# Pipes and Cisterns Quiz Set 019

### Question 1

Tap M can fill a cistern in 18 mins. And, a tap N can empty it in 11 mins. In how many minutes will the cistern be emptied if both the taps are opened together when the tank is \$13/17\$th already empty?

A

\$6{78/119}\$ mins.

B

\$7{85/118}\$ mins.

C

\$5{68/121}\$ mins.

D

\$9{60/121}\$ mins.

Soln.
Ans: a

1 filled cistern can be emptied in \${18 × 11}/{18 - 11}\$ mins. So \$1 - 13/17\$ = \$4/17\$ filled cistern can be emptied in \${18 × 11}/{18 - 11}\$ × \$4/17\$ = \${792/119}\$, which is same as: \$6{78/119}\$ mins.

### Question 2

A tank is filled in 13 minutes by three taps running together. Tap A is twice as fast as tap B, and tap B is twice as fast as tap C. How much time will tap A take to fill the tank?

A

91 mins.

B

92 mins.

C

90 mins.

D

93 mins.

Soln.
Ans: a

Let the time taken by tap A be x mins. Then 13 minutes work of all the taps should add to 1. So we have, \$13 × 1/x + 13 × 2/x + 13 × 4/x\$ = 1, which is same as \$13 × 7/x\$ = 1. Solving, we get x = 91 mins.

### Question 3

Pipe A can fill a cistern in 30 minutes, while the pipe B can fill it in 20 minutes. They are alternately open for 1 minute. How long will it take the cistern to fill completely?

A

24 mins.

B

25 mins.

C

23 mins.

D

26 mins.

Soln.
Ans: a

Let the total time taken be 2x minutes. Both X and Y run for x mins. So \$(x/20 + x/30)\$ = 1. Solving for x, we get x = 12, which gives 2x = 24.

### Question 4

A tank is filled in 6 minutes by three taps running together. Times taken by the three taps independently are in an AP[Arithmetic Progression], whose first term is a and common difference d. Then, a and d satisfy the relation?

A

a3 - 18a2 - ad2 + 6d2 = 0.

B

a3 - 12a2 + ad2 + 6d2 = 0.

C

a3 - 6a2 - ad2 + 6d2 = 0.

D

a3 - 30a2 + ad2 + 6d2 = 0.

Soln.
Ans: a

Let the times taken by the three taps be a - d, a and a + d. Then 6 minutes work of all the taps should add to 1. So we have, \$6 × 1/{a - d} + 6 × 1/a + 6 × 1/{a + d}\$ = 1, which is same as a3 - 18a2 - ad2 + 6d2 = 0.

### Question 5

A tank is filled in 3 minutes by three taps running together. Times taken by the three taps to independently fill the tank are in an AP[Arithmetic Progression]. If the first tap is a leakage tap and the second tap takes 1 minute to fill the tank, then, the common difference of the AP can be?

A

2.

B

3.

C

1.

D

4.

Soln.
Ans: a

Let the times taken by the three taps be 1 - d, 1 and 1 + d. The time taken by the first tap will be negative because it is a leakage tap. Then 3 minutes work of all the taps should add to 1. So we have, 3 × \$(1/{1 - d} + 1/1 + 1/{1 + d})\$ = 1, which is same as \$2/{1 - d^2} + 1\$ = \${1/3}\$. Solving we get d = ±2.