__Quantitative Aptitude Practice Questions for IBPS Clerk – Set 23__**1. P can finish a work in 18 days. Q can finish the same work in 15 days. Q started the work, worked for 10 days and left the job. How many days does P alone need to finish the remaining work?**

a) 13

b) 8

c) 6

d) 5

**2. 40 pens and 80 pencils were purchased for Rs. 520. If the average price of a pencil was Rs. 2.00, find the average price of a pen.**

a) Rs. 10

b) Rs. 11

c) Rs. 9

d) Cannot
be determined

e) None of
the above

**3. The principal is 25 times its simple interest. What are the rate of interest and time if they are equal?**

a) 2 years,
2%

b) 3 years,
3%

c) 4 years,
4%

d) 10
years, 10%

e) None of
these

**4. P sold a pendant to Q at a loss of 20%. Q sold it to R at a loss of 30%. At what price should R sell it to S to earn a profit of Rs. 40?**

a) Rs. 144

b) Rs. 156

c) Rs. 80

d) Rs. 120

e) Cannot
be determined

**5. Ramesh mixes 15 kg of sugar purchased at the rate of Rs. 10 per kg with 25 kg of sugar purchased at the rate of Rs. 12 per kg. At what rate per kg, should Ramesh sell the mixture to get a profit of Rs. 3 per kg?**

a) Rs.
14.00

b) Rs.
12.25

c) Rs.
14.25

d) Rs.
12.50

e) None of
these

**6. By how much the fourth proportional to 26, 143 and 68 will be more than third proportional to 49 and 63?**

a) 198

b) 263

c) 293

d) 374

e) 455

**7. A 150 metre-long train passes a bridge at 72 km per hour in 27 seconds. The length of the bridge is**

a) 150 m

b) 410 m

c) 390 m

d) 540 m

e) 350 m

**8. Gibbs invested an amount of Rs. 4000 in compound interest for 3 years at 10% per annum rate of interest. How much money should he invest in simple interest so as to earn same amount of interest at same rate in 4 years?**

a) Rs. 3000

b) Rs. 3310

c) Rs. 3620

d) Rs. 4420

e) Rs. 3760

**9. A boat takes 30 hours for travelling downstream from point A to point B and coming back to point C midway between A and B. If the velocity of the stream is 5 km/hr and the speed of the boat in still water is 10 km/hr, what is the distance between A and B?**

a) 146km

b) 150km

c) 180km

d) 190km

e) None

**10. Weights of two friends P and Q are in the ratio 4:5. If P’s weight is increased by 10% and total weight of P and Q become 82.8 kg, with an increases of 15%. By what percent did the weight of Q has to be increased?**

a) 19%

b) 22%

c) 17.5%

d) 12.5%

e) None

**Directions (11 – 15): What approximate value should come in place of question mark (?) in the following questions?**

**11. (5 / 6) × [(2 / 9) ÷ (4 / 9)] ÷ (6 / 7) = ?**

a) 0.44

b) 0.32

c) 0.49

d) 1.6

e) 0.35

**12. 15% of 62.5 + 10% of 4.80 = ?**

a) 14.18

b) 15.20

c) 10

d) 16

e) 18

**13. 543.28 ÷ 55 = ?**

a) 4

b) 8

c) 10

d) 12

e) 9

**14. [(3 / 0.8) × (11 / 0.2)] ÷ [(28 / 3) × (21 / 5)] = ?**

a) 3

b) 5

c) 7

d) 9

e) 13

**15. (1.1)^2 + (3.2)^2 + (3.0)^2 = ?**

a) 20

b) 22

c) 24

d) 25

e) 27

**Solutions:**

**1. C)**∵ P can finish a work in 18 days.

∴ Work done
by P in 1 day = 1/18

∵ Q can
finish a work in 15 days.

∴ Work done
by Q in 1 day = 1/15

∵ Q worked
for 10 days.

∴ Work done
by Q in 10 days = 10/15 = 2/3

∴ Remaining
work = 1 – 2/3 = 1/3

Number
of days in which P can finish the remaining work = (1/3)/(1/18) = 6 days

**2. C)**We know,

Average
of quantities = sum of all quantities/no. of quantities

40
pens and 80 pencils were purchased for Rs. 520.

The
average price of a pencil was Rs. 2.00. So, the total price of 80 pencils = Rs.
80 × 2 = Rs. 160

The
total price of 40 pens = Rs. (520 – 160) = Rs. 360

∴ The
average price of a pen = Rs. 360/40 = Rs. 9

**3. A)**Let the simple interest be ‘a’.

∵ principal
is 25 times simple interest, principal = 25a

Let
the time be ‘t’ years. Since the time and the rate of interest are equal,

∴ Rate of
interest = t%

We
know that,

Simple
Interest = (P × R × T)/100

Where,
P = principal Amount, R = % Rate of interest, T = time period in years

Here,
P = 25a; R = t; T = t; SI = a

∴ a = (25a×t×t)/100

⇒ t

^{2}= 4
⇒ t = 2
years

⇒ Rate of
interest = t = 2%

**4. E)**R earns a profit of Rs. 40. To know the price at which R sold to S, we must know the price at which R bought from Q. But we know only the loss percentages at which pendant was sold from P to Q to R. Actual value at which Q sold pendant to R cannot be uniquely found with this data.

**5. C)**C.P. of the 15 kg sugar @ Rs. 10 per kg = 15 × 10 = Rs. 150

C.P
of the 25 kg sugar @ Rs. 12 per kg = 25 × 12 = Rs. 300

Average
cost price of 1 kg mixture = (150+300)/(15+25) = 450/40 = 11.25

∴ Ramesh
sells the mixture to get a profit of Rs. 3 per kg

⇒ Selling
price per kg = 11.25 + 3 = 14.25 per kg

**6. C)**Let Fourth proportional to 26, 143 and 68 be P.

⇒ 26/143 =
68/P

⇒ P = 374

Let
Third proportional to 49 and 63 be Q.

⇒ 49/63 =
63/Q

⇒ Q = 81

∴ Difference
= 374 – 81 = 293

**7. C)**Speed of the train = 72km/hr

Converting
the speed of train into m/sec = 72 × 5/18 = 20m/sec

Time
taken to cross the bridge = 27 sec

Distance
covered by train to cross the bridge = 27 × 20 = 540 m

Length
of the bridge = Distance covered by the train – length of the train

⇒ Length of
the bridge = 540 – 150

⇒ Length of
the bridge = 390 m

Hence,
the length of the bridge is 390 m

**8. B)**Compound Interest = Principal × [(1+R/100)^T−1]

Interest
earned = 4000 × ((1 + (10/100))

^{3}-1) = 4000 × 0.331 = 1324
To
earn same amount of simple interest at 10% per annum rate of interest in 4
years,

Simple
interest = (Principal × Rate × Time)/100

⇒ 1324 =
(Principal × 10 × 4)/100

⇒ Principal
= 132400/40 = 3310

∴ Rs. 3310
should be invested in simple interest.

**9. C)**Downstream speed= 10+5 = 15

Upstream
speed = 10-5 = 5

Now
total time is 30 hours

If
distance between A and B is d, then distance BC = d/2

Now
distance/speed = time, so

d/15
+ (d/2)/5= 30

Solve,
d = 180 km.

**10. A)**

10……………..x

……….15

x-15
: 15-10

Now,
given ratio of P and Q’s weight = 4:5

Hence,
(x-15)/(15-10) = 4/5

x
= 19%.

**11. C)**? = (5 / 6) × [(2 / 9) ÷ (4 / 9)] ÷ (6 / 7)

=
(5 / 6) × [ 1 / 2 ] × (7 / 6) = (35 / 72)

=
0.486 ≈ 0.49

**12. C)**? = [(15 × 62.5) / 100] + [(10 × 4.80) / 100]

=
9.375 + 0.48 = 9.855 ≈ 10

**13. C)**? = 543.28 / 55 = 9.8778 = 10

**14. B)**?= (33 / 0.16) × [5 / (28 × 7)] = (165 / 31.36) = 5.26 ≈ 5

**15. A)**? = 1.21 + 10.24 + 9 = 20.45 ≈ 20