__Quantitative Aptitude Practice Questions for IBPS Clerk – Set 9__**1. Kate has some oranges with her. She eats half of her oranges every day. After eating the oranges, she is given two more oranges. After 3 days, she is left with 16 oranges. How many oranges she had in beginning?**

(a) 32

(b) 62

(c) 66

(d) 100

**2. The sum of the ages of a father and his son is 4 times the age of the son. If the average age of the father and the son is 24 years, what is the son’s age?**

(a) 14 years

(b) 16 years

(c) 12 years

(d) Data inadequate

(e) None of these

**3. A circular wire of diameter 28 cm is folded in the shape of a rectangle whose sides one the ratio of 1: 2 find the approximate area enclosed by the rectangle.**

(a) 308 cm

^{2}
(b) 426.3cm

^{2}
(c) 402.8cm

^{2}
(d) 323.3cm

^{2}
(e) 488.32 cm

^{2}**4. The average height of 20 boys is 160 cm. on a particular day, it was found that four students were absent and the average height of remaining 16 boys is 156 cm. find the average height of absent boys?**

(a) 174 cm

(b) 172 cm

(c) 166 cm

(d) 153 cm

(e) 176 cm

**5. A merchant buys Rs. 20,000 worth of goods. On the way 40% of the items are damaged. He is forced to sell them at a 10% loss. What profit % should he make on the rest of the items to make an overall profit of 20%?**

(a) 20% profit

(b) 25% profit

(c) 35% profit

(d) 40% profit

(e) 32% profit

**6. Rs. 430 are divided amongst A, B, C so that 5 times A’s share, six times B’s share and nine times C’s share are all equal. Find C’s share.**

(a) Rs. 130

(b) Rs. 150

(c) Rs. 170

(d) Rs. 175

(e) Rs. 100

**7. Raju borrows an amount of Rs. 5000 from Ravi at 10 per cent per annum and he lends it to Laxman at 12 per cent per annum interest rate. Find the gain of Raju (in rupees) in a period of 3 years is:**

(a) 300

(b) 500

(c) 600

(d) 700

(e) 800

**8. The amounts on a certain principal for 3 years and for 4 years at a certain rate of CI are Rs. 18250.50 and Rs. 20988.10. What is the rate of interest?**

(a) 10%

(b) 12%

(c) 13%

(d) 15%

(e) None of these

**9. Wayne sells rice. He has two varieties of rice, one with a cost price of Rs. 60 per kg and other with a cost price of Rs. 80 per kg. To earn more profit, he mixes both of them in equal amounts and sells them at price of higher valued variety. How much profit does he earn?**

(a) 12.5%

(b) 14.28%

(c) 15%

(d) 16.64%

(e) 18.8%

**10. P bought and sold a chair to Q so that the ratio of selling price to cost price is 7:4. Q sold the chair to R to earn a profit of 20%. R sold it to T at Rs. 5250 and earned a profit of 25%. For how much did P buy the chair?**

(a) Rs. 2000

(b) Rs. 2625

(c) Rs. 2400

(d) Rs. 2780

(e) Cannot be determined

**11. What would be the selling price of an article, if a shopkeeper allows two successive discounts of 10% each on the marked price of Rs. 120 ?**

(a) 96 Rs

(b) 84 Rs

(c) 97.2 Rs

(d) 100 Rs

(e) None of these

**12. 6 table spoons, 9 tea spoons and 12 Spatulas can fill up a glass with water in 27 hours. A table spoon contributes to double the volume that the tea spoon does and a Spatula contributes to half the volume that a tea spoon does. How many Spatulas alone can fill up the same glass in 18 hours?**

(a) 27

(b) 144

(c) 81

(d) 108

(e) None of these

**13. A cistern is filled by two taps A and B, in 4 hours and 6 hours respectively, and is emptied by a waste pipe C in 3 hours. When the cistern is half full, A and B are closed and C is opened; after an hour, B is turned on ; after half an hour more, A is turned on. In what time after C is first opened, does the cistern become full ?**

(a) 6 ½ hrs

(b) 12 ½ hrs

(c) 16 ½ hrs

(d) 20 ½ hrs

(e) None of these

**14. The speed of a boat in still water is 6 km/hr and the speed of the stream is 1.5 km/hr. A man rows to a place at a distance of 22.5km and comes back to the starting point. Find the total time taken by him.**

(a) 8 Hr

(b) 9 Hr

(c) 10 Hr

(d) 11 Hr

(e) None of these

**15. The two trains are coming towards each other and cross a signal pole in 30 seconds and 20 seconds respectively and they cross each other in 24 seconds. Find the ratio of the speed of these two train.**

(a) 2 : 3

(b) 3 : 5

(c) 4 : 5

(d) 3 : 4

(e) None of these

**Solutions:**

**1. D)**Suppose Kate has T oranges in beginning.

She eats half of her
oranges and then gets two more oranges daily.

If she has N oranges on a
day, she will have [(N/2) + 2] at end of day.

Number of oranges after 1
day = (T/2) + 2 = 0.5T + 2

Number of oranges after 2
days = ((0.5T + 2)/2) + 2 = 0.25T + 3

Number of oranges after 3
days = ((0.25T + 3)/2 ) + 2 = 0.125T + 3.5

This should be equal to
16.

⇒ 0.125T + 3.5 = 16

⇒ T = 12.5/0.125 = 100

∴ Kate had 100 oranges in beginning.

**2. C)**Let the age of father and son be x years and y years respectively.

The sum of the ages of a
father and his son is 4 times the age of the son.

So, we can write,

x + y = 4y

⇒ x – 3y = 0 ………..(i)

If the average age of the
father and the son is 24 years then the summation of the age of father and son
= 24 × 2 = 48 years.

So, we can write,

x + y = 48 …………(ii)

Subtracting the eq.(i)
from the eq.(ii), we get,

(x + y) – (x - 3y) = 48 –
0

⇒ x + y – x + 3y = 48 ⇒ 4y = 48 ⇒ y = 48/4 = 12

∴ The son’s age = 12
years.

**3. B)**given that, diameter of a circular wire 28 cm.

then, radius of the
circular wire = diameter/2 = 28/2 = 14 cm

Circumference of wire =
2πr = 2 × 22/7 × 14 = 88cm

Let the length and
breadth of rectangle be x and 2x respectively,

∴ Perimeter of rectangle = 2(x + y) = 2 (x + 2x) = 6x

According to question
perimeter of rectangle would be equal to the circumference of circular wire,

So, 6x = 88

⇒ x = 88/6 = 14.6cm

∴ Length of rectangle x = 14.6 cm

Then breadth of rectangle
2x = 2 × 14.6 = 29.2 cm

∴ Area of rectangle = length × breadth = 14.6 × 29.2 = 426.3 cm

^{2}**4. E)**Given, Average height of 20 boys = 160 cm

∴ Average height of 20 boys = sum of total height of 20
boys/20 = 160 cm

Then, sum of their
heights = 160 × 20 = 3200
cm
…(i)

It is given that, average
height of 16 boys = 156 cm

∴ Average height of 16 boys = sum of total height of 16
boys/16 = 156 cm

Then, sum of their height
= 156 × 16 = 2496
cm
…(ii)

Subtract eq. (ii) from
(i), we get the sum of height of 4 boys:

Sum of height of 4 absent
boys = sum of heights of 20 boys – sum of height of 16 boys

= 3200 – 2496 = 704 cm

Therefore, average height
of 4 boys = 704/4 = 176 cm

**5. D)**40% of items are damaged ⇒ 40% of 20000 worth items are damaged = 0.40 × 20000 = Rs.8000 worth damaged goods and remaining 12000 worth normal goods.

Cost price of damaged
goods = Rs.8000

Loss of 10% is incurred
on damaged goods = 10% of 8000 is loss = Rs. 800 loss.

⇒ Selling price of damaged goods = 8000 – 800 = Rs.7200

Overall profit = 20% ⇒ 20% of 20000 is profit = 0.20 × 20000 = Rs.4000 is overall profit.

Overall selling price of
goods = 20000 + 4000 = Rs.24000

Selling price of damaged
goods + Selling price of normal goods = Overall selling price

⇒ Selling price of normal goods = 24000 – 7200 = Rs.16800

Cost price of normal
goods = Rs.12000

Profit made on normal
goods = 16800 – 12000 = Rs.4800

∴ Profit percentage on normal goods = (4800/12000) × 100 = 40%

**6. E)**Total money which is divided = Rs. 430

As per the given
information:

5 × A’s share = 6 × B’s
share = 9 × C’s share = k

∴ A’s share = k/5

∴ B’s share = k/6

∴ C’s share = k/9

∴ total amount = k/5 + k/6 + k/9

**7. A)**We know that SI = PTR/100

Where, SI = Simple
interest, P = Principal Amount, R = % rate of interest, T = Time in years

According to the given
information,

Raju borrows money from
Ravi, and lends the same amount to Laxman at higher interest rate.

∴ Gain of Raju = Simple Interest earned from Laxman – Simple interest given to Ravi

Here, For both transactions:
P = 5000, T = 3;

Rate of interest: R

_{1}= 12 for Laxman and R_{2}= 10 for Ravi
∴ Gain of Raju = [(P × R1 × T)/100] – [(P × R2 × T)/100] =
[(5000 × 12 × 3)/100] = [5000 × 3 × (12 – 10)]/100 = (5000 × 3 × 2)/100 =
Rs.300

Hence Raju has total gain
of Rs.300

**8. D)**We Know that, A = P (1 + r/100)^t

For 1

^{st}case, A = 18250.50 and t = 3years
18250.50 = P (1 +
r/100)^3 ----------- (i)

For 2

^{nd}case, A = 20988.10 and t = 4years
20988.10 = P (1 +
r/100)^4 ----------- (ii)

Dividing eq (ii) by eq
(i), we get

1.15 = (1 + r/100)^(4 –
3) => 1.15 = 1 + r/100 => r/100 = 0.15 => r = 15%

**9. B)**To make one kg of mixture, Wayne will mix half kg of Rs. 60 per kg rice and half kg of Rs. 80 per kg rice.

Cost price = (1/2) × Rs.
60 + (1/2) × Rs. 80 = Rs. 70

Wayne sells the mixture
at price of higher valued variety.

Selling price = Rs. 80

We know, Selling Price =
Cost Price × (1 + Profit%/100)

=> 80 = 70 × (1 +
PP/100)

⇒ Profit percentage = 100 × (80/70 - 1) = 14.28%

**10. A)**Suppose P bought the chair at Rs. S.

P bought and sold a chair
to Q so that the ratio of selling price to cost price is 7:4.

⇒ P sold the chair at Rs. (7S/4), or Rs. 1.75S.

We know, Selling Price =
Cost Price × (1 + PP/100)

Q sold the chair to R to
earn a profit of 20%.

⇒ Selling price for Q = Cost price for R = 1.75S × (1 + 20/100) = 2.1S

R sold it to T at Rs.
5000 and earned a profit of 25%.

⇒ Selling price for R = Cost price for T = 2.1S × (1 + 25/100) = 2.625S

We are given that T
bought it for Rs. 5250.

⇒ 2.625S = 5250

⇒ S = Rs. 2000

∴ P bought the chair for Rs. 2000.

**11. C)**When two discounts of 10% each are allowed on the marked price of Rs. 120:

⇒ Required selling price = 120 × 90/100 × 90/100 = 972/10 =
97.2

**12. C)**Consider a tea spoon alone can do 1 piece of work (fill up 1 volume) in n hours.

Thus, work done by a tea
spoon in 1 hour = 1/n

∴ Based on given data,

Work done by 1 table
spoon in 1 hour = 2/n,

Work done by 1 Spatula in
1 hour = 1/2n

⇒ Work done by 6 table spoons, 9 tea spoons and 12 Spatulas in
1 hour = (12/n), (9/n) and (6/n) respectively.

⇒ Work done in 27 hours = 27 × (12/n + 9/n + 6/n)

⇒ 27 (12/n + 9/n + 6/n) = 1 => n = 729 ------ (i)

⇒ Work done by 1 Spatula in 1 hour = 1/1458

Let the number of
Spatulas required to complete the work in 18 hours = m

∴ Work done by m Spatulas in 1 hour = 1/ 18

⇒ (1/1458) × m = 1/18

∴ m = 81

Thus, a total of 81
Spatulas are required to do the job in just 18 hours.

**13. B)**Time taken by A and B to fill the tank half = m/4 + m/6

⇒ 5m/12 = 1/2

⇒ m = 6/5 hours = 1.2 hrs

Now, we have half the
tank and 1.2 hours already gone.

Now, for next 1 hour,

⇒ Tank emptied =(1/2) – (1/3)

= 1/6 remaining in the
tank

Now, for next 1/2 hour, B
and C are opened.

Then part of tank
remaining = 1/6 – 1/6 + 1/12 = 1/12

⇒ Now the tank is 1/12 filled and 1.2 + 1 + 0.5 = 2.7 hours
have passed

Now, we have all the 3
taps running for k hours.

⇒ 1/12 + k (1/4 + 1/6 – 1/3) = 1

⇒ 1/12 + k(1/12) = 1

⇒ k = 11

⇒ Hence time after which c is opened = 11 + 1 + 0.5 = 12.5

**14. A)**Upstream speed = speed of boat – speed of current

∴ upstream speed = 4.5 kmph

Downstream speed = speed
of boat + speed of current

∴ downstream speed = 7.5 kmph

Total time = time in
upstream travel + time in downstream travel

But time = distance/speed

∴ total time taken = 22.5/4.5 + 22.5/7.5

∴ total time taken = 8 hours

Thus the total time taken
is 8 hours

**15. A)**Let’sassume that the train which passes the signal pole in 30 seconds runs at speed of x m/s and the other train, which passes the signal in 20 seconds runs at speed of y m/s.

When a train passes a
pole, it travels a distance equal to its own length.

∴ length of the first train = time taken × speed = 30x m

∴ length of the second train = time taken × speed = 20y m

When two trains cross
each other, each of them travels a distance equal to the sum of their lengths,
at relative speed.

If two trains are going
toward each other their relative speed = Sum of their speed = (x + y) m/s

These two train cross
each other in 24 seconds.

∴ While crossing each other, distance traveled = time taken × relative speed

∴ 30x + 20y = 24 × (x + y) = 24x + 24y

⇒ 30x + 20y = 24x + 24y ⇒ 30x – 24x = 24y – 20y ⇒ 6x = 4y ⇒ x/y = 4/6 = 2 : 3