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Pipes & Cistern Questions and Answers

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11.

A cistern is normally filled in 8 hours but takes another 2 hours longer to fill because of a leak in its bottom. If the cistern is full, the leak will empty it in

Exam Year(s)

ISRO Assistant - 2023,

Answer: D

The leak would empty the full cistern in 40 hours.

Without the leak the cistern fills in 8 hours; with it, in 10. The leak's rate is the difference between the two filling rates:

\[ \frac{1}{8} - \frac{1}{10} = \frac{5 - 4}{40} = \frac{1}{40} \]

The leak empties 1/40 of the cistern per hour, so 40 hours to empty it. 16, 20 and 25 hours do not satisfy this; for example, a 20-hour leak would give a net rate of 1/8 - 1/20 = 3/40 and a filling time of 13.3 hours.

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12.

Three pipes A, B and C can fill a tank in 6 hours. After working at it together for 2 hours, C is closed and A and B can fill the remaining part in 7 hours. The number of hours taken by C alone to fill the tank is

Exam Year(s)

MPPSC State Service - 2022, MPPSC Forest Service - 2022,

Answer: C

C alone takes 14 hours.

Working together, A, B and C fill 1/6 of the tank an hour, so in 2 hours they fill 1/3, leaving 2/3 for A and B:

\[ \begin{aligned} A + B &= \frac{2/3}{7} = \frac{2}{21} \text{ per hour} \\ C &= \frac{1}{6} - \frac{2}{21} = \frac{7 - 4}{42} \\ &= \frac{3}{42} = \frac{1}{14} \text{ per hour} \end{aligned} \]

So C alone fills the tank in 14 hours. Check: A and B alone would take 21/2 = 10.5 hours, and 1/10.5 + 1/14 = 1/6, the combined rate given. 10, 12 and 16 do not satisfy this.

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13.

A pipe can fill a tank in 6 hours, while another can empty it in 9 hours. If both are opened together, in how many hours will the tank be filled ?

Exam Year(s)

Tripura Public Service Commission - 2026,

Answer: A

The tank will be filled in 18 hours.

Work out how much of the tank each pipe handles in one hour. The filling pipe adds 1/6 of the tank and the emptying pipe removes 1/9, so the net filling per hour is:

\[ \frac{1}{6} - \frac{1}{9} = \frac{3 - 2}{18} = \frac{1}{18} \]

One-eighteenth of the tank fills each hour, so the whole tank takes 18 hours.

The trap is to add the rates (which gives 3.6 hours) or to subtract the times (which gives 3 hours); neither is offered. 12, 10 and 15 hours would each need a faster net rate than 1/18.

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14.

12 buckets of water fill a tank when the capacity of each tank is 13.5 litres. How many buckets will be needed to fill the same tank, if the capacity of each bucket is 9 litres?

Answer: C

Capacity of the tank = (12*13.5) litres

= 162 litres

Capacity of each bucket = 9 litres.

So we can get answer by dividing total capacity of tank by total capacity of bucket.

Number of buckets needed = (162/9) = 18 buckets

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15.

Three pipes A, B and C are connected to a tank. Working alone, they require 10 hours, 20 hours and 30 hours respectively. After some time, A is closed and after another 2 hours, B is also closed. C works for another 14 hours so that the tank gets filled completely. Find the time (in hours) after which pipe A was closed.

Answer: C

Let the capacity of the tank be LCM (10, 20, 30) = 60
=> Efficiency of pipe A = 60 / 10 = 6 units / hour
=> Efficiency of pipe B = 60 / 20 = 3 units / hour
=> Efficiency of pipe C = 60 / 30 = 2 units / hour
Now, all three work for some time, say ‘t’ hours.
So, B and C work for 2 more hours after ‘t’ hours and then, C works for another 14 hours.
=> Combined efficiency of pipe A, pipe B and pipe C = 11 units / hour
=> Combined efficiency of pipe B and pipe C = 5 units / hour
 
So, we have 11 x t + 5 x 2 + 14 x 2 = 60
=> 11 t + 10 + 28 = 60
=> 11 t = 60 - 38
=> 11 t = 22
=> t = 2
 
Therefore, A was closed after 2 hours.

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16.

A pump can fill a tank with water in 2 hours. Because of a leak, it took \( 2\tfrac{1}{3} \) hours to fill the tank. The leak can drain all the water of the tank in:

Answer: D

Work in rates: a pipe that fills a tank in \( n \) units of time fills \( \tfrac{1}{n} \) of it per unit, and rates add (an emptying pipe counting negative).

Without the leak the pump fills the tank at \( \tfrac{1}{2} \) per hour; with it, the effective rate is \( \tfrac{3}{7} \) per hour.

\[ \frac{1}{2} - \frac{1}{L} = \frac{1}{7/3} = \frac{3}{7} \]

\[ \frac{1}{L} = \frac{1}{2} - \frac{3}{7} = \frac{7 - 6}{14} = \frac{1}{14} \]

\[ L = 14 \text{ hours} \]

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17.

If a pipe can fill a tank in 6 hours, find the part of tank it fills in one hour.

Answer: A

If the whole tank takes 6 hours, then in one hour the pipe fills

\[ \frac{1}{6} \text{ of the tank} \]

This reciprocal is the basic move in every pipes-and-cisterns problem: times cannot be added, but rates can.

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18.

Pipes A and B fill a tank in 5 and 6 hours respectively. Pipe C empties the tank in 12 hours, if you open all the three pipes together this tank will be filled in?

Answer: A

It is given that Pipe A,B and C work simultaneously.

Pipe A and B are filling the tank and Pipe C is emptying it.

In 1 hour volume of tank filled by Pipe A: 1/5th

In 1 hour volume of tank filled by Pipe B: 1/6th

In 1 hour volume of tank emptied by Pipe C: 1/12th

Working together Pipe A B and C fill x volume of tank.

=>1/x = (1/5)+(1/6)-(1/12)

=>x=60/17 hours

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19.

A cistern can be filled in 6 hours by taps P and Q. If tap R also joins them, then cistern is filled in 5 hours. Tap P can fill the cistern at twice the rate of tap Q. In what time tap Q and R fill the cistern?

Answer: C

Work in rates: a pipe that fills a tank in \( n \) units of time fills \( \tfrac{1}{n} \) of it per unit, and rates add (an emptying pipe counting negative).

P and Q together fill in 6 hours, and P works at twice Q's rate:

\[ 3q = \frac{1}{6} \quad\Longrightarrow\quad q = \frac{1}{18}, \qquad p = \frac{1}{9} \]

Adding R brings the time to 5 hours:

\[ r = \frac{1}{5} - \frac{1}{6} = \frac{1}{30} \]

\[ q + r = \frac{1}{18} + \frac{1}{30} = \frac{5 + 3}{90} = \frac{4}{45} \]

\[ t = \frac{45}{4} = 11.25 \text{ hours} \]

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20.

Taps A and B can fill a bucket in 12 minutes and 15 minutes respectively. If both are opened and A is closed after 3 minutes, how much further time would it take for B to fill the bucket?

Answer: A

Work in rates: a pipe that fills a tank in \( n \) units of time fills \( \tfrac{1}{n} \) of it per unit, and rates add (an emptying pipe counting negative).

For the first 3 minutes both taps run:

\[ 3\left(\frac{1}{12} + \frac{1}{15}\right) = 3 \times \frac{5 + 4}{60} = \frac{27}{60} = \frac{9}{20} \]

B alone must supply the remaining \( \tfrac{11}{20} \):

\[ t = \frac{11}{20} \times 15 = 8.25 \text{ min} = 8 \text{ min } 15 \text{ s} \]

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