We are providing a number of problems on the Pipes and Water tank topic with simple solutions. You can quickly understand and learn more about the Pipes and Water Tank topic by following the below diverse problems on Worksheet on Pipes and Water Tank. Practice the below worksheet and get a good knowledge of it. The basic equations to remember are

(i) If Pipe A can fill the water tank in ’X’ hours and Pipe B can fill the water tank in ‘Y’ hours. If we open Pipe A and Pipe B at a time, then the time to fill the water tank by Pipe A and Pipe B is \(\frac{XY}{X + Y}\) hours.

(ii) If Pipe A can fill the water tank in ‘X’ hours and Pipe B can empty the water tank in ‘Y’ hours. So, the water tank is filled in \(\frac{XY}{X – Y}\) hours by Pipe A and Pipe B.

Do Check:

- Worksheet on Work Done in a Given Period of Time
- Worksheet on Time Required to Complete a Piece of Work

## Pipes and Water Tank Worksheets

1. A pipe can fill the cistern in 14 hours. Due to leakage in the cistern, it is filled in 16 hours. After the cistern is full, in how much time will it be emptied due to leakage?

## Solution:

As per the given information, A Pipe can fill the cistern in 14 hours. That is a = 14 hours.

So, A Pipe can fill the cistern in one hour = \(\frac{1}{a}\) = \(\frac{1}{14}\).

Let us assume that, total time is taken by the leakage pipe to empty the cistern is ‘x’.

So, leakage pipe can empty the cistern in 1 hour = \(\frac{1}{x}\).

Because of leakage, the cistern is filled in 16 hours. That is b = 16 hours.

So, cistern is emptied in 1 hour due to leakage = \(\frac{1}{b}\) = \(\frac{1}{16}\).

Leakage pipe can empty the cistern in one hour = A Pipe can fill the cistern in one hour – Cistern is emptied in 1 hour due to leakage.

\(\frac{1}{x}\) = \(\frac{1}{14}\) – \(\frac{1}{16}\).

\(\frac{1}{x}\) = \(\frac{8 -7}{112}\).

\(\frac{1}{x}\) = \(\frac{1}{112}\).

The total time to empty the cistern by leakage is 112 hours.

Therefore, the cistern is emptied due to the leakage in 112 hours.

2. A cistern can be filled by a tap in 10 hours and emptied by an outlet pipe in 12 hours. How long will it take to fill the cistern if both the taps and pipe are opened together?

## Solution:

The given details are A cistern has two pipes, one is inlet and another one is outlet pipe.

A cistern can be filled by a tap in 10 hours by an inlet pipe. That is X = 10 hours.

So, an inlet pipe can fill the cistern in 1 hour = \(\frac{1}{X}\) = \(\frac{1}{10}\)th part of the cistern.

An outlet pipe can empty the cistern in 12 hours. That is Y = 12 hours.

So, an outlet pipe can empty the cistern in 1 hour = \(\frac{1}{Y}\) = \(\frac{1}{12}\)th part of the cistern.

If the Inlet pipe and outlet pipe are opened together, then the cistern is filled in an hour is

\(\frac{1}{X}\) – \(\frac{1}{Y}\) = \(\frac{1}{10}\) – \(\frac{1}{12}\).

= \(\frac{Y – X}{XY}\) = \(\frac{12 – 10}{120}\).

= \(\frac{Y – X}{XY}\) = \(\frac{2}{120}\).

= \(\frac{Y – X}{XY}\) = \(\frac{1}{60}\).

So, By opening the inlet and an outlet pipe at a time, the cistern is filled in 60 hours.

Therefore, Cistern is filled in 60 hours by an inlet and outlet pipe together.

3. A tank can be filled by one tap in 10 hours and by another in 14 hours. How long will it take to fill if both the taps are opened simultaneously?

## Solution:

The given details are One tap can take the time to fill the tank is X = 10 hours.

So, one tap can fill the tank in one hour = \(\frac{1}{X}\) = \(\frac{1}{10}\).

Another tap that can take the time to fill the tank is Y = 14 hours.

So, another tap can fill the tank in one hour = \(\frac{1}{Y}\) = \(\frac{1}{14}\).

If both the taps are opened simultaneously, then the tank is filled in one hour is

\(\frac{1}{X}\) + \(\frac{1}{Y}\) = \(\frac{1}{10}\) + \(\frac{1}{14}\).

\(\frac{X + Y}{XY}\) = \(\frac{10 + 14}{(10)(14)}\).

\(\frac{X + Y}{XY}\) = \(\frac{24}{140}\).

\(\frac{X + Y}{XY}\) = \(\frac{6}{35}\).

Total time to fill the tank by two pipes is equal to \(\frac{XY}{X + Y}\) = \(\frac{35}{6}\).

Therefore, the tank is filled with both pipes in 6 hours.

4. A pipe can fill the tank in 2 hours. Due to leakage at the bottom, it is filled in 3 hours. When the tank is full, in how much time will it be emptied by the leak?

## Solution:

As per the given information, A Pipe can fill the tank in 2 hours. That is a = 2 hours.

So, A Pipe can fill the tank in one hour = \(\frac{1}{a}\) = \(\frac{1}{2}\).

Let us assume that, total time taken by the leakage pipe to empty the tank is ‘x’ hours.

So, leakage pipe can empty the tank in 1 hour = \(\frac{1}{x}\).

Because of leakage, the tank is filled in 3 hours. That is b = 3 hours.

So, tank is emptied in 1 hour due to leakage = \(\frac{1}{b}\) = \(\frac{1}{3}\).

Leakage pipe can empty the tank in one hour = A Pipe can fill the tank in one hour – Tank is emptied in 1 hour due to leakage.

\(\frac{1}{x}\) = \(\frac{1}{2}\) – \(\frac{1}{3}\).

\(\frac{1}{x}\) = \(\frac{3 -2}{6}\).

\(\frac{1}{x}\) = \(\frac{1}{6}\).

The total time to empty the tank by leakage is 6 hours.

Therefore, the tank is emptied due to the leakage in 6 hours.

5. A water tank can be filled by two taps A and B in 6 hours and 10 hours respectively. The full tank can be emptied by the third tap in 5 hours. If all the taps are opened at the same time, in how much time will the empty tank be filled up completely?

## Solution:

As per the given information, Tap A can fill the water tank in 6 hours. That is X = 6 hours.

Tap A can fill the water tank in 1 hour = \(\frac{1}{X}\) = \(\frac{1}{6}\).

Tap B can fill the water tank in 10 hours. That is Y = 10 hours.

Tap B can fill the water tank in 1 hour = \(\frac{1}{Y}\) = \(\frac{1}{10}\).

Tap C can empty the water tank in 5 hours. That is Z = 5 hours.

Tap C can empty the water tank in 1 hour = \(\frac{1}{Z}\) = \(\frac{1}{5}\).

Three taps are opened at the same time. Then the water tank is filled in one hour is equal to

= \(\frac{1}{X}\) + \(\frac{1}{Y}\) – \(\frac{1}{Z}\) = \(\frac{1}{6}\) + \(\frac{1}{10}\) + \(\frac{1}{5}\).

= \(\frac{YZ + ZX + XY}{XYZ}\) = \(\frac{(10)(5) + (6)(5) + (6)(10)}{(5)(6)(10)}\).

= \(\frac{50 + 30 + 60}{300}\).

= \(\frac{140}{300}\).

= \(\frac{7}{15}\).

So, three taps together can fill the water tank in one hour is equal to \(\frac{7}{15}\)th part of the water tank.

Three taps can take time to fill the water tank is

\(\frac{XYZ}{XY + YZ + ZX}\) = \(\frac{15}{7}\) = 2.1 hours.

Therefore, Tap A, Tap B, and Tap C together can take 2 hours of time to fill the water tank.

6. 10 taps together can fill the water tank in 40 minutes. If 2 taps are going out of order, then calculate the time to fill the water tank by the remaining taps?

## Solution:

As per the given details, 10 taps together can take time to fill the water tank is equal to 40 minutes. That is 10 × 40 ——(1).

If 2 taps are going out of order, then the remaining taps are 10 – 2 = 8 taps.

Let us assume, 8 taps can take the time to fill the water tank as ‘P’ minutes. That is 8 × P —–(2).

By Equating the equation (1) and (2). We will get

10 × 40 = 8 × P.

P = \(\frac{400}{8}\).

P = 50 minutes.

Therefore, 8 taps together can fill the water tank in 50 minutes.

7. Three pipes are connected to the water tank. Pipe C can empty the water tank in 20 minutes. If the leakage Pipe C is closed, then Pipe A and Pipe B can fill the water tank in 10 minutes and 15 minutes respectively. Find the time to fill the water tank by three taps together?

## Solution:

The given details are Pipe A can take time to fill the water tank = 10 minutes.

Pipe A can fill the water tank in one minute = \(\frac{1}{A}\) = \(\frac{1}{10}\).

Pipe B can take time to fill the water tank = 15 minutes.

Pipe B can fill the water tank in one minute = \(\frac{1}{B}\) = \(\frac{1}{15}\).

If Pipe is closed, then the water tank is filled by pipe A and pipe B in one hour is

\(\frac{1}{A}\) + \(\frac{1}{B}\) = \(\frac{1}{10}\) + \(\frac{1}{15}\).

= \(\frac{A + B}{AB}\) = \(\frac{10 + 15}{(10)(15)}\).

= \(\frac{25}{150)}\).

= \(\frac{1}{6}\).

So, Pipe A and Pipe B can take 6 hours of time to fill the water tank. (when pipe C is closed).

If Pipe C is open, then it can take time to empty the water tank is = 20 minutes.

Pipe C can empty the water tank in one hour = \(\frac{1}{C}\) = \(\frac{1}{20}\).

Three pipes are open at the same time, then water is filled in one hour is

= \(\frac{A + B}{AB}\) – \(\frac{1}{C}\).

= \(\frac{1}{6}\) – \(\frac{1}{20}\).

= \(\frac{20 – 6}{(6)(20)}\).

= \(\frac{14}{120}\).

= \(\frac{7}{60}\).

The water tank is filled in \(\frac{60}{7}\) minutes by three taps together.

Therefore, three taps can take 8.5 minutes of time to fill the water tank together.

8. A Pipe supplies 4 liters of water per minute into a tank. The tank is emptied in 5 hours because of a leak at the bottom of the tank. A full tank with the pipe open is emptied by the leak in 10 hours. Find the capacity of the tank?

## Solution:

As per the information, the Leakage pipe can empty the tank in 5 hours.

So, Leakage pipe can empty the tank in 1 hour = \(\frac{1}{5}\).

By opening the pipe, the tank is emptied in 10 hours.

So, the tank is emptied in one hour is \(\frac{1}{10}\).

The rate of filling of the tank in one hour is

\(\frac{1}{5}\) – \(\frac{1}{10}\).

= \(\frac{2 – 1}{10}\).

= \(\frac{1}{10}\).

The pipe can fill the tank in 10 hours. That is 600 minutes.

As per the details, the pipe supplies 4 liters of water per minute.

The tank is filled in 600 minutes. Then the capacity of the tank is 600 × 4 = 2400 liters.

Therefore, the capacity of the water tank is 2400 liters.

9. Two Pipes are connected to the water tank. Pipe A and Pipe B can fill the water tank in 25 minutes and 20 minutes respectively. Both pipes are opened simultaneously, after 5 minutes pipe A is closed. 5 minutes after that pipe B is closed and pipe A is reopened. How much more time does it takes to fill the water tank?

## Solution:

As per the details, Pipe A can fill the tank in 25 minutes. Pipe A can fill the water tank in 1 minute is = \(\frac{1}{25}\).

Pipe B can fill the tank in 20 minutes. Pipe B can fill the water tank in 1 minute is = \(\frac{1}{20}\).

Pipe A and Pipe B can fill the water tank in 1 minute is

\(\frac{1}{25}\) + \(\frac{1}{20}\) = \(\frac{4 + 5}{100}\).

= \(\frac{9}{100}\).

First 5 minutes, both the pipes are opened. Then the tank is filled by two pipes is

= \(\frac{9}{100}\) × 5.

= \(\frac{9}{20}\) ——-(1).

In the next 5 minutes, water tank is filled by Pipe B. That is,

= \(\frac{1}{20}\) × 5.

= \(\frac{1}{4}\) ——-(2).

By adding equations (1) and (2). We will get

\(\frac{9}{20}\) + \(\frac{1}{4}\).

\(\frac{9 + 5}{20}\).

\(\frac{14}{20}\).

The remaining part is 1 – \(\frac{14}{20}\) = \(\frac{6}{20}\)= \(\frac{3}{10}\).

\(\frac{3}{10}\)th part of the water tank is filled by Pipe A.

Pipe A can fill the water tank in 25 minutes.

So, 25 × \(\frac{3}{10}\) = \(\frac{15}{2}\) = 7.5 minutes.

Therefore, Pipe A can take 7.5 minutes of extra time to fill the water tank.

10. Pipe A can fill the water tank in 8 hours. Pipe B can empty the water tank in 24 hours. Pipe A is opened at 6 ‘o’ clock after 1 hour Pipe B is opened and Pipe B is opened for 10 minutes for every one hour up to 10 ‘o’ clock. Calculate the time taken by Pipe A and B to fill the water tank?

## Solution:

As per the given information, Pipe A can take the time to fill the water tank is 8 hours.

So, Pipe A can fill the water tank in one hour = \(\frac{1}{8}\).

Pipe B can take time to empty the water tank is 24 hours.

So, Pipe B can empty the water tank in one hour = \(\frac{1}{24}\).

60 minutes — \(\frac{1}{24}\).

10 minutes — ?

(10 ×\(\frac{1}{24}\) ) / 60 = \(\frac{1}{144}\) in 10 minutes.

So, from 6 ‘ o’ clock to 7 ‘o’ clock water, \(\frac{1}{8}\)th part of the water tank is filled by pipe A.

The remaining part is 1 – \(\frac{1}{8}\) = \(\frac{7}{8}\)th part of the water tank is empty.

In next 1 hour (7 ‘o’ clock to 8 ‘o’ clock), the water gets filled = Tank filled by Pipe A in one hour – Tank gets emptied by Pipe B in one hour

= \(\frac{1}{8}\) – \(\frac{1}{144}\).

= \(\frac{18 – 1}{144}\).

= \(\frac{17}{144}\).

So, in one hour \(\frac{17}{144}\)th part of the water tank is filled.

So to fill the \(\frac{7}{8}\)th of the empty part of the water tank.

= \(\frac{7}{8}\) ×\(\frac{144}{17}\).

= \(\frac{126}{17}\).

= 7

Total required time is 7 + 1 = 8 hours. (Pipe B empty the tank for 15 minutes for every 1 hour up to 4 hours).

The tank is empty at 6 ‘o’ clock and after 8 hours the water tank filled by Pipe A and Pipe B. That is 2 ‘o’ clock.

So, the water tank is filled at 2 ‘o’ clock by Pipe A and Pipe B together.