# Pipes and Cisterns Quiz Set 006

### Question 1

A tank is filled in \$1{2/33}\$ 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

6.

B

7.

C

5.

D

8.

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 \${35/33}\$ minutes work of all the taps should add to 1. So we have, \${35/33}\$ × \$(1/{1 - d} + 1/1 + 1/{1 + d})\$ = 1, which is same as \$2/{1 - d^2} + 1\$ = \${33/35}\$. Solving we get d = ±6.

### Question 2

A tank is (2/5)th filled with water. When 44 liters of water are added, it becomes (8/9)th filled. What is the capacity of the tank?

A

90 liters.

B

100 liters.

C

110 liters.

D

120 liters.

Soln.
Ans: a

Let x be the capacity in liters. \${2x}/5 + 44 = {8x}/9\$. Solving, x = 90 liters.

### Question 3

One tap can fill a tank 2 times faster than the other. If they together fill it in 9 minutes, how much time does the slower alone take to fill the tank?

A

27 mins.

B

3 mins.

C

4 mins.

D

5 mins.

Soln.
Ans: a

Let the one minute work of the taps be 1/x and 2/x. We have \$1/x + 2/x = 1/9\$, which gives x = 3 × 9 = 27 mins.

### Question 4

A tank is filled in 11 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 - 33a2 - ad2 + 11d2 = 0.

B

a3 - 22a2 + ad2 + 11d2 = 0.

C

a3 - 11a2 - ad2 + 11d2 = 0.

D

a3 - 55a2 + ad2 + 11d2 = 0.

Soln.
Ans: a

Let the times taken by the three taps be a - d, a and a + d. Then 11 minutes work of all the taps should add to 1. So we have, \$11 × 1/{a - d} + 11 × 1/a + 11 × 1/{a + d}\$ = 1, which is same as a3 - 33a2 - ad2 + 11d2 = 0.

### Question 5

Tap X can fill the tank in 11 mins. Tap Y can empty it in 5 mins. In how many minutes will the tank be emptied if both the taps are opened together when the tank is \$8/10\$th full of water?

A

\$7{1/3}\$ mins.

B

\$12{1/2}\$ mins.

C

\$6{1/3}\$ mins.

D

\$6{1/5}\$ mins.

Soln.
Ans: a

1 filled tank can be emptied in \${11 × 5}/{11 - 5}\$ mins. So 8/10 can be emptied in \${11 × 5}/{11 - 5}\$ × \$8/10\$ = \${22/3}\$, which is same as: \$7{1/3}\$ mins. 