# Pipes and Cisterns Quiz Set 004

### Question 1

Two pipes, A and B, can fill a cistern in 11 and 6 mins respectively. There is a leakage tap that can drain 12 liters of water per minute. If all three of them work together, the tank is filled in 6 minutes. What is the volume of the tank?

A

132 liters.

B

133 liters.

C

131 liters.

D

45 liters.

Soln.
Ans: a

Work done by the leakage in 1 min is \$1/11 + 1/6 - 1/6\$ = \${1/11}\$. This work is equivalent to a volume of 12 liters. So, the total volume is 12 × 11 = 132 liters.

### Question 2

A tank is (3/8)th filled with water. When 68 liters of water are added, it becomes (4/5)th filled. What is the capacity of the tank?

A

160 liters.

B

170 liters.

C

180 liters.

D

190 liters.

Soln.
Ans: a

Let x be the capacity in liters. \${3x}/8 + 68 = {4x}/5\$. Solving, x = 160 liters.

### Question 3

Tap X can fill the tank in 18 mins. Tap Y can empty it in 9 mins. In how many minutes will the tank be emptied if both the taps are opened together when the tank is \$5/17\$th full of water?

A

\$5{5/17}\$ mins.

B

\$6{11/16}\$ mins.

C

\$3{16/19}\$ mins.

D

\$7{8/19}\$ mins.

Soln.
Ans: a

1 filled tank can be emptied in \${18 × 9}/{18 - 9}\$ mins. So 5/17 can be emptied in \${18 × 9}/{18 - 9}\$ × \$5/17\$ = \${90/17}\$, which is same as: \$5{5/17}\$ mins.

### Question 4

Tap X can fill the tank in 16 mins. Tap Y can empty it in 6 mins. In how many minutes will the tank be emptied if both the taps are opened together when the tank is \$6/16\$th full of water?

A

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

B

\$5{3/4}\$ mins.

C

\$1{6/7}\$ mins.

D

\$4{5/7}\$ mins.

Soln.
Ans: a

1 filled tank can be emptied in \${16 × 6}/{16 - 6}\$ mins. So 6/16 can be emptied in \${16 × 6}/{16 - 6}\$ × \$6/16\$ = \${18/5}\$, which is same as: \$3{3/5}\$ mins.

### Question 5

Two pipes, A and B, can fill a bucket in 17 and 9 mins respectively. Both the pipes are opened simultaneously. The bucket is filled in 7 mins if B is turned off after how many minutes:

A

\$5{5/17}\$ mins.

B

\$6{11/16}\$ mins.

C

\$3{16/19}\$ mins.

D

\$7{8/19}\$ 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 \$7/17\$ + \$x/9\$ = 1. Solving, we get x = \${90/17}\$, which is same as: \$5{5/17}\$.