
send
4.
saving = object of the preposition / traveling = direct
object
5.
was sent
5.
gossiping = indirect object
6.
am sending
Exercise 15
Exercise 11
1.
to skate = subject
1.
can understand - action
2.
to enjoy = direct object
2.
is going - action
3.
to win = predicate nominative
3.
can be held - action
4.
have seen - action
4.
to marry = direct object
5.
to kill = appositive
Exercise 16
Exercise 12
1.
to sit in judgment = subject
1.
roared - intransitive complete (no receiver of the
action)
2.
to waste time in class = subject
3.
to party/to sleep in = subjects
2.
was - intransitive linking (captain is a predicate
nominative)
4.
to build a home/(to) raise a family = direct objects
5.
to be home soon = direct object
6.
to play cricket/to visit friends = predicate nominatives
7.
to save money for a rainy day = subject
8.
to go to college/to study management = predicate
nominatives
Exercise 13
9.
to be rewarded for one’s efforts = direct object
1.
failing
10.
to live happily = object of the preposition
2.
bordering
Exercise 17
3.
to consult
1.
broken modifying spoke
4.
searching
2.
smiling modifying face
5.
to match
3.
frightened modifying child
4.
growling modifying dog
5.
squeaking modifying wheel
3.
will be needed - transitive passive (be is the helping
verb and dress receives the action)
4.
did forget - transitive active (title receives the action
and is the direct object)
5.
has - transitive active (camera receives the action
and is the direct object)
Explanations:
Fundamentals of Grammar
MBA
Test Prep
Page: 127
Exercise 18
1.
are = verb, you = subject, important = predicate
adjective modifying you, too = adverb modifying
important, to help the poor = adverb infinitive phrase
modifying important, poor = direct object to the verbal
to help, the = adjective modifying poor
2.
had upset = verb, child = subject, the = adjective
modifying child, crying = participle modifying child,
everyone = direct object, in the room = prepositional
phrase modifying everyone, in = preposition, room =
object of the preposition, the = adjective modifying
room
3.
4.
5.
6.
7.
8.
9.
jumped = verb, he = subject, across the gap =
prepositional phrase modifying jumped, across =
preposition, gap = object of the preposition, the =
adjective modifying gap, without knowing the distance
= prepositional phrase modifying jumped, without =
preposition, knowing = gerund used as the object of
the preposition, distance = direct object of the verbal
knowing, the = adjective modifying distance
is = verb, exercising = gerund used as the subject,
good = predicate adjective modifying exercising, for
everyone = prepositional phrase modifying good, for
= preposition, everyone = object of the preposition
loves = verb, Jasmine = subject, to dance constantly
= noun infinitive phrase used as a direct object,
constantly = adverb modifying to dance
is = verb, teasing by your friends = gerund phrase
used as the subject, by your friends = prepositional
phrase modifying teasing, by = preposition, friends =
object of the preposition, your = adjective modifying
friends, hard = predicate adjective modifying teasing,
to take = adverb infinitive modifying hard
fled = verb, people = subject, the = adjective modifying
people, fearing reprisal = participial phrase modifying
people, reprisal = direct object to the verbal fearing,
from the city = prepositional phrase modifying fled,
from = preposition, city = object of the preposition, the
= adjective modifying city
is = verb, eating out = gerund phrase used as a subject,
out = adverb modifying the verbal eating, thing =
predicate nominative, the = adjective modifying thing,
to do tonight = adjective infinitive phrase modifying
thing, tonight = adverb modifying to do
do know = verb, I = subject, n’t = adverb modifying do
know, whether/or = correlative conjunction, to tell
him/to keep quiet = noun infinitive phrases used as
direct objects, him = direct object to the verbal to tell,
quiet = adverb modifying to keep
Page: 128
10.
should be done = verb, job = subject, our/next =
adjectives modifying job, to run to the shop = noun
infinitive phrase used as an appositive, to the shop =
prepositional phrase modifying to run, to = preposition,
shop = object of the preposition, the = adjective
modifying shop, quickly = adverb modifying should be
done.
Test 1
1.
Both “to spend” and “spending” could be correct
2.
to have gone
3.
Both “to call” and “calling” are correct.
4.
Answer E. (the participial phrases in A and B correctly
modify Rahul.)
5.
The overloaded car gathered speed slowly.
6.
The opening participial phrase is misplaced because
it is intended to modify him, not the proposal. A possible
revision would be: Espousing a conservative point of
view, he was bothered by the proposal for more
spending on federal social programs.
7.
When I return to Mumbai next year, I will be very
happy.
8.
Rahul goes to school every day.
9.
Sapna is visiting her family right now.
10.
I studied/was studying Economics in 1994.
Test 2
1.
He has spoken/has been speaking French since he
was a child.
2.
Raj had visited many places before he came here.
3.
We saw terrible things back then.
4.
Sometimes I still have dreams like I did twenty years
ago
5.
Japan had never had democracy until 1945
6.
The father will call the family together if he thinks
there is disharmony.
7.
When I was young, I never cooked because my
parents had two servants.
8. d
‘hearing’ is not used in present continuous. It is ‘hear’
even in continuous tense — ‘I hear her’.
MBA
Test Prep
Solution Book-2
9. b
The verb with amounts of money and time periods is
singular. Therefore, ‘twenty thousand’ is required.
5. a
I go. Simple present is correct here rather than
continuous tense to indicate habit.
6. e
‘ate … left’ the parallelism in the sentence is correct.
7. a
Present perfect continuous is correct here.
Test 3
8. e
The sentence in indirect speech is correct.
1. a
‘a’ is the right option. In ‘A’ it should be ‘off’ not ‘of’. In
‘B’ the correct word will be ‘seemed’ not ‘seemingly’.
And in ‘D’ it should be ‘at’ not ‘in’.
9. a
‘with aplomb’ is correct.
‘e’ is the correct option. B and C are correct because
in both cases ‘its’ does not refer to Alan Davey, but to
the company that is not mentioned.
11. a ‘Reach’ doesn’t take a preposition.
10. e ‘ordered for three cups of tea’ is correct.
2. e
3. b
‘b’ is the right option. In ‘A’ it should ‘was’ and not
‘were’; in ‘B’ it should ‘than’ and not ‘then’; and in ‘D’ it
should be ‘conservatives’’ and not ‘conservative’s’.
4. a
In ‘B’ the correct verb should be began because ‘..as
I nosed..’ happens simultaneously with the rest,
therefore past perfect is not appropriate in this case.
5. d
It should be ‘changing ..’
6. e
In ‘C’ it should be ‘attracts us..’ and in ‘E’ it should be
‘combine’ because it talks about a habitual action.
7. c
In C it should be ‘you’ll bump..’; ‘have bump’ is incorrect.
8. a
In ‘A’ it should be ‘helped’ because the sentence is
referring to past action.
9. a
10. c (c) is the correct way of asking a question.
Test 2
1. a
In A the correct helping verb is ‘is’ and not ‘was’,
because the whole sentence is in present tense. In E
instead of ‘gives’ it should be ‘given’ because the act
is in past tense.
2. b
In B it should be ‘...will be’ and not ‘...have to be…’
because the sentence is indicating towards the future
action, besides ‘have’ is used for a plural subject.
3. d
In A it should be ‘rears’ since the sentence is talking
about a routine action.
4. e
In A, it should be only ‘happened’ and not ‘will have
happened’; since the statement is referring to action
that has already taken place.
5. c
In A it should be ‘I have’ because ‘will be followed’ is
an incorrect expression. In D it should be ‘gathered’
and not ‘gathers’, since the statement talks about past
action.
6. a
In A it should be ‘to encompass’. In E it should be
‘dampen’ and not ‘dampens’.
7. e
In E ‘does’ is the incorrect word. It should be ‘do’.
8. d
In ‘B’ it should be ‘..have been’, since the sentence is
in present tense.
9. b
In B the correct tense will be present, so it should be
‘They work…’.
In ‘A’ it should be ‘of making’ and in ‘C’ ‘being’ is not
required.
10. e In ‘A’ it should be ‘did..’ and not ‘do’ because the
statement is in past tense. In ‘C’ it should be ‘like..’ and
not ‘alike’.
Chapter 9
Test 1
1. a
In (a), instead of ‘using’ it should be ‘use’.
2. d
Rather than ‘precipitate’ it should be ‘precipitating’
because the action talked about is in present
continuous.
3. a
Instead of ‘giving’, it should be ‘given’ because ‘the
efforts’ have already been made therefore past tense
should be used.
10. a In A, since ‘may’ is already there, ‘will’ is not required;
and since the idea is related to probability, ‘will’ is too
strong a word to be used here.
4 . d ‘Being’ is not required here.
Explanations:
Fundamentals of Grammar
MBA
Test Prep
Page: 129
Test 3
9.
Here is (change to are)the three books you wanted.
1. b
Option B is the correct answer because ‘by’ should
precede ‘involving’ to give a proper meaning to the
sentence.
10.
Five hundred rupees is/are all I am asking.
'Has been' is the correct expression here since the
context suggests that the work has got over first
now. Thus present perfect should be used here.
1 d.
2. d
3. e
Option E is correct in all respects of the verb usage,
‘becomes’ is the correct usage.
4. c
correct usage of the verb is ‘is’.
5. b
Correct usage is ‘ motivated’ which means to provide
with a motive. In ‘C’ the meaning of the sentence is
entirely changed.
6. c
Option C is correct with respect to the form of the
verb as well the subject, since the sentence is in past
tense.
Test 2
2.
The majority of the Parliament is/are Congressmen.
(The majority of the Parliament… what follows the of
is singular. Hence singular verb)
3. c
(‘who’ refers to what is immediately before it
‘students’, hence ‘are’)
4. d
5.
The original document, as well as subsequent copies,
was/were lost.
6. c
(what follows the ‘of’ is plural) ‘... are going to the
polls.”
7.
Almost all of the magazine is/are devoted to
advertisements.
7. d
Option D is correct because once the results come
out then only analyzing discoveries can the made
thus the use of past tense.
8.
Here is/are Manish and Mandar.
8. e
Only option is E is correct in its usage of the verbs
‘are’ and ‘explore’.
9.
Taxes on interest is/are still deferrable.
Correct usage is ‘have’.
10. b (five rupees is singular, hence is)
9. a
10. d Option D is the correct answer becaue have should
precede ‘held and hold’ which is a phrase here.
Chapter 10
Test 1
Test 3
1. d
In D it should be ‘have’ since it is referring to both
graffiti and its street art cousins.
2. b
The correct verb is ‘come’ because the subject is
‘irreverence’ which is singular.
3. b
In B ‘carries’ is incorrect. It will be ‘carry’ since the
subject i.e. is in plural.
1.
In the newspaper, an interesting article appeared.
2.
Across the road lived her boyfriend.
3.
Around every cloud is a silver lining.
4. d
It says ‘Mr. Lott, along with Speaker …’, therefore in D
it should be ‘was ..’ because the subject is ‘Mr. Lott’.
4.
Neither he nor his brother are (change to is) capable
of such a crime.
5. e
In E it should be ‘that are ...’, since the subject referred
to is ‘techniques’.
The teacher or student is going to appear on stage
first. (No change required)
6. b
It should be ‘there is ...’ since it is referring to ‘good as
well’.
The mother duck, along with all her ducklings, swim
(change to swims) so gracefully.
7. c
In C it should be ‘a hacker could get into…’.
7.
Each of those games is exciting. (No change required).
8. c
In C the correct helping verb is ‘has’ and not ‘have’
since the subject referred to is ‘No one’.
8.
The file, not the documents, were (change to was)
misplaced.
5.
6.
Page: 130
MBA
Test Prep
Solution Book-2
9. e
In E it should be ‘is’ since the referred subject is ‘each’.
10. d In D it should be ‘nail scissors were..’ and in E it
should be ‘channel was..’ because here the subject
is ‘breach’; ‘some’ here has a different contextual
meaning.
Unit – 5
Exercise 5
1.
Either and the first any are pronouns, this and the
second any are adjectives.
2.
The first each and their are adjectives, and the
second each is a pronoun.
3.
This and no one are pronouns.
4.
Both and one are pronouns, and many and neither
are adjectives.
5.
What and neither are pronouns, and your is an
adjective.
Chapter 11
Exercise 1
1.
Heavy, red, fifty.
2.
My, two, my, graduation
Exercise 6
3.
That, small, Indian, the, next, fresh
1.
4.
Little, black, the, well-dressed
2.
Happy, three, frolicking, their, his
5.
Old, wood, several, discarded, packets
3.
Exciting, most
4.
Flooded, terrible
5.
Kaushik, hot, exhausting
Exercise 2
1.
red
2.
Those, brown
3.
Two, what, many, their
4.
Third, city, another
5.
That, good, this
Exercise 3
Our, first, many, strong
Exercise 7
1.
jolly, jollier, jolliest
2.
honest, more honest, most honest
3.
dim, dimmer, dimmest
4.
friendly, friendlier, friendliest
5.
little, less or lesser or littler, least or littlest (Little when
referring to amount uses less, lesser and least; when
referring to size uses littler and littlest.)
1.
New
2.
Stormy, spring, flash
3.
New, UK
Exercise 8
4.
December, dangerous Himachal
1.
interesting, more interesting, most interesting
2.
critical, more critical, most critical
3.
splendid, more splendid, most splendid
4.
delicious, more delicious, most delicious
5.
outstanding, more outstanding, most outstanding
Exercise 4
1.
Soaking
2.
Broken, crying
3.
Great, giving
4.
Laughing
5.
Eager, torn
Explanations:
Fundamentals of Grammar
MBA
Test Prep
Page: 131
Exercise 9
Test 1
1.
many, more, most
1. a
2.
ill, worse, worst
3.
much, more, most
4.
perfect - cannot be compared since there is no
more perfect or most perfect.
In B ‘dried’ is not appropriate, since it cannot be used
as a predicative adjective here and it does not match
‘clean’ and ‘safe’. Similarly C is incorrect because
‘cleaned’ is not appropriate. In D ‘and’ is not required
after ‘dry’ because when using adjectives in a string
we can separate them by commas. Therefore correct
way of writing will be ‘dry, clean and safe’. In E
‘effected’ is not the right adjective to be used.
5.
bad, worse, worst
2. e
In A ‘relenting’ is not the right adjective because it
changes the meaning of the sentence. In B
‘relentlessly’ is incorrect because it is an adverb and
not an adjective and contextually we require an
adjective here to qualify the noun ‘increase’. In C
‘sophisticated’ should come before ‘systems’ since it
is qualifying ‘systems’ and is used as an attributive
adjective. In D ‘sophistication’ is a noun and not an
adjective therefore incorrect.
3. c
In A instead of ‘Kings eye’ it should be King’s eye’
because here possessive case is required. In B the
apostrophe is missing from ‘family’s’. In D instead of
‘family’s’, ‘families’ is given. In E ‘king eye’ is incorrect.
4. a
In B ‘pushy’ should be preceded by ‘and’ because
only two adjectives have been used here. In C ‘which
seems’ should be followed by ‘surprising’ which is an
adjective and it qualifies the unwritten noun in the
sentence that is ‘love’; whereas ‘surprisingly’ is an
adverb which cannot qualify a noun. In D ‘velvety’ is
not an appropriate word because ‘brocade’ is a noun
here thus requires a noun ‘velvet’ as the other word
in the pair, whereas ‘velvety’ is an adjective; though
‘Velvet and Brocade’ has been used here as a ‘Noun
adjunct’ (refer to Practice Exercise 2). In E ‘velvetand-brocade’ should precede ‘backdrop’ rather than
following it since the former is qualifying the latter.
5. c
In A, ‘original’ is an appropriate word because the
context requires an adjective. In B ‘in probing his
opponent’s defences’ should follow ‘duke’s skill lay..’.
In D it should be ‘persuasively’ and not ‘persuasive’
because the former qualifies the verb ‘argues’ thus it
is an adverb whereas the latter is an adjective and
cannot qualify a verb.. In E, ‘when a weakness
opened up..’ should follow ‘quickly and committing
troops’.
6. a
In B ‘enjoyed fervently the support ..’ does not make
any sense; rather it should be ‘the fervent support’
because ‘fervently’ is an adverb and ‘fervent’ is an
adjective. In C ‘of Americans of all colours’ should
come after ‘’fervent support’ or else it will not be clear
as to preposition ‘of’ is referring to what. In D ‘billed as
a contest’ is referring to ‘a fight’ therefore it should
follow ‘a fight’. In E, ‘similar’ is not an appropriate
Exercise 10
1.
worst
2.
hungrier
3.
shortest
4.
best
5.
happiest
Exercise 11
1.
a wife and doctor
2.
a girl and a boy
3.
a green and red
4.
a rock star or a lawyer
5.
a bat and a ball
Exercise 12
1.
this or that
2.
this or that
3.
these or those
4.
these or those
5.
these or those
Page: 132
MBA
Test Prep
Solution Book-2
adjective. Instead ‘same’ should be used. If at all ‘similar’
has to be used it should be preceded by ‘a’ because
it can be any.
7. d
In A rather than ‘irrelevantly’ it should be irrelevant,
since the former is an adverb and an adverb cannot
qualify a noun. In B ‘irrelevance’ is used incorrectly
because it is a noun and the context requires an
adjective. In C ‘publicly’ is incorrect. In E ‘nuisance’
should follow ‘irrelevant’.
8. a
In B a comparative adjective is not required, therefore
‘crisper’ is incorrect. In C ‘admiring’ is incorrect
because an adverb (admirably) is required here which
is qualifying the adjective ‘self- deprecating’. In D ‘both
emotional and professional’ is placed incorrectly. It
should come after ‘..life’. In E ‘self-deprecatingly’ is an
adverb which is inappropriate since ‘vignettes’ is a
noun which requires an adjective.
9. d
In A ‘populist’ is an incorrect adjective since the
statement talks about ‘sovereignty’. In B ‘with the ideal
of popular sovereignty’ is placed incorrectly; it should
come in the end. In C ‘habitual’ is incorrect since it is
an adjective and the sentence requires a noun that is
‘habit’. Again in E ‘popularly’ is an adverb and the
context requires an adjective not an adverb.
10. c A doesn’t require an adverb ‘extraordinarily’; rather it
requires an adjective ‘extraordinary’. In B
‘demographic’ is placed incorrectly before
‘extraordinary’ but the ‘growth’ in this case pertains
to ‘demography’. In D ‘in which settler pressure’ should
precede ‘boosted by extraordinary demographic
growth’, because the second phrase qualifies the
‘..settler pressure..’. In E ‘demography’s’ is incorrect
because it is a noun and an adjective is required out
here since ‘growth’ is a noun here.
5. c
In ‘C’ ‘unsightly’ should precede ‘beggars’ since it is
modifying ‘beggar’.
6. e
In ‘A’, ‘exuberance’ is a noun whereas the context
demands an adjective, therefore it should ‘exuberant’.
In ‘E’ rather than ‘huge’ it should be ‘hugely’ because
‘welcome’ is an adjective out here qualifying ‘start’.
‘Huge’ is also an adjective therefore it cannot qualify
another adjective whereas ‘hugely’ can since it is an
adverb.
7. e
In ‘A’ ‘particularly’ is an adverb and therefore
inappropriate. It should be ‘particular’ which is an
adjective. In ‘E’ ‘extensive network’ is meaningless.
Here ‘network’ will be treated as a verb rather than a
noun and ‘extensive’ will become an adjective by
adding ‘-ly’ to it. Thus it will be ‘network extensively’.
8. b
In ‘B’ ‘most absolute’ is incorrect because ‘absolute’ in
itself is an absolute term and thus cannot have any
superlative form. In ‘D’ ‘red facedly’ is an adverb
therefore inappropriate out here. It should be ‘red
faced’.
9. d
In ‘D’ the appropriate adjective will be ‘irrational’. Since
in ‘E’the statement is talking about ‘all the new sitcoms’
it should be ‘least favorite’. If it were a comparison
between ‘two’ then it would have been ‘less’.
10. d In D it should be impetuousness since the context
demands a noun and not an adjective. In E, it should
be ‘an Elvis Presley wig’. An apostrophe with Presley
will change the meaning. It will mean that the wig
actually belongs to Elvis Presley whereas the context
implies that the wig should be of Elvis Presley style.
Test 3
1. a
In ‘A’ it should be ‘cleverer’ since here a comparison is
made between what the Robots were and what they
are now.
2. c
‘no more longer’ is incorrect because longer will not
take ‘more’.
‘A’ is incorrect because it’s probably not a good idea
to use this construction with an adjective that is already
a negative: “He is less unlucky than his brother,”
although that is not the same thing as saying he is
luckier than his brother. In ‘C’ when we are already
using ‘harder’, ‘more’ is not required.
3. e
In ‘E’ ‘tempting’ should precede ‘special’ since ‘special’
describes the offer, and ‘tempting’ is the nature of the
‘special offer’.
4. d
Rather than the adjective ‘apparent’ an adverb
apparently should by used.
3. a
In ‘B’ ‘more ardent’ is incorrect because ‘ardent’ cannot
have any comparatives or superlatives.
5. a
In ‘A’ ‘harder’ is the correct form of adjective since the
sentence refers to ‘this time’, which implies a
‘comparison’ is being made.
4. a
In C, ‘party’s preference’ is incorrect, rather it should
be ‘party preference’ otherwise the meaning will be
entirely changed. In d, ‘running-down’ is an incorrect
phrase. In should be ‘run- down’.
6. d
In ‘D’ rather ‘solid linking’, it should be ‘solidly linked’.
Test 2
1. e
2. d
In ‘A’ it should be ‘rural poor have..’ and not ‘has’.
Similarly in ‘C’ ‘the rich..are..’; since collective
adjectives are taken as plural.
Explanations:
Fundamentals of Grammar
MBA
Test Prep
Page: 133
7. d
In ‘D’ using more with ‘humility’ and ‘grace’ is incorrect,
since these two words do not require to be used
with comparatives.
Exercise 4
1.
not, correctly (both words modify the verb did do)
8. c
Rather than professionalised it should be professional.
2.
9. a
It should be ‘most reliable’ and not ‘reliable most’.
never, so, deeply (never and deeply modify the
verb was pleased; so modifies deeply telling how
much)
10. b In ‘B’ ‘freer’ is an incorrect; ‘free’ is a more appropriate
adjective.
3.
actually, almost (actually modifies the verb likes;
almost modifies every telling how much)
Chapter 12
4.
recently, n’t (recently modifies the verb found; n’t
modifies the verb would help)
Exercise 1
5.
not, tomorrow, very (not and tomorrow modify the
verb will go; very modifies scary telling how much)
1.
quickly (how)
2.
too (how much), softly (how)
Exercise 5
3.
soon (when), yesterday (when)
1.
n’t, very, quickly, foolishly
4.
then (when), there (where)
2.
suddenly, quietly
5.
why (why), so (how much), often (when)
3.
laughingly, happily, together
Exercise 2
4.
rapidly, accurately
1.
Daily, very
5.
today, tomorrow
2.
totally
Exercise 6
3.
hourly, rapidly
1.
wisely
4.
suddenly
2.
seldom
5.
unusually, outside
3.
often
4.
now, again
5.
now
Exercise 3
1.
completely modifies the adjective exhausted telling
how much, quickly modifies the verb was pulled telling
how, aboard modifies the verb was pulled telling
where
2.
once/twice modify the verb has called telling when
3.
usually/slowly modify the verb work telling how,
rather modifies the adverb slowly telling how much
4.
5.
surely/n’t modify the verb were telling how, very
modifies the predicate adjective expensive telling
how much
Exercise 7
1.
n’t (when/how), often (when), here, (where),
before, (when). They all modify the verb have
stopped.
2.
faithfully (how), carefully (how). They both modify
the verb does.
3.
sometimes (when), always (when), highly (how
much). Sometimes modifies the verb say. Always
modifies the verb have been. Highly modifies the
adjective critical.
4.
yesterday (when), by, (where), once (when), twice
(when). They all modify the verb came.
greedily modifies the verb had telling how, too
modifies the adjective much telling how much
Page: 134
MBA
Test Prep
Solution Book-2
5.
there (where), very (how much), safely (how). There
and safely modify the verb lay. Very modifies the
adverb safely.
5.
6.
today (when), rather (how much). Today modifies
the verb seemed. Rather modifies the adjective
restless.
Exercise 12
Exercise 8
often modifies the verb go, too modifies the adverb
far, and far modifies the verb go
1.
sure
2.
surely
3.
surely
4.
sure
5.
surely
1.
not, now
2.
yesterday, very
3.
never, there, before
4.
too
Exercise 13
5.
always
1.
well
2.
surely
Exercise 9
3.
very
1.
today, tomorrow, technically
4.
good
2.
n’t, alone
5.
really
3.
where, yesterday
6.
badly
4.
too, again
7.
really
5.
finally, away
Test 1
Exercise 10
1.
any
2.
can
3.
ever
4.
have
5.
any
Exercise 11
1.
extremely modifies the adjective tired, and very
modifies the adjective sore
2.
yesterday and hardly modify the verb had completed,
very modifies the adjective hard, and rudely modifies
the verb was interrupted
3.
gradually modifies the verb reached, and before
modifies the verb had climbed
4.
just modifies the adverb now, now modifies the verb
remembered, and rather modifies the adjective
important
Explanations:
Fundamentals of Grammar
1. e
Since none of the sentence/s or part/s thereof have/
has any error.
2. b
In 'A' the adverb to be used is 'doubly'. It doesn't
require an adjective since 'irrational' is an adjective
and 'doubly' as an adverb will qualify it. On the contrary
option C will have an adjective i.e. 'huge' and not
'hugely', which qualifies the noun 'business
opportunity'.
3. e
'B' is correct since 'downright' can be used as both
adjective and adverb.
4. c
In 'D' 'evenly' is incorrect because it changes the
meaning of the sentence. The correct adverb will be
'evenly'.
5. d
In B 'far' is an adverb that is describing the adverb
'worse' and structurally it should come before 'worse'
and not after it. In c since the context requires an
adjective and both 'obsessive' and 'obsessed' are
adjectives. But the right choice here will be 'obsessed'
that means 'having an obsession', whereas
'obsessive' means 'excessive in degree'. Here with
achievement 'obsessed' parents will be appropriate
MBA
Test Prep
Page: 135
6. e
7. d
In D instead of 'humanly' it should be 'human'. There is
no word such as 'humanly'.In E 'nonetheless' should
come after 'robots', because 'nonetheless' is an adverb
and not an adjective. Usually an adjective precedes
the noun.
11. b 'Discounting' is a verb and the context requires a
noun adjunct. Therefore 'discount' is a better option.
In D 'anyways' is not appropriate since it is not used
in standard English usage. The correct adverb will be
'anyway'.
Test 3
8. a
In 'A' 'deep' is an adjective and changes the meaning
of the sentence therefore it should be replaced by an
adverb 'deeply' which qualifies the verb 'held' here.
9. b
In a 'very unique' is incorrect because 'unique' in itself
is a superlative adjective and 'very' will not make it
more so. In c the context demands an adverb since it
describes the verb 'explaining' therefore 'originally'
should replace 'original'.
12. c 'although' is not a required because 'yet' has
already been used in the sentence.
1. b
'a' is incorrect because 'yet' is incorrect in this
sentence because it means 'for the present'; whereas
'still' means 'nevertheless'. Option c is also incorrect
because the placing of 'nevertheless' is incorrect.
Option d inappropriate because 'albeit' means
'notwithstanding' and though is not required, it doesn't
make any sense in the sentence. In 'e' the use of 'yet'
is incorrect.
2. a
In 'a' 'such' which is an adjective,, means 'of the kind'.
In 'b' 'as' with 'such' changes the meaning. In 'c' 'still' is
incorrect because it doesn't make any sense here.
Again in 'd' 'yet' is inappropriately placed after 'but'. In
'e' 'as such' again changes the meaning of the
sentence.
3. e
In 'a' instead of 'more old' it should be 'older'. In 'b' a
comparative form is required therefore 'older' is
required. In c 'reassured' is incorrect because the
context requires an adverb. In d rather than
'reassuring' an adverb is required thus 'reassuringly'
is appropriate.
4. c
In 'a' 'enormous' is used which implies that the meaning
of the sentence will convey that markets have become
enormous, though the stress in the sentence is on
the 'growth'. 'Enormously' is apt because it is an
adverb, describing the growth. In b 'possibly' is an
adverb and inappropriate in the given context. In d
'vigorously' an adverb is incorrectly used. Since the
word is talking about the 'capital markets'- a noun, it
can be qualified only by an adjective. In e 'enormously'
should follow 'grown'.
5. b
In 'a' instead of 'poor' it should be 'poorly' because it is
describing the verb 'enforced'. In 'c' 'wretched' will
be correct because wretchedly is an adverb and
thus inappropriate since the context is dealing with
'China disclosures' which is a noun.
10. b In C 'fully' is incorrect because the context requires
an adjective since it has to describe a noun (account). Therefore 'full' is the right word and it should
come before account.
Test 2
1. c
Correct adverb will be 'parallel'.
2. a
It should be 'bump along..' because using 'alongside'
here will imply that the surface is also bumping along.
3. e
The use of an adverb 'deeply' is incorrect out here.
The context requires the comparative form of an adjective.
4. d
The correct adjective in this part will be 'worse' which
is the comparative form of 'bad' and in this sentence
the comparison is made with 'useless'.
5. d
It should be 'moving quickly' rather than 'moving quick'.
'Quickly' an adverb will modify 'moving' which is a
verb.
6. e
Rather than 'relative' which is an adjective, 'relatively'
(an adverb) will be more appropriate
7. b
In part (b), 'helpful' (an adjective) should be preceded
by 'politically'(an adverb).
8. c
'Rather' itself means 'somewhat', therefore 'somewhat' is not required here.
9. a
It should be 'rapidly' because 'rapid' is an adjective
and cannot qualify a verb i.e. 'developing'.
10. e Rather than an adverb the context requires an
adjective; therefore it should be 'official' and not
'officially'; since it is qualifying the noun 'neglect'.
Page: 136
In d 'poor' is incorrect because it changes the meaning
of the sentence and it is also placed incorrectly. In e
'enforcing' is incorrect because the 'laws' talked about
have already been in past tense.
6. a
In b 'awful' is an adjective and changes the meaning
of the statement. In c 'awfully' should be preceded by
'an', because the former is an adverb which is
qualifying an adjective 'long'. In d 'worried' changes
the meaning. It appears that the 'signs' are 'worried'.
In e 'awfully' is not placed properly.
MBA
Test Prep
Solution Book-2
7. e
In 'a' 'urbane' is inappropriate because though an
adjective its meaning is 'having the polish and suavity'
whereas the correct word should be 'urban'. For
similar reasons 'b' is incorrect. In 'c' 'costing' is
inappropriate. In 'd' 'costlier' should precede 'fuel'.
8. d In 'a' 'fast' is inappropriate because here Yuan is
competing against dollar. The requirement is that of
the comparative form of adjective. In 'b' 'instead of' is
inappropriate. Because instead (an adverb) means
'as a substitute' whereas 'instead of' means 'in lieu
of'. The context here demands 'instead'. In c 'annually'
an adverb is incorrect since it cannot qualify a noun
i.e. rate. Therefore the correct word is 'annual'. In e
'despite' is an incorrect adverb to be used.
8. b
Only option (b) is correct in all respects, the subject is
the type of Indian architecture in central India, an
example of misplaced modifier.
9. a
Only option (a) correctly modifies the subject in
question, an example of misplaced modifier.
10. c The sentence is correct in all respects, the subject is
the new experimental device, a case of misplaced
modifier.
Test 2
1.
'Smaller' doesn't make any sense in option b because
there is no comparison being made. In c 'opposing'
does not make any sense. 'Opposite' will give a
meaning to the sentence. In d 'complete' is an adjective,
whereas the context requires an adverb. Similarly in
e 'utter' is incorrect. It should be 'utterly'.
Looking through the telescope, we could see Venus
clearly in the night sky.
2.
Flying out the window, the papers were grabbed by
him.
OR
He grabbed the papers as they flew out the window.
10. c In 'a' 'seldom' is incorrectly placed. It should be before
'dissuade us'. Similarly in 'b' 'seldom' is placed between
'dissuade' and 'us'. In d also seldom is incorrectly
placed. And in e 'seldomly' i s an incorrect word.
3.
Dhas arrived with the keys as I was waiting outside.
OR
While I was waiting outside with the keys, Dhas
arrived.
OR
While I was waiting outside, Dhas arrived with the
keys.
(The ambiguity in the original is removed in all these
sentences. First and third sentences are preferred.)
4.
While walking on the grass he was bitten by a snake.
5.
I tried calling half a dozen times to tell you about the
Career Launcher Seminar.
OR
I called half a dozen times to tell you about the Career
Launcher Seminar.
(The ambiguity in ‘tried calling’ is eliminated in the second
sentence. Choose this over the first sentence if both
are given as options)
6.
Dhas manged to finish the soup although it was
extremely spicy.
7.
While walking across the street, she was surrounded
by them and was robbed of her purse.
OR
She was surrounded by the and was robbed of
her purse while walking across the street.
8.
In her lunch box, she has some cake (that) she baked.
9.
I really/very glad to be of help to you.
10.
The baby smells very sweet.
9. a
Chapter 13
Test 1
1. b
Only option (b) is correct. Option (b) clearly defines
the reason for going on trial and develops the
consequences in the correct order, a misplaced
modifier.
2. a
Option (a) conveys the meaning of the sentence
completely, defines the efficacy of houses 'with'
central air conditioning.
3. c
Option (c) is correct. The sentence describes the
'mirror with the brass inlaid figures', only option (c)
conveys the meaning correctly.
4. b
Only (b) is the correct option. All other options make it
seem that Federer has blue stripes, a case of
misplaced modifier.
5. d
Only (d) conveys the complete meaning of the
sentence. Here the subject is the rider who was
thrown while jumping the obstacle, and example of
dangling modifier.
6. b
Only option (b) is meaningful and conveys the
complete meaning, an example of misplaced modifier.
7. e
Only option (e) is correct, place the adjectives after
the words they modify, the air is the subject which is
being described here.
Explanations:
Fundamentals of Grammar
MBA
Test Prep
Page: 137
Test 3
Exercise 3
1.
No change required.
1.
of the new book modifies “title”/ about morals modifies
“book”
2.
No change required.
2.
3.
Life in the city is exciting, but life in the countryside is
better.
on the planning commission modifies “work”/ of ideas
and concepts modifies “kinds”
3.
4.
Drive more slowly as work is in progress.
on the west side modifies “houses”/ of town modifies
“side”
5.
No change required.
4.
in the next room modifies “man”
6.
Speak a little more slowly or you will not be
understood.
5.
of the citizens modifies “few”
7.
When he spoke at a press conference on Saturday
night, the Home Minister acknowledged the role played
by the men who subdued the gunman.
Exercise 4
1.
with sharp thorns modifies “tree”/ beside the wall
modifies “grew”
8.
To improve company morale, the consultant
recommended three things.
2.
above the people modifies “soared”/ on the field
modifies “people”
9.
In reviewing the company’s policy, the board identified
three areas of improvement.
3.
of the village modifies “owner”/ past the house
modifies “rode”
10.
Baked, boiled, or fried, potatoes make a welcome
addition to almost any meal.
4.
by its tracks modifies “followed”/ in the snow modifies
either “tracks”(telling which tracks) or “followed”
(telling where we followed it)
5.
over the fence / into some bushes modify “tumbled”
6.
of wreckage modifies “tons”/ after the tsunami
modifies “were left”
7.
over a hill / through a beautiful valley modify
“wound”
Unit – 6
Chapter 14
Exercise 1
1.
on the wall, of the house
2.
in the shade, of the apple tree, of the jobs, for the
day
3.
over the mound, behind the farm, into the street.
4.
but you, from home, with parental permission
5.
around the yard, for miles, except junk
Exercise 5
1.
from her desk modifies “took” telling where / of one
modifies “edition” telling which / of the classics
modifies “one” telling what kind
2.
in the display case modifies “was placed” telling
where / in the corner modifies “case” telling which /
of the library modifies “corner” telling which
3.
of mysteries and detective stories modifies “books”
telling what kind / in the library modifies “are found”
telling where
4.
about magic modifies “story” telling what kind / in
our literature book modifies “appears” telling where
5.
to the solution modifies “clues” telling which / of the
mystery modifies “solution” telling which
Exercise 2
1.
in, with
2.
up, on, through
3.
by, at
4.
with, near
5.
from
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MBA
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Solution Book-2
6.
by my uncle modifies “stories” telling which / about
Sherlock Holmes modifies “stories” telling what kind
5. a
‘Naval gazing’ is the correct expression. It means
indulge in self-absorbed pursuits.
7.
of ancient Pompeii modifies “wall” telling which / by
an ordinary peasant modifies “was discovered” telling
how
6. b
‘Crystal-ball gazing’ is the correct expression. It
involves ‘the use of a crystal ball, a seer stone or
other crystal as a divine tool through which one seeks
to receive visions or information about the future.
7. d
‘Sell himself’ is the right expression. Other phrasal
verbs used in the options have different meanings.
‘Sell off’ means to get rid of by selling, often at reduced
prices. ‘Sell out’ means to betray one’s cause or
colleagues. ‘Sell down’ means to betray the true trust
or faith of.
I cannot put up with this sort of English
8. e
‘Cast a shadow’ is the correct idiom.
2.
I don’t know where she will end up.
9. a
3.
It’s the most curious book I have ever run across.
‘Lo and behold’ is the correct idiom which means look
and behold; an expression of surprise and
amazement.
4.
No change required.
5.
India became free on 15th Aug. 1947.
6.
India is independent for more than 50 years.
Test 3
7.
India is free since 1947.
1.
Cut the pizza into six pieces.
8.
Where did you get this?
2.
No change required.
3.
Where did he go?
9.
If we split it evenly among the three of us, no one will
be unhappy.
4.
Where did you get this?
10.
You can’t just walk into the class without permission.
5.
I will go later.
6.
Cut it into small pieces.
7.
We will arrive on the fourth of next month.
8.
No change required.
9.
No change required.
10.
No change required.
11.
Tanya entered the room.
12.
She dived into the pool.
Test 1
1.
This is the sort of English that I cannot put up with.
OR
Test 2
1. e
2. a
3. c
4. d
The correct idiom is ‘light at the end of the tunnel’,
which means seeing some hope after period of
despair.
‘Step down’ means to resign from a high post or to
reduce, especially in stages. Therefore it is correct
contextually. Other options are incorrect. ‘Step aside’
means “to resign from a post, especially when being
replaced.”
‘Chipped away at’ is the right way of writing. It means
to reduce or make progress on something
incrementally. ‘Chipped in’ means too contribute money
or labor and doing that to ‘that legislation’ is somewhat
illogical.
‘With a straight face’ is the right way of writing, which
means ‘to do or say something confidently without a
trace of guilt’.
Explanations:
Fundamentals of Grammar
10. a ‘Faced down’ is correct because it means to confront
boldly or intimidate.
Test 4
1. e
It means ‘Criticism’.
2. a
‘keeping an eye’ is the correct idiom.
3. b
‘Let loose’ means without any restriction (free).
4. e
‘Shell-shocked’ is the correct idiom which means
‘stunned’.
MBA
Test Prep
Page: 139
5. c
‘At a low ebb’ is the correct expression which means
a bad state.
6. e
‘Wrapping up’ means to complete or stop doing
something.
7. d
‘Middle of the road’ means having a balanced approach,
between the two extremes.
8. a
‘Horse-trading’ takes place in parliament when
members change sides.
9. c
‘Lame duck’ means a person or company that is in
trouble and needs help.
4. b
It means ‘to make an effort to understand and deal
with a problem or situation’. come/get off your high
horse- to stop talking as if you were better or more
clever than other people.
come/go along for the ride- to join in an activity without playing an important part in it. come/go cap in
hand- to ask someone for money or help in a way
which makes you feel ashamed.
come/go down in the world- to have less money and
a worse social position than you had before.
5. d
Cut the Gordian knot means to deal with a difficult
problem in a strong, simple and effective way.
cut some slack- to allow someone to do something
that is not usually allowed, or to treat someone less
severely than is usual.
cut the (umbilical) cord- to stop needing someone
else to look after you and start acting independently
cut the ground from under- to make someone or their
ideas seem less good, especially by doing something
before them or better than them.
Cut the crap!- an impolite way of telling someone to
stop saying things that are not true or not important.
6. b
It implies something that you say which means that
failing to do something when you almost succeeded
is no better than failing very badly.
miss a trick- to not fail to notice and use a good
opportunity. miss out- to fail to use or enjoy an
opportunity miss the boat- to be too late to get
something that you wantmiss the point- to fail to
understand what is important about something
7. a
still and all- despite that. still waters run deepsomething that you say which means people who
say very little often have very interesting and
complicated personalities. sting in the tail- an
unpleasant end to something that began pleasantly,
especially a story or suggestion stink up- to make a
place smell unpleasant; to do something very badly.
stitch in time- something that you say which means it
is better to deal with a problem early before it gets too
bad
8. e
Sticks and stones may break my bones (but words
will never hurt me)- something that you say which
means that people cannot hurt you with bad things
they say or write about you. sink like a stone- to fail
completely
leave no stone unturned- to do everything that you
can in order to achieve something or to find someone
or something.
set in stone- firmly established and very difficult to
change get blood from a stone- to do something very
difficult
9. a
hang the cost/expense - if you say that you will do or
have something and hang the cost, you mean that
you will spend whatever is necessary.
at all costs -if something must be done or avoided at
10. d ‘Out in the cold’ means exposed and vulnerable. In
option A, ‘cord’ is the correct word and not ‘cords’.
Test 5
1. a
In A-Correct usage is 'in' business and in D it should
be 'in' the share market values.
2. e
A is incorrect, should be 'In' January, B is incorrect
should be was found 'in' a street, C is incorrect should
be 'in' his family's care.
3. d
B is incorrect should be 'think of…' C is incorrect
should be 'in a thing…'
4. c
C is incorrect, should be 'outcome of…' , D is incorrect,
should be 'us into'.
5. e
A is incorrect should be 'in your neighbourhood…'
and B is incorrect should be 'across the globe…'.
6. c
A is incorrect, use the preposition 'by' instead of from,
D is incorrect, use the preposition 'for' instead of
through.
7. b
B is incorrect, use 'of the health' instead of 'with…', C
is incorrect use 'in time…' instead of 'from time…'
8. a
A is incorrect, should be 'ultimate form of….', D is
incorrect should be ' appeal as strongly to …'
9. c
A is incorrect should be 'moved by', D is incorrect
should be 'through this…"
10. d A is incorrect, should be 'of' instead of 'from', C is
incorrect, should be 'in earthquakes…'.
Test 6
1. e
It should be ‘blink of ..’.
2. e
It should be winning streak.
3. a
It should be ‘pigeonholed..’.
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MBA
Test Prep
Solution Book-2
all costs, it must be done or avoided whatever
happens.
cost (someone) a pretty penny -to be very expensive.
cost (someone) an arm and a leg (informal) to be very
expensive.
count the cost - to start to understand how badly
something has affected you
10. b copper-bottomed- a copper-bottomed plan,
agreement, or financial arrangement is completely
safe. cordon bleu- cordon bleu cooking is food which
is prepared to the highest standard and a cordon
bleu cook is someone who cooks to a very high
standard corner the market- to become so successful
at selling or making a particular product that almost no
one else sells or makes it. cotton-picking- something
that you say before a noun to express anger corridors
of power- the highest level of government where the
most important decisions are made.
Chapter 15
6.
Either accept our conditions or leave.
7.
We rested until the storm was over and we felt better.
(until is an adverb)
Test 1
1. c
Rather than 'in lieu of', the appropriate conjunction
will be 'because', since the sentence is talking about
the reason that why 'pipeline for public offerings has
dried up.'
2. a
In A the correct conjunction should be 'however' rather
than 'wherever' because the latter doesn't make any
sense in the given context.
3. b
'Moreover' is an incorrect conjunction out here
because the sentence talks about a contradiction.
Therefore 'but ' should replace 'moreover'.
4. c
'Due to' should be replaced by 'since' because the
latter gives the meaning that 'due to the tiny town',
which has no meaning.
5. a
The word 'for' is most often used as a preposition, of
course, but it does serve, on rare occasions, as a
coordinating conjunction. Other options change the
meaning of their respective sentence.
6. e
The correlative conjunction 'not only' is always
coupled with 'but also'.
7. d
The appropriate conjunction will be 'as long as..'
8. c
In A 'as' is following 'though' which is incorrect. In B
'like' has been incorrectly used. 'Like' as a conjunction
means 'as if', therefore doesn't make sense here.
Similarly in D, 'though' and 'yet' are used together thus
rendering the sentence incorrect. Same problem is
with E.
9. a
Other options change the meanings of their respective
sentences.
Exercise 1
1.
and - joining Akbar/I
2.
but - joining slow/strong
3.
or - joining Jatin/Lalit
4.
nor - joining like/appreciate
5.
or - joining You/I
Exercise 2
1.
neither-nor
2.
not only-but also
3.
either-or
4.
whether-or
5.
both-and
Exercise 3
10. e Other options change the meanings of their respective
sentences.
1.
Cox and Kings is open today so we’re going to buy
our tickets to Australia. (so is an adverb)
2.
As he read the letter he laughed. (There is no
conjunction As is an adverb: He laughed as he read
the letter)
1. c
Instead of 'neither' it should be 'nor'.
2. d
The correct conjunction will be 'otherwise'.
3.
So he told me but I didn’t believe him.
3. a
The appropriate conjunction will be 'despite'. With 'yet',
'its size' doesn't make any sense.
4.
She did not reply, nor did she make any gesture.
4. d
5.
We ran from the building when we noticed the time.
(when is an adverb)
In this part the conjunction 'nor' is incorrect. In this
case with 'whether' 'or' should be coupled.
Explanations:
Fundamentals of Grammar
Test 2
MBA
Test Prep
Page: 141
5. d
Here 'such' should be followed by 'as' or else it does
not make any sense.
6. a
Rather than alongside the conjunction 'along with' will
be more appropriate.
7. a
Rather than 'even', the appropriate conjunction here
will be 'although'.
8. e
'So' is an incorrect adverb here. It should be replaced
by 'like'.
9. b
'Wherever' doesn't make any sense here. It should be
replaced by 'however'.
Unit – 7
Chapter 16
Test – 1
1
I like hiking, skiing, and snowboarding. (A verb has
been mentioned with gerunds. So, ‘to snowboard’
should be changed to ‘snowboarding’.)
2.
A low-fat diet, most experts agree, is the best way to
reduce artery blockage and achieve weight loss. (The
writer describes the second benefit of a low-fat diet
as “the achievement of weight loss”—a noun phrase.
Clearly, following the first infinitival phrase with
another to create “to reduce artery blockage and to
achieve weight loss” improves clarity.)
3.
When one takes the CAT, it’s perfectly natural to be a
little nervous, irritable and sweaty. (“Nervous” and
“irritable” are adjectives. “Sweaty” is also an adjective,
but, because it is followed by “palms,” the third element
is a noun phrase modified by an adjective. Parallelism
could be achieved simply by removing “palms.”)
4.
Eating huge meals, snacking between meals, and
exercising too little can lead to obesity. A noun here
has been used with gerunds. So, exercise should be
changed to exercising.)
5.
Our coach is a former champion batsman, but now
he is overpaid, overweight, and over forty. (The fact
of being a champion batsman cannot be mentioned in
the same breath with negative attributes like overpaid,
obese and over fifty.)
6.
Mustaine likes people who have integrity and
character.(The verb ‘have’ should modify both nouns.)
7.
I like editing books more than just reading them.
(Here, an infinitive is paired with a gerund. A gerund
should be paired with a gerund.)
8.
Career Launcher needs teachers who are ambitious,
self-motivated, and dedicated. (Here, a verb form has
been given with one adjective and one noun. The
parallelism can be achieved by converting all three
words into adjectives)
9.
As an artist, he drew, painted, and sculpted. (For the
parallelism to be there, all three activities should be
mentioned as verbs.)
10.
Aggression and melancholy are behaviours that many
steroid-users exhibit.
(In this sentence the parallel elements should both be
nouns that function as the subject of the sentence.
So they have to be in the same grammatical form.)
10. c The appropriate conjunction here will be 'as much
as'.
Test 3
1. b
In A 'like' is an inappropriate adverb. It should be replaced by 'as'. In D rather than 'therefore', 'and' will be
more appropriate.
2. b
In B, 'henceforth' will be replaced by 'while', since
the sentence is talking about the ongoing mood of
pessimism.
3. b
In C, rather than 'then', 'than' should be used because
comparison is being made here. In E 'unless' is inappropriate because it means 'except', whereas the
sentence talks about 'till'. Therefore 'until' will be appropriate.
4. c
In E, 'even' should be followed by 'before' or 'when',
rather than 'if'.
5. d
In C 'though even' is incorrect. Either it should be
'though' or 'even though'. In D 'while' is inappropriate.
It should be replaced by 'and'.
Usage notes: Among some conservatives there is a
traditional objection to the use of though in place of
although as a conjunction. However, the latter (earlier all though) was originally an emphatic form of the
former, and there is nothing in contemporary English
usage to justify such a distinction.
6. d
In E, rather than 'nor' it should be 'or'.
7. e
In B, 'despite' is incorrect because it is always followed by 'of'. It should be replaced by 'instead'. In E,
rather than 'or' should be 'nor' because the previous
sentence has already discussed something in negative therefore 'nor' can appropriately follow it; since
the very first sentence says 'But this is not it'.
8. b
In D 'whether' is inappropriate and should be replaced
by 'if'.
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MBA
Test Prep
Solution Book-2
Test 2
1. e
There is no error in the statement. The first part of the
sentence before hyphen (-) is the main clause and
rest are the ‘elements’ (subordinate clauses) that the
main clause includes. All of them are parallel- ‘balance’,
‘age’ and ‘wearing’. ‘Wearing’ is correct because it is
a noun (gerund) here and not a verb.
2. b
Following the principles of parallelism, ‘to keep’ should
be match by ‘to supply’.
3. c
Again according to the rules of parallelism, in a string
of nouns should have all singulars or all plurals. Since
we have ‘whispers’, ‘tingles’ and ‘shocks’; ‘shout’
should also be in plural.
4. d
Instead of using an infinitive ‘to herd’; a gerund should
be used that is ‘herding’; because there is a string of
gerunds used in the sentence, which are ‘singing’,
‘irritating’ and ‘barking’. Secondly ‘at’ cannot be followed by ‘to’.
5. d
It should be ‘selling’ and not ‘to sell’.
6. d
To make the structure parallel ‘who will benefit’ should
be matched with ‘to what extent’ because ‘and the
extent’ does not have the component of questioning
though the context requires that.
7. d. In D rather than ‘consolidating’ it should be ‘consolidate’ keeping the parallel structure in view because
the verbs should match when in a string.
8. e
9. b
Both should be followed by ‘manufacturers and dealers’ to keep the parallel structure at its best.
All the verbs in the sentence are in simple past; thus
it doesn’t make any sense to use past continuous in
the middle of the sentence. (Opossums- a prehensile-tailed marsupial, of the eastern U.S., the female
having an abdominal pouch in which its young are
carried: noted for the habit of feigning death when in
danger.)
10. e ‘Running’ is a gerund here whereas the requirement
is that of an infinitive to match ‘to have’ and ‘to buy’.
3. b
In C ‘to accept’ is incorrect. It should be replaced by
‘accepting’
4. a
In E ‘the crucible …’ should be preceded by ‘as’, so
that the parallel structure of the statement is intact.
The correlative ideas here should be joined by ‘either
as the …or as the’.
5. b
This tradition began in ancient Rome and continues
into modern times. (‘Begun’ is a participial adjective
while ‘continues’ is an active verb.)
6. b
I acquired my wealth by investing carefully, working
hard and finding a rich father-in-law. (Here, a noun
has been mixed with a pair of gerunds.)
7. e
We can either drive to Shimla or fly to Chennai.
(Correlative conjunctions, here ‘either . . . or’, introduce
clauses that must be parallel.)
8. c
I do not like hot water or cold milk. (Here, unequal
structures are being used to explain equal ideas. Converting ‘milk that is cold’ to ‘cold milk’ rectifies the problem.)
9. e
Please write a long story, a short novel and an epic
poem. (Again, unequal structures are being used to
explain equal ideas. Converting ‘a poem that is epic’ to
‘an epic poem’ rectifies the problem.).
Chapter 17
Exercise 1
1.
Most graciously,
2.
Dear Madam: (a business letter)
3.
B-52, Okhla Industrial Estate, New Delhi-110020?
4.
1 March, 2006, at
5.
(no comma needed - only one part)
6.
(no comma needed - only one part)
7.
Charu Gera, Sr., South Extension, New Delhi
8.
New Delhi, India
Exercise 2
Test 3
1. a
2. d
In C, rather than ‘running’ it should be ‘runs’ since the
context talks about universal truth thus has to be
simple present tense following the rules of parallelism.
Similarly in D, instead of ‘cleared’ it should be ‘to clear’.
In B ‘will have’ is incorrect because it is not parallel
with ‘has launched’ and ‘have fought’. Since the frame
of reference pertains to present perfect, the insertion
of ‘will have’ is totally out of context here.
Explanations:
Fundamentals of Grammar
1.
Football, basketball, track, and tennis require running.
2.
The numbers 8, 16, 32, and 48 are called even
numbers.
3.
Eat, drink, and make merry, for you will soon die.
4.
I like shopping, my friend likes dining, and the family
likes activities.
MBA
Test Prep
Page: 143
5.
Working hard, saving some money, and providing for
a family should be important for a father.
6.
I saw him run up the mountain, jump off the cliff, and
land in a pine tree.
7.
He was from Mahableswar and she was from
Kanyakumari.
8.
She likes to sing, to play the piano, and to read novels.
9.
The search party looked along the road, up the hill,
and down the alleys for clues.
Exercise 5
1.
Chandan is tall; his brother is short.
2.
He knocked several times; no one came to the door.
3.
The siren blew loudly; I rushed to the window; the
police raced past as I looked out.
4.
I waited several hours for you; you did not return; I
became concerned.
5.
My sister loves mysteries; my brother likes technical
manuals.
Exercise 3
Exercise 6
1.
Ila, indeed, is a good mother.
2.
I hope, Vishal, that you don’t go to jail.
3.
My son-in-law Salim will be able to vote in the coming
election. (a closely related appositive or use commas
around Salim if you thought it was a noun of address)
My son-in-law, Salim, will be able to vote in the coming
election.
4.
Oh, Girish, I hope that you, on the other hand, will be
happy with your decision, your move to Europe.
5.
We sat in the shade beneath a broad green tree,
Lara.
6.
It was a lovely, happy, memorable time.
7.
I know, after all, you will be successful.
8.
1.
Since you asked my opinion, I will tell you; and I hope
you will listen well.
2.
Although he is highly qualified, he is not dependable;
and I am afraid to hire him.
3.
Because Preeti is absent a great deal, she has a hard
time keeping up; but she is willing to work overtime.
4.
Although I prefer English, I know that math is important;
and I will work hard in both classes.
5.
When you arrive on the train, take a taxi to the metro
station; or I can meet you at the platform.
Exercise 7
1.
I am looking for the poem “The Path Not Taken”; I need
it tomorrow.
Mr. Dinesh Bobby, the boy next door, has been fighting
with your brother Harish. (Harish is a closely related
appositive)
2.
Ram sings bass; Hari, tenor.
3.
I have visited U.S.A., Australia, Canada and Bhutan.
9.
Of course, we could hear immediately that you, after
all, will be going to Santos, a great city in Brazil.
4.
I will steal, cheat, and lie for you; but I will not kill for
you.
10.
Well, Vishal, I hope to see you, by the way, in Goa on
our return from our vacation, a trip to Mumbai.
5.
There was a sudden noise; everything stopped
immediately.
6.
Although we may need more time, I believe we will be
victorious; and I believe you feel that way, too.
7.
We can trust him implicitly; nevertheless, we should
not be careless.
Exercise 4
1.
“I wish the election was over,” said Farah.
2.
“Will they finish this week?” asked Farheen.
3.
Avinash added, “It is becoming a joke!”
8.
The house looked like what we wanted; on the other
hand, we had not been inside.
4.
“We can now see that every vote counts,” concluded
Nidhi.
9.
“Yes, we know that we should vote every time,”
commented Rajiv.
I had food, clothing, and furniture; but I didn’t have my
family.
10.
He was such a “bore”; I couldn’t stand him.
5.
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MBA
Test Prep
Solution Book-2
Exercise 8
5.
The cow’s udder was cut from jumping the
neighbour’s fence.
6.
Rahul and Sanjeev’s store will be open on Diwali
7.
Everybody else’s help will be appreciated by my
mother’s family.
1.
The boy’s bike is in the back yard.
2.
Gaurav’s car was in the accident yesterday.
3.
Mr. Gupta’s talk was the best yet.
4.
What happened to that horse’s leg?
8.
Just two days’ work will finish this room.
5.
That woman’s umbrella is blowing away in the wind.
9.
Anika’s and Rani’s costumes were the prettiest of
everyone’s.
10.
The women’s and girls’ ages were revealed to
everyone. (could be girl’s)
Exercise 9
1.
These women’s hats are sold in this store.
2.
The children’s party was a great success.
Exercise 12
3.
The mice’s tracks were everywhere in the dust.
1.
we’re it’s you’ve who’s hasn’t
4.
We followed the two deer’s tracks in the snow.
2.
I’ll I’m she’ll she’ll I’ll
5.
The geese’s flight was smooth and graceful.
3.
I’ve we’ll they’re aren’t didn’t
4.
he’s you’ll you’re isn’t hadn’t
Exercise 10
5.
wasn’t haven’t couldn’t we’d they’ll
The men’s and boys’ boots were all mixed together.
(separate ownership)
6.
shouldn’t doesn’t there’s they’ve you’d
2.
Shashi’s mother lives next door to us.
7.
weren’t wouldn’t that’s I’d won’t
3.
The dog’s growl scared the baby in the neighbor’s
yard.
4.
Both Vinod’s and Shyam’s hair is red. (separate
ownership)
5.
Meera’s and Seeta’s mother came to the performance.
(joint ownership)
1.
6.
The babies and the children’s fun ended with the
parents’ return. (joint ownership)
7.
The men’s hoods covered their faces.
8.
The coop was covered with several chickens’
feathers.
9.
I could hardly hear the puppy’s bark.
10.
The wolves’ howls came sharply to the deer’s ears.
Do not confuse the contractions (it’s, who’s, they’re,
you’re) with the possessive pronouns (its, whose,
their, your).
Exercise 13
1.
It’s about time you started looking for your shoes.
2.
They’re coming at about nine for their children.
3.
Its mouth was sore because it’s chewing all the time.
4.
Whose briefcase will you be using for your papers?
5.
You’re going to be late, but who’s going to be on time?
Exercise 14
1.
It used to be that one had to be twenty-one to vote.
2.
When adding thirty-four and forty-two, you get
seventy-six.
3.
One hundred thirty-seven people were killed in that
crash.
4.
The sixty-fourth running of that race was cancelled
due to weather.
5.
Many more privileges come to people who are sixtyfive or older
Exercise 11
1.
Could I buy fifty rupees’ worth of sweets for the
kids?
2.
Somebody’s shoes have been left in the living room.
3.
His shoes are here, but where are yours?
4.
His aunt’s nephew will be on television with Ahuja’s
group.
Explanations:
Fundamentals of Grammar
MBA
Test Prep
Page: 145
Exercise 15
1.
We will invite Sushma—she is the new girl next door—
to our party.
2.
The dog slid on the vinyl—his nails acting like skates—
and crashed into the trash can.
3.
When our stockpile was sold—indeed, dumped for
surplus—all our sales were compromised.
4.
Today has been—but I will not bore you with my
troubles.
Let me tell you about—watch where you are going!
5.
Exercise 16
1.
We fished (or should I say mashed worms) in the
murky pond.
10.
Correct. or colours;
Test 2
1.
“How,” I asked, “can you always be so forgetful?”
2.
The girl who is standing there is his fiancée.
3.
Correct.
4.
Finish your job; it is imperative that you do.
5.
You may, of course, call us anytime you wish.
6.
You signed the contract; consequently you must
provide us with the raw materials.
7.
“Stop it!” I said. “Don’t ever do that again.”
2.
They listened to the teacher’s stories (they were very
dull) which gave some background for the book.
8.
Because of his embezzling, the company went
bankrupt.
3.
Gauri and Prial (you remember them) moved to a new
house last week.
9.
Correct.
10.
4.
Even though he was not qualified (according to his
transcripts), he knew more than most of the others.
Nature lovers will appreciate seeing whales, sea lions,
and pelicans.
5.
Another possibility (the possibilities seem endless)
was suggested by a person at the back of the room.
Test 3
1. c
Uses the correct punctuation marks for distinction.
2. a
Put a colon to introduce a list.
3. d
Only commas are required to separate the item list,
the colon is incorrect here; do not use apostrophes
for sweets, marshmallows and toffees.
Test 1
1.
The girl’s (girls’) vitality and humor were infectious.
2.
New clients’ accounts showed an 11 percent
increase in sales.
3.
These M.D.s’ credentials are excellent. (These M.D.s
– plural)
4. e
Semicolon is used to separate sentences which are
closely connected in thought.
4.
Several M.D.s (Or M.D.’s) agreed that one bacterial
strain caused many of the symptoms.
5. b
No interrogation mark is used after an indirect question, commas are not required.
5.
You asked for forgiveness; he granted it to you.
6. e
6.
We ask, therefore, that you keep this matter
confidential.
Use comma to separate co-ordinate clause in a compound sentence.
7. b
The order was requested six weeks ago; therefore I
expected the shipment to arrive by now.
Use comma after adverbial phrases of absolute construction.
8. c
I need a few items at the store: tissues, a bottle opener,
and some milk.
Use comma to separate words/ phrases/clause
inserted into the body of a sentence.
9. d
Use inverted commas for reported speech and put
commas for short pauses.
7.
8.
9.
I needed only three cards to win, namely, the ten of
hearts, the jack of diamonds, and the king of hearts.
(or win; namely,)
Page: 146
10. b Use commas to mark off phrases in apposition.
MBA
Test Prep
Solution Book-2
Unit – 8
3. c
B is incorrect, the correct phrase is 'akin to' , akins as
word does not exist, C is incorrect, correct usage is
'of' after the word silhouette.
4. b
A is incorrect should be 'stop' instead of stops, error
of subject verb agreement. B is incorrect; the correct
phrase is day-to- day activities.
Chapter 18
Test 1
1. d
Options B, D & E are correct. A is incorrect as the
correct phrase is 'as much as' , C is incorrect as the
correct verb to be used is 'there are..'.
5. c
2. a
Only A& B are correct, C is incorrect because it should
be 'recent months', D is incorrect because it should
be 'at the moment', E is incorrect because it should be
'pants'.
A is incorrect should be 'about' instead of 'over', C is
incorrect should be 'the' public not 'a' public. E is incorrect uses a comparative 'greater', correct usage
is 'great'.
6. c
B is incorrect, use 'dreamt ' instead of 'dream' in keeping
with the tense of the sentence, C is incorrect uses
'which' instead of 'that', which is mostly used for
inanimate objects.
7. d
A is incorrect should be 'resonant' in the adjective
form rather than 'resonance' in the noun form, C is
incorrect, use 'have' after Patriarchs as the correct
form of subject verb agreement.
3. d
C is incorrect because the correct usage is 'their', E is
incorrect an article should precede the word leader,
should be 'a leader'.
4. d
C uses incorrect parallelism, should be 'transporting',
E use the article 'the' before China which is incorrect.
5. b
A is incorrect should be 'Greeks 'since it's addressing
the entire race, D is incorrect, the verb should be
'seek', E is incorrect because it should be 'than' not
then.
8. c
A is incorrect, incorrect usage of the preposition 'upon'
which means on something, something elevated,
correct usage is 'on'. B should be 'scores' instead of
'score', should agree with plural verb.
6. a
A is incorrect, should be 'built', error of tense, C is
incorrect should be 'on Monday' as the deed is already
done.
9. a
D is incorrect, requires an article before 'user's' , E is
incorrect due to the incorrect usage of the phrase
'real time'.
7. e
A is incorrect should be 'in business', C is incorrect
due to incorrect subject verb agreement, and use
'has 'here.
10. e C is incorrect, should be the contraction 'it's' instead
of 'its', D is incorrect, uses the verb 'is' instead of
'are'.
8. e
B has incorrect usage of the pronoun should be 'in'
instead of 'at', C is incorrect should be 'candidates
instead of the singular 'candidate'- should agree with
the verb, E is incorrect because the collective usage
of 'skills; is being discussed here and should agree
with the verb 'are'.
9.c
C is incorrect, should be 'a good ...' the defining article
is missing, D is incorrect should be 'friends' in the
plural.
10. a B is incorrect, should be 'setting' instead of set: error
of parallelism, D is incorrect should be 'said' instead
of 'says' incorrect tense usage, E is incorrect because
the article 'a' is missing before the word negative.
Test 2
1. b
A is incorrect, use the adverb form 'increasingly' here,
in sentence B the correct usage is 'eight times' and in
E the correct form is 'it's' instead of the possessive
form 'its'.
2. a
A is incorrect, use 'between' instead of 'among', D is
incorrect, use the article 'the' before the word 'plant'.
Explanations:
Fundamentals of Grammar
Chapter 19
1.
Eating insects does far less damage. For one thing,
the habit could help to protect crops. Some 30 years
ago the Thai government, struggling to contain a plague
of locusts with pesticides, began encouraging its
citizens to collect and eat the insects. Officials even
distributed recipes for cooking them. Locusts were
not commonly eaten at the time, but they have since
become popular. Today some farmers plant corn just
to attract them. Stir-frying other menaces could help
reduce the use of pesticides.
But insect populations vary with the seasons, and it
is hard to control the amount on offer at a given time.
“There is very little knowledge or appreciation of the
potential for managing and harvesting insects
sustainably,” notes Patrick Durst, a Bangkok-based
senior forestry officer at the FAO. Those looking for a
reliable source of protein might prefer to farm them.
Protein makes up a high proportion of most insects’
weight. That makes them much more efficient at
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Test Prep
Page: 147
converting feed to protein than livestock. For example,
a cow yields only 10lb (4.5kg) of beef for every 100lb
of feed it eats, whereas the same amount of feed
would produce tens times as much cricket.
2.
members. Some (or many) existing members have
been setting a bad example for them.
Nor was the fifth enlargement a simple matter of
countries governed by former dissidents accepting
the democratic embrace of the West. Plenty (or most
or many) of ex-communists smoothly relabelled
themselves and hung on to power across the block.
Brussels is full of talk about “backsliding” to describe
the way that politicians in the new member countries
forgot, or actively undermined, reforms that the EU
demanded during accession negotiations. Corruption
and organised crime blight many of the newcomers.
Parliaments and ministerial suites shelter too many
bad men.
While IT managers are understandably leery about
recommending the iPhone for company employees,
individual owners are going to be sneaking their prized
possession into the office to access e-mail and
corporate information. Bosses, too, who become
enamoured of the iPhone will be insisting the firm
support it. Against their better judgment, many IT
managers will doubtless comply.
No question, Apple’s innovative design has sent a
wake-up call to smartphone makers everywhere.
From tiny Palm, the struggling smartphone pioneer, to
Research In Motion (RIM), the BlackBerry
manufacturer, and the dozen or so handset makers
who build smartphones based on Microsoft’s
Windows Mobile, as well as the 800-pound gorilla of
the business, Nokia—all have been forced to react to
the iPhone’s touch screen, intuitive software and
iconic appeal, not to mention its rapid penetration of
the market.
All this has led some to suggest that enlargement
happened too soon, and that many of these problems
could have been avoided by waiting until the accession
countries were better prepared. This report will argue
the opposite: that enlargement came in the nick of
time. Inside the candidate countries the first victims of
further delay would have been reformers who for
years had been pushing painful changes as vital for
achieving EU membership. Had the public started to
doubt that entry was fairly imminent, the drive for
reforms would have been undermined.
Apple has sold more than 5m iPhones over the past
year. Practically all have been bought by individuals
at their own expense, rather than their employers’.
The iPhone now accounts for 19% of the smartphone
market in the United States. That’s ahead of Palm’s
13%, but still well behind RIM’s 45%.
Over the past few months, competing 3G smartphones
with touch screens and a host of features have been
coming thick and fast. Sprint has started to offer
Samsung’s Instinct, which seeks to trump the iPhone
with a higher download speed, better video, picture
messaging, navigation and applications, plus a battery
that can be removed.
3.
For the existing member countries, three big reasons
would have made enlargement far more difficult if it
had come any later than it did. These can be
summarised as migration, money and Moscow.
4.
The banks’ problems feed back into the credit markets.
For if the banks cannot raise as much capital as they
need, they may feel they have to sell assets. And that
may mean off-loading corporate and asset-backed
bonds, even if the banks must accept fire-sale prices.
The ABX index of asset-backed bonds is below the
level that it sank to in March.
In truth, EU firms have been investing heavily in central
and eastern Europe since soon after the Berlin Wall
came down, and Italy was home to about 350,000
Romanian migrants before Romania joined the union.
Yet public fears about Polish plumbers and other
bogeymen are real enough. Even though German
exporters have flourished by selling to the new
member states, 63% of Germans, according to
Eurobarometer, think that enlargement is making Europe
as a whole less prosperous.
Analysts tend to agree that spreads are much higher
than are needed to compensate investors for the
likely default rate. According to Moody’s, a ratings
agency, the default rate over the past 12 months on
global speculative, or junk, bonds was just 2%; the
only company outside America to walk away from its
debt was a Bulgarian steelmaker. Jim Reid, a credit
strategist at Deutsche Bank, quips that, if the spreads
on investment-grade debt accurately reflect default
rates, everyone will end up living in caves.
Some (0r most) of the newcomers have not helped
their cause since joining. Nasty populists have done
well in elections in several countries, and Romania,
Bulgaria, Slovakia and the Czech Republic have
shown prejudice against the Roma too. But then
prejudice, bad government, corruption and organised
crime are not the exclusive preserve of the new
But spreads often fail to reflect the likelihood of
defaults. This was true during the credit boom, when
they fell to historically low levels, as investors chased
anything with a yield higher than government bonds.
And it is true now because, although an investor
could make money by buying corporate bonds and
Page: 148
MBA
Test Prep
Solution Book-2
1776. He found Hume in good cheer, pressed him on
life after death but could not shake the great man of
his disbelief. The thought of annihilation at death, Hume
told Boswell, caused him no more uneasiness than
the thought of non-existence before birth.
holding them for five years, few are able to follow
such a strategy. The big risk investors face is buying
too soon. Traditional bond managers fear losing their
clients if they have another poor quarter. Hedge-fund
managers may find that the prime brokers who lend
them money will cut off their funding if their positions
deteriorate.
Making a success out of such material requires style
and wit. Simon Critchley’s “The Book of Dead
Philosophers” shows leaden playfulness. His tone
can be portentous or giggly, as if he is unsure who
he is writing for or why. The reader will “die laughing”,
he says. But most of the deaths he describes on his
padded-out list of 190 philosophers are banal, and
virtually none is funny. The real trouble is that Mr
Critchley, a professor at the New School of Social
Research in New York, cannot decide whether he is
writing about philosophers’ own deaths, exemplary
deaths or philosophers’ thoughts on death in general.
Just like last summer, the risk is of a downward spiral,
in which nervous investors sell bonds, driving prices
down, only increasing the general nervousness. The
system is still deleveraging, at least according to the
latest figures from America—which show bank loans
contracting at an annualised rate of 8% over the 13
weeks to June 25th.
To make matters worse, the fundamentals for the
corporate-bond market have worsened since last
August. Expectations for corporate profits are being
revised down, as margins come under pressure from
slower growth and higher commodity prices. Headline
inflation is well above target, so central banks are
unlikely to ride to the rescue with interest-rate cuts;
indeed, many prefer to tighten monetary policy. This
is one sequel where the scriptwriters are going to
have to work extra hard to come up with a happy
ending.
5.
IT came to Seneca by his own hand in a warm bath
after falling foul of the emperor Nero, to Boethius by
order of the Ostrogothic King Theodoric and to
Giordano Bruno for affronting the Inquisition.
Pneumonia saw off both Francis Bacon when he
tried to deep-freeze a chicken with snow and René
Descartes on having to rise before dawn in a Swedish
winter to instruct Queen Christina. In modern times,
Nietzsche’s friend Paul Rée fell off an Alp and Moritz
Schlick, a logical positivist, was shot to death by a
deranged student.
An eye-catching list perhaps. But quite
unrepresentative. Save when at work making
distinctions and arguing relentlessly, philosophers are
much like everyone else. They lead on the whole
worthwhile but humdrum lives. Most die of old age in
their beds. Philosophers’ deaths make an unpromising
theme for a book.
A handful of philosophers, it is true, have died
philosophically, by which people usually mean they
have figured in memorable accounts of exemplary
ends. Socrates was one. Unjustly condemned, he
could have left Athens for exile. But, as Plato recounts,
he chose city over self, virtue over expediency, and
drank the hemlock, surrounded by friends.
Another philosopher who had a philosophical end in
this sense was David Hume. James Boswell, a famous
diarist, visited him as he was dying in Edinburgh in
Explanations:
Fundamentals of Grammar
6.
All of these aspects of Noguchi’s career will be
explored in an exhibition opening Friday at the
Yorkshire Sculpture Park, near Wakefield in England.
The stars of the show are a hundred or so of the
paper and bamboo Akari light sculptures that he began
making in the early 1950s, and that became his bestknown work. Lovely to look at and surprisingly robust,
the Akari lights not only fuse Noguchi’s Japanese and
American influences, but art and design,
craftsmanship and industry. They were also the
catalyst for economic regeneration of a declining
Japanese industry and, last but not least, their
dramatically shaped mulberry-bark paper shades emit
a very beautiful light.
The Akari project came about by chance, after Noguchi
went back to Japan in 1950. By then his father was
dead, and the Japanese welcomed him as a famous
American artist. He visited the city of Gifu, where the
traditional candlelit paper lantern industry was
declining dramatically as more and more Japanese
homes introduced electricity. The mayor asked
Noguchi how to revive it.
Noguchi’s solution was to modernize the old paper
lanterns. Settling in an ancient teahouse with his thenwife, the Japanese movie star Yoshiko Yamaguchi,
he designed a series of lamps powered by electricity,
rather than candles. For the shades, he used the
silky Mino paper that had been made in a nearby
village from locally grown mulberry bark since the
eighth century, but replaced the recently adopted wire
frames with traditional bamboo. The design process
was traditional, too. Noguchi began by making a
wooden mold in the shape of the finished shade and
wound fine strands of bamboo around it. Strips of
Mino paper were glued to the bamboo, and the mold
removed once the glue had dried. A slender metal
structure was designed to hold the bulb and support
the shade, both at the top and the bottom, where it
seemed to float above the floor on spindly legs.
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Test Prep
Page: 149
7.
Like Tempelhof, the Beijing air terminal boasts a
sweeping concourse that evokes the glamour of air
travel while enclosing a surprisingly intimate interior.
But Foster pushes the ideal of mobility to a new
extreme. Guided by twinkling lights embedded in the
terminal’s ceiling, arriving visitors glide up ramped
floors and across broad pedestrian bridges before
spilling out onto the elevated concourse. From there
they can disperse along a fluid network of roads,
trains, subways, canals and parks whose tentacles
extend out through the region.
9.
“This is the big issue we had to deal with all the time,”
he says. “The danger is that by collaborating with the
subjects on the painful retrieval of these memories,
you somehow make them say things. So, yes, there’s
always a risk of something.
This sprawling web has completely reshaped Beijing
since the city was awarded the Olympic Games seven
years ago. It is impossible not to think of the enormous
public works projects built in the United States at
midcentury, when faith in technology’s promise
seemed boundless. Who would have guessed then
that this faith would crumble for Americans, paving
the way for a post-Katrina New Orleans just as the
dream was being reborn in 21st-century China at 10
times the scale?
Yet your sense of marvel at China’s transformation is
easily deflated on the drive from the airport. A banal
landscape of ugly new towers flanks both sides.
Many of those towers are sealed off in gated
compounds, a reflection of the widening disparity
between affluent and poor. Although most of them
were built in the run-up to the Olympics, the poor
quality of construction makes them look decrepit and
decades old.
8.
There may be recession closing in around the world,
but nobody has told the major soccer clubs. This may
be the “closed season” of big European leagues, but
the players, the teams, the organizers are trading at
full pelt.
Any moment now, Ronaldinho, the wonderfully gifted
Brazilian who has obliged Barcelona to sell him
because he has frequented parties more than he has
played games over the past season, will get the
transfer his agent brother has negotiated. When the
bartering stops, AC Milan will announce its acquisition
of Ronaldinho for a “cut price” •15 million, or
$23.9 million.
It will grant his wish to play in the Olympics, and then
will start paying his $10 million a season salary. Milan
will build its attacks through the much coveted Brazilian
trio of Kaká, Ronaldinho and the teenager Pato.
That’s show business. That’s where soccer has
taken over from Hollywood as the industry that trades
the talents of men for the length of contract bartered
between them.
To paraphrase Sam.uel Goldwyn, the owners
sometimes pay too much but consider the players
worth it.
Page: 150
Some might say, of course, that when history loses
its focus on the representation of straight facts, it
ceases to be history and becomes historical fiction.
Empathy, after all, is hardly a neutral subject in history
circles, and has long been held responsible in some
quarters for the dumbing-down of school history
teaching. Is Figes not wary of inviting such charges
to be brought to his own work? Does not the active
engagement of his researchers and himself in the
generation of the sources compromise their findings?
“But you’ve got to remember that, without some kind
of empathy, the retrieval process wouldn’t take place
at all. The important thing is to prepare the ground so
that the subjects can tell their stories in their own
words. Those interviews, in several cases, lasted
days, with many pauses where the discussion
became too painful for the subject. But it’s only when
you start discussing the content of the interview
afterwards that you come to put your own intentions
on the material.
10.
I wasn’t expecting such a tender kiss on our first
date, but April was determined. She had already
signalled her intentions by swimming past me in the
pool, sliding her smooth skin next to mine, flipping
over to give me a tantalising glimpse of her white
tummy. But I knew better than to be flattered. It wasn’t
me she was interested in but the dead sardine in my
hand.
She nuzzled my cheek, her chin as slippery as a wet
mushroom, then reared effortlessly out of the water.
I slipped the fish past her rows of peg-like teeth and
watched her plunge back beneath the surface... April,
if you were in any doubt, is not a mermaid but a fouryear-old dolphin. She lives in a marine park called
Xcaret on Mexico’s Caribbean coast and swimming
with her and her friends in a large sea-pen is just one
of the many grin-provoking magical experiences on
offer within the wild haven of the park.
Xcaret, pronounced Escaray, is a sustainable tourist
development set up in 1990 on a picturesque inlet
about 45 miles south of Cancún. Its far-reaching aim
was to conserve both the flora and fauna of the area
- some 4,000 species can be found on the 80-hectare
site - and to promote the area’s colourful culture and
Mayan heritage. All of which makes for a bizarre
hybrid of serious conservation zone and thrill-seekers’
paradise. It’s what might happen if Mickey Mouse
suddenly became an eco-warrior.
MBA
Test Prep
Solution Book-2
