# Pipes and Cisterns Quiz Set 015

### Question 1

Two pipes, A and B, can fill a bucket in 13 and 14 mins respectively. Both the pipes are opened simultaneously. The bucket is filled in 6 mins if B is turned off after how many minutes:

A

\$7{7/13}\$ mins.

B

\$9{1/4}\$ mins.

C

\$5{2/3}\$ mins.

D

\$9{2/15}\$ mins.

Soln.
Ans: a

Let B be closed after it has been filling for x minutes. Work done by pipes A and B should add to 1. So \$6/13\$ + \$x/14\$ = 1. Solving, we get x = \${98/13}\$, which is same as: \$7{7/13}\$.

### Question 2

Two pipes can together fill a cistern in 4 minutes. How long does the slower alone take if the speeds of the pipes are in the ratio 4 : 1?

A

20 mins.

B

21 mins.

C

19 mins.

D

22 mins.

Soln.
Ans: a

Let the time taken by the slower pipe alone be x. Then 4 × \$(1/x + 4/x)\$ = 1. Solving for x, we get x = 4 × 5 = 20 mins.

### Question 3

Two pipes, A and B, can fill a bucket in 19 and 5 mins respectively. Both the pipes are opened simultaneously. The bucket is filled in 11 mins if B is turned off after how many minutes:

A

\$2{2/19}\$ mins.

B

\$3{5/18}\$ mins.

C

1 mins.

D

\$4{13/21}\$ mins.

Soln.
Ans: a

Let B be closed after it has been filling for x minutes. Work done by pipes A and B should add to 1. So \$11/19\$ + \$x/5\$ = 1. Solving, we get x = \${40/19}\$, which is same as: \$2{2/19}\$.

### Question 4

Two taps A and B can fill a tank in 17 and 68 minutes respectively. Both the taps are turned on at the same time. After how many minutes should B be turned off so that the tank can be filled in 14 minutes?

A

12 mins.

B

13 mins.

C

11 mins.

D

15 mins.

Soln.
Ans: a

Let B be closed after x mins. Then sum of works done by A and B = 1. \$14/17 + x/68 = 1\$. Solving, we get x = 12.

### Question 5

Two taps X, Y and Z can fill a tank in 4, 18 and 7 minutes respectively. All the taps are turned on at the same time. After how many minutes is the tank completely filled?

A

\$2{26/113}\$ mins.

B

\$3{29/112}\$ mins.

C

\$1{24/115}\$ mins.

D

\$5{16/115}\$ mins.

Soln.
Ans: a

Let the time be x mins. Then sum of works done by X, Y and Z = 1. \$x/4 + x/18 + x/7 = 1\$. Solving, we get x = \$2{26/113}\$. Or use the shortcut \${abc}/{ab + bc + ca}\$. Another thing, instead of solving the entire calculation, you can keep an eye on the options to find the nearest answer. 