Pipes and Cisterns Quiz Set 012

Question 1

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 2

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 3

A bucket can be filled by a tap in 2 minutes. Another tap on the same bucket can empty it in 12 mins. How long will it take to fill the bucket if both the taps are opened together?

A

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

B

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

C

1 mins.

D

\$3{6/7}\$ mins.

Soln.
Ans: a

Net part filling in one hour is \$1/x - 1/y\$ = \$(y - x)/(xy)\$. So complete filling occurs in \$(xy)/(y - x)\$ = \${2 × 12}/{12 - 2}\$ = \${12/5}\$, which is same as: \$2{2/5}\$ mins.

Question 4

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}\$.

Question 5

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. 