# Pipes and Cisterns Quiz Set 008

### Question 1

A tank is filled in \$1{2/13}\$ 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

4.

B

5.

C

3.

D

6.

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 \${15/13}\$ minutes work of all the taps should add to 1. So we have, \${15/13}\$ × \$(1/{1 - d} + 1/1 + 1/{1 + d})\$ = 1, which is same as \$2/{1 - d^2} + 1\$ = \${13/15}\$. Solving we get d = ±4.

### Question 2

A tank is filled in 13 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 - 39a2 - ad2 + 13d2 = 0.

B

a3 - 26a2 + ad2 + 13d2 = 0.

C

a3 - 13a2 - ad2 + 13d2 = 0.

D

a3 - 65a2 + ad2 + 13d2 = 0.

Soln.
Ans: a

Let the times taken by the three taps be a - d, a and a + d. Then 13 minutes work of all the taps should add to 1. So we have, \$13 × 1/{a - d} + 13 × 1/a + 13 × 1/{a + d}\$ = 1, which is same as a3 - 39a2 - ad2 + 13d2 = 0.

### Question 3

A city tanker is filled by two large pipes, X and Y, together in 42 and 28 minutes respectively. On a certain day, pipe Y is used for first half of the time, and both X and Y are used for the second half. How many minutes does it take to fill the tank?

A

21 mins.

B

22 mins.

C

20 mins.

D

23 mins.

Soln.
Ans: a

Let the time taken be x. Y is running for x mins, and X for x/2. So \$(x/28 + x/{2 × 42})\$ = 1. Solving for x, we get x = 21 mins.

### Question 4

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 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.