Pipes And Cisterns Questions & Answers

Question 21

Medium

There are three taps namely A, B, and C are connected to a tank. Tap A can fill a tank in 16 hours and tap C can empty it in 40 hours. Tap A and B is opened to fill the tank with tap C to empty it. The tap is completely filled in 20 hours. In what time tap B can fill the tank completely if opened alone?

A 60 hours

B 72 hours

C 80 hours

D 90 hours

Correct Answer: C

Explanation:

Let tank capacity = LCM(16, 40, 20) = 80 units. Rate of A = 80 / 16 = +5 units/hour. Rate of C = -80 / 40 = -2 units/hour. Net rate when A, B, and C operate together = 80 / 20 = +4 units/hour. Equation: Rate(A) + Rate(B) + Rate(C) = Net Rate => 5 + Rate(B) - 2 = 4 => 3 + Rate(B) = 4 => Rate(B) = 1 unit/hour. Time taken by B alone = 80 / 1 = 80 hours.

Question 22

Easy

One pipe can fill a tank three times as fast as another pipe. If together the two pipes can fill the tank in 36 minutes, then the slower pipe alone will be able to fill the tank in :

A 108 minutes

B 120 minutes

C 144 minutes

D 160 minutes

Correct Answer: C

Explanation:

Let rate of slower pipe = 1 unit/min. Rate of faster pipe = 3 units/min. Combined rate = 1 + 3 = 4 units/min. Total tank capacity = 4 × 36 = 144 units. Time taken by slower pipe alone = 144 / 1 = 144 minutes.

Question 23

Easy

12 pumps working 6 hours a day can empty a completely filled reservoir in 15 days. How many such pumps working 9 hours a day will empty the same reservoir in 12 days?

A 8 pumps

B 10 pumps

C 12 pumps

D 15 pumps

Correct Answer: B

Explanation:

Using the chain rule formula M1 × D1 × H1 = M2 × D2 × H2: Here M1 = 12 pumps, H1 = 6 hours/day, D1 = 15 days. M2 = ?, H2 = 9 hours/day, D2 = 12 days. 12 × 15 × 6 = M2 × 12 × 9 => 1080 = M2 × 108 => M2 = 1080 / 108 = 10 pumps.

Question 24

Medium

Two fill pipes A and B can fill a tank in 15 and 12 hours respectively. Pipe B alone is kept open for ¾ of time and both pipes are kept open for remaining time. In how many hours, the tank will be full?

A 8 hours

B 10 hours

C 12 hours

D 9.5 hours

Correct Answer: B

Explanation:

Let total time required be T hours. Pipe B is open for (3/4)T alone, and both A and B are open for remaining (1/4)T. Thus, Pipe B is open for the entire duration T, while Pipe A is open for (1/4)T hours. Work equation: Rate(B) × T + Rate(A) × (T/4) = 1. (1/12)T + (1/15) × (T/4) = 1 => T/12 + T/60 = 1 => (5T + T)/60 = 1 => 6T / 60 = 1 => T / 10 = 1 => T = 10 hours.

Question 25

Easy

A tap can fill a swimming pool in h hours. What part of the pool is filled in y hours?

A h / y

B y / h

C y * h

D 1 / (y * h)

Correct Answer: B

Explanation:

By unitary method: Part of pool filled in h hours = 1. Part of pool filled in 1 hour = 1/h. Therefore, part of pool filled in y hours = y × (1/h) = y / h.

Pipes and Cistern Practice Questions