Quantitative Aptitude Practice Papers for IBPS Clerk Mains ...

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Quantitative Aptitude Practice Papers for IBPS Clerk Mains 2019

(Solutions)

S1. Ans.(b)

Sol. No. of cars sold by Audi in 2nd quarter = 42

100× 20000 = 8400

No. of cars sold by Audi in 4th quarter = 8400+3

8× 8400 = 11550

Required sum= 29+38

100× 25000 + 8400 + 11550 = 36700

S2. Ans.(a)

Sol. No. of Cars sold by Ford in 1st quarter =29

100× 25000 = 7250

No. of cars sold by Ford in 4th quarter = 7250+1750 = 9000

Required % = 9000

34000× 100

=450

17= 26

8

17%

S3. Ans.(d)

Sol. No. of cars sold by Audi in 2nd quarter = 42

100× 20000 = 8400

Let total no. of cars sold by Audi in whole year = x

∴8400 =28x

100

x = 30000

No. of cars sold by Audi in 4th quarter = 30000−20000 = 10000

Required no. of cars= 0.34× 20000 + 10000 = 16800

S4. Ans.(e)

Sol. No. of cars sold by ford in 2nd quarter = 38

100× 25000 = 9500

Total no. of cars sold by Audi in 1st and 3rd quarter = (34+24

100) × 20000 = 11600

Required % =11600−9500

11600× 100 = 18.10%

S5. Ans.(c)

Sol. No. of cars sold by audi in quarters 2nd and 3rd = (34+42)

100× 20000

= 15200

no. of cars sold by ford in 4th quarter=12

11×

33

100× 25000 = 9000

no. of cars sold by ford in quarters 3rd and 4th= (33

100× 25000) + 9000

= 8250+9000=17250

Required answer= 17250-15200=2050

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S6. Ans.(d) Sol. Given,

(A + B) =72

5 days

B + C = 72

7 days

ATQ, (A + B)8 days + (B + C)4 days + (C)1 days = Total work 8×5

72+

4×7

72+

1

C= 1

5

9+

7

18+

1

C= 1

17

18+

1

c= 1

1

c= 1 −

17

18

1

c=

1

18

C = 18 days

B =7

72−

1

18

B =7−4

72

B = 24 days

A =5

72−

1

24

A = 36 days Total work = 72 units (LCM of days taken by A, B & C) Efficiency of A = 2 unit/day Efficiency of B = 3 units/day Efficiency of C = 4 units/day

New efficiency of C =4

2 = 2 units/day

Required days =72

(2+3+2)= 10

2

7 days

S7. Ans.(c) Sol. Let the efficiency of one man be M unit/day, one woman be W unit/day and that of one child be C unit/day ATQ, 12 × 10 × M = 15 × 12 × W=20 × 15 × C ⇒ 2M = 3W = 5C Total work = 12 × 10 × M = 120M units. In 5 days, work done by men = (12 × 5 × M) = 60M units. Remaining work = 60M units Now 9 women work for 5 days

Work done by them = 2

3M × 9 × 5 = 30M units

Remaining work to be done by children = 60M – 30M = 30M units This work to be done in 5 days

Per day work by children = 30M

5= 6M units

Required children = 5

2M× 6M = 15

15 children are required to complete the remaining work in 5 days.


S8. Ans.(e)

Sol. Juice obtained from first machine is 90

100× 1000 gm = 900 gm

900 × 1.05 ml = 945 ml

Pulp remain from Ist machine = 1000 – 900 = 100 gm

Amount of juice obtained from second machine = 40

100× 100 = 40gm = 40 × 1.05 ml = 42 ml

Pulp remained from IInd machine = 100 – 40 = 60 gm

Juice obtained by shopkeeper from IIIrd machine

= 16

2

3

100× 60

= 50

300× 60 = 10 gm

Juice obtained = 10 × 1.05 = 10.5 ml

Total juice obtained from 1kg of orange is

= 945 + 42 ml + 10.5 ml

= 997.5 ml

S9. Ans.(b)

Sol. Let initial speed of Aman is x km/hr.

Distance travelled by him in 2 hours is 2x km.

While distance travelled by Rajiv in these 2 hours is 75 × 2 = 150 km

When both of them meet, Aman had travelled a distance of 108 km.

Distance travelled by Aman with 25% increase in his speed = (108 – 2x) km

And his increased speed = x(125

100) =

5x

4 km/hr

If Aman had travelled 108 km, Rajiv had travelled 192 km.

192 – 150 = 42 km after 2 hours

Time taken by Rajiv to travel 42 km

= 42

75 =

14

25 hours

This is equal to time taken by Aman to travel (108 – 2x) km

(108–2x)5

4x

= 14

25

108– 2x =14×5x

25×4

108 = 7

10x + 2x

27

10x = 108 ⇒ x = 40 km/hr.

Therefore increased speed of Aman was = 5

4× 40 = 50 km/hr

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S10. Ans.(a)

Sol. Let length of train – A & train – B be ‘l1’ & ‘l2’ m respectively.

And let speed of train – A & train – B be ‘x’ & ‘y’ m/s respectively.

ATQ,

x+y = 𝑙1+𝑙2

8

8 (x+y) = 𝑙1 + 𝑙2 …(i)

And, 𝑙2

8.4= 𝑦

𝑙2 = 8.4y …(ii)

And, 𝑙1+90

12= 𝑥

𝑙1= 12x –90 …(iii)

On solving (i), (ii) & (III), we get:

8x + 8y = 12x–90 +8.4y

4x+0.4y = 90 …(iv)

Now,

x–y=15×5

18

⇒ 6x-6y=25 …(v)

On solving (iv) & (v), we get:

𝑦 =50

3 𝑚 /𝑠

x = 125

6 m / s

Put value of y in (ii):

𝑙2 = 8.4 ×50

3

𝑙2=140 m

Put value of x in (iii):

𝑙1 = 12 ×125

6 –90 = 160 m

Required ratio = 160

140 = 8 : 7

Solutions (11-15):

Monday Tuesday Wednesday Thursday Friday

Hero 180 150 250 150 180

Bajaj 160 220 200 180 140

Honda 200 200 300 250 200

540 570 750 580 520

S11. Ans.(b)

Sol. 540

750= 18 ∶ 25


S12. Ans.(a)

Sol. Total number of bikes produced by Bajaj from Monday to Friday

= 160+220+200+180+140 = 900

S13. Ans.(e)

Sol. Required average= 200+200+300+250+200

5=

1150

5= 230

S14. Ans.(c)

Sol. No. of bikes produced on Tuesday and Thursday is same i.e. 150

S15. Ans.(c)

Sol. Maximum number of bikes produced = 750 , on Wednesday.

S16. Ans.(d)

Sol. Let total chocolate in box be 10c. Munch chocolate be a and kit-kat chocolate be b .

Probability of selecting Munch chocolate =𝑎

10𝑐=

3

10

a = 3c .(i)

Probability of selecting Kit Kat chocolate =𝑏

10𝑐=

2

5

b = 4c …(ii)

ATQ,

a + b + 6 = 10c

⇒ 3c + 4c + 6 = 10c

⇒ c = 2

So, total chocolate = 20

Required probability = 1 − 6𝐶3+ 8𝐶3+ 6𝐶3

20𝐶3

= 1 −(20+56+20)

1140 =

1044

1140=

87

95

S17. Ans.(b)

Sol. ATQ

For tank A —

ATQ

For tank B —

6 (x – 24) + 3 (x – 24) – 4 (x – 24) = 90

(𝑥 − 24) × 5 = 90

𝑥 = 42 𝑚𝑖𝑛𝑢𝑡𝑒𝑠

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Total quantity of tank B = 90 × 4 = 360 liter

Alternatively (P + Q – M) for 42 minutes, means each pipe for 14 minutes—

All three in 14 minutes

(P + Q – M) = 14 × 6 + 14 × 3 – 14 × 4 = 70 liter

𝐹𝑖𝑙𝑙𝑒𝑑 𝑝𝑜𝑟𝑡𝑖𝑜𝑛 =70

360=

7

36

S18. Ans.(e)

Sol.

ATQ—

(𝐴 + 𝐵)𝑥 + (𝐴 + 𝐵 – 𝐶) (5𝑥+24

5) = 72

7𝑥 + 5 (5𝑥+24

5) = 72

12x = 48

x = 4

(A + B + C) work for

= 4 + 44

5 = 8

4

5𝑑𝑎𝑦𝑠

S19. Ans.(c)

Sol. cost price of article = Rs 600

Let marked price of article be Rs 100x.

ATQ

600 − 100𝑥 × (1 −140

300) = 100𝑥 − (1 −

20

100) − 600

600 −160

3𝑥 = 80𝑥 − 600

400

3𝑥 = 1200

𝑥 = 9

∴ marked price of article = Rs 900

S20. Ans.(e)

Sol. Speed of boat in downstream = 24 = a + b

Where a = speed of boat in still water and b = Speed of stream

Now a =4b

⇒ 24 = 5b ⇒ b= 24

5 and 𝑎 = 4 ×

24

5=

96

5

Let, total distance covered is ‘x’ km

ATQ,

6 =𝑥

2×24+

𝑥×5

2×72

𝑥 =144×6

8= 108 𝑘𝑚

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S21. Ans.(e)

Sol. W = X + 12.5 = Z − 4 ⟹ Z − X = 16.5

60% − 45% = 16.5

100% = 110

Marks of each, X = 49.5, Y = 55, Z = 66, W = 62

So, none of the statements is required

S22. Ans.(b)

Sol. From A and C

Let share of Q & R be Rs. 3x & Rs. 7x respectively

7x − 3x = 100

x = 25

Share of R = 7x = Rs. 175

Clearly, statements A and C together are required to answer

S23. Ans.(d)

Sol. From statements A, B & C

Period of investment of Ramu = 12 months

Period of investment of Shyam = 12 − 2 = 10 months

Profit ratio = (12000 × 12) ∶ (9000 × 10) = 24 ∶ 15

Profit of Shyam = 15

9× 1500 = Rs. 2500

Clearly, all statements are required to answer

S24. Ans.(b)

Sol. From A & B

Let marked price be Rs. 100x

SP = 82

100× 100x = Rs. 82x

CP = 100

125× 100x = Rs. 80x (when no discount was provided)

Cost price of book = 80x

82x× 516.6 = Rs. 504

From A and C

Let marked price be Rs. 100x

SP = 90

100× 100x = Rs. 90x (when only 10% discount provided)

CP = 100

112.5× 90x = Rs. 80x

SP (actual) = 82

100× 100x = Rs. 82x (when 18% discount was provided)

Cost price of book = 80x

82x× 516.6 = Rs. 504

Clearly, statement A and either statement B or C are required to answer


S25. Ans.(b)

Sol. From B & C

Speed of train = 97.5

5= 19.5 ms−1 = 70.2 kmph

Required time = 567

70.2 = 8.07 hours or 485 min

Time of arrival = 7: 20 PM

Clearly, statement B and C together are required to answer

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Quantitative Aptitude Practice Papers for IBPS Clerk Mains ...

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