Hello Aspirants. Welcome to Online Quantitative Aptitude Section with explanation in AffairsCloud.com. Here we are creating question sample in **Pipes & Cistern **which is common for all competitive exams. We have included Some questions that are repeatedly asked in bank exams !!!

**Two taps can separately fill the tank in 10m and 15min respectively. They fill the tank in 12 minutes when a third pipe which empties the tank is also opened. What is the time taken by the third pipe to empty the whole tank?**

A) 14 minutes

B) 15 minutes

C) 12 minutes

D) 20 minutes

E) 16 minutes

**C) 12 minutes**

Explanation:

1/10 + 1/15 – 1/x = 1/12

Solve, x = 12

**Two pipes A and B can fill a tank in 12 hours and 15 hours respectively. If they are opened on alternate hours with pipe A opened first, then in how many hours the tank will be full?**

A) 13 hrs

B) 14 1/2 hrs

C) 12 hrs

D) 12 1/2 hrs

E) 10 2/3 hrs

**D) 12 1/2 hrs**

Explanation:

A = 12 hours, B = 15 hours

Total work = LCM(12,15) = 60

So efficiency of A = 60/12 = 5, efficiency of B = 60/15 = 4

2 hrs work of (A+B) = 5+4 = 9

2*6(12) hours work of (A+B) = 9*6 = 54

So remaining work = 60-54 = 6

Now A’s turn at 13th hour, he will do remaining work(6) in 6/12 hr

So total 12 1/2 hrs

**Pipes P and Q can fill the tank in 24 minutes and 32 minutes respectively. Both piped are opened together. To have the tank full in 18 minutes, after how many minutes the pipe P must be closed?**

A) 22 minutes

B) 21 minutes

C) 15 minutes

D) 12.5 minutes

E) 10.5 minutes

**E) 10.5 minutes**

Explanation:

P is to be closed before 18 minutes, let it is closed after x minutes, then Q worked for all 18 minutes. So,

(1/24)*x + (1/32)*18 = 1

Solve, x = 10.5**Three pipes, A, B and C are opened to fill a tank such that A and B cam fill the tank alone in 36 min. and 45 min. respectively and C can empty it in 30 min. After 6 minutes the emptying pipe is closed. In how many minutes the tank will be full in this way?**

A) 20

B) 25

C) 18

D) 24

E) 30

**D) 24**

Explanation:

Let the tank full in x minutes, then A and B opened for x minutes and C for 6 minutes.

(1/36 + 1/45)*x – (1/30)*6 = 1

(1/20)*x = 6/5

Solve, x = 24

**A and B are pipes such that A can empty the tank in 60 minutes and B can fill in 30 minutes. The tank is full of water and pipe A is opened. If after 18 minutes, pipe B is also opened, then in how much total time the tank will be full again?**

A) 32 minutes

B) 29 minutes

C) 36 minutes

D) 23 minutes

E) 18 minutes

**B) 36 minutes**

Explanation:

Emptying pipe A is opened first for 18 minutes, so in 18 minutes the part of tank it has emptied is (1/60)*18 = 9/30

Now filling pipe is also opened, now since only 9/30 of the tank is empty so 9/30 is only to be filled by both pipes, let it take now x minutes, so

(1/30 – 1/60)*x = 9/30

Solve, x= 18

So total = 18+18 = 36 minutes [total time is asked – 18 minutes when emptyimh pipe was only opened, 18 minutes when both were operating.]**Two pipes A and B can alone fill a tank in 20 minutes and 30 minutes respectively. But due to a leak at the bottom of tank, it took 3 more minutes to fill the tank. In how many hours, the leak can alone empty the full tank?**

A) 60

B) 30

C) 48

D) 56

E) 72

**A) 60**

Explanation:

A and B can fill tank in (1/20 + 1/30) = 1/12 so 12 minutes

But it took 3 more minutes, this means the tank got full in 12+3 = 15 minutes

So (1/20 + 1/30 – 1/x) = 1/15

Solve, x = 60**Pipes A and B can fill a cistern in 15 hours together. But if these pipes operate separately A takes 40 hours less than B to fill the tank. In how many hours the pipe A will fill the cistern working alone?**

A) 60

B) 20

C) 40

D) 15

E) 25

**B) 20**

Explanation:

Let A takes x hours, then B = (x+40) hours

1/x + 1/(x+40) = 1/15

Solve, x = 20**Three pipes A, B and C can fill the cistern in 10, 12, and 15 hours respectively. In how much time the cistern will be full if A is operated for the whole time and B and C are operated alternately which B being first?**

A) 10 hours 32 minutes

B) 6 hours

C) 5 1/2 hours

D) 5 7/10 hours

E) 9 2/11 hours

**D) 5 7/10 hours**

Explanation:

In first hour, part of cistern filled is (1/10 + 1/12) = 11/60

In second hour, part of cistern filled is (1/10 + 1/15) = 1/6

So in 2 hours, part of cistern filled is 11/60 + 10/60 = 21/60 = 7/20

now in 2*2 (4) hours, part of cistern filled is (7/20)*2 = 14/20 = 7/10

now in the 5th hour, A+B’s turn which fill 11/60 in that hour, but the cistern remaining to be filled is (1 – 7/10) = 3/10, since 3/10 is more than 11/60, so after 5th hour remaining part to be filled is 3/10 – 11/60 = 7/60

now in 6th hour, (A+C)’s turn, it will fill remaining 7/60 in (7/60)*(6/1) = 7/10

so total 5 7/10 hours**A cistern is 1/4th full. Two pipes which fill the cistern in 15 minutes and 20 minutes respectively are opened simultaneously. After 5 minutes, a third pipe which empties the full cistern in 30 minutes is also opened. In how many minutes the cistern will be full?**

A) 6 1/2

B) 2

C) 5

D) 7

E) 8

**D) 7**

Explanation:

Since 1/4th is already filled, 3/4th is to filled now.

So

(1/15 + 1/20)*(5+x) – (1/30)*x = 3/4

(7/60)*5 + (7/60 – 1/30)*x = 3/4

(5/60)*x = 2/12

Solve, x = 2 mins

So total 7 minutes**Pipes A, B and C which fill the tank together in 6 hours are opened for 2 hours after which pipe C was closed. Find the number of hours taken by pipe C to fill the tank if the remaining tank is filled in 7 hours.**

A) 16

B) 14

C) 20

D) 22

E) Cannot be determined

**B) 14**

Explanation:

1/A + 1/B + 1/C = 1/6

Now given that first all open for 2 hours, then C closed and A+B completes in 7 hours, so

(1/A + 1/B + 1/C) *2 + (1/A + 1/B)*7 = 1

Put 1/A + 1/B = 1/6 – 1/C

(1/6 – 1/C + 1/C) *2 + (1/6 – 1/C)*7 = 1

2/6 + 7/6 – 7/C = 1

Solve, C = 14

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