Problems on Pipes and Water Tank is provided with the different types of problems. Follow the complete concept and learn more about the pipes and water tank topic. Different methods are given to solve a single problem here. Therefore, check out various problems with methods and choose the best process to solve them. We have also provided answers and explanations for the problems.

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## Word Problems on Pipes and Cisterns

**Example 1.**

Two pipes are connected to the water tank or cistern. Pipe A and B can fill the water tank in 15 minutes and 20 minutes respectively. If the two pipes are opened at a time, then find the time to fill the water tank?

**Solution:**

As per the given information, Pipe A and Pipe B are connected to the water tank or cistern.

Pipe A can take the time to fill the water tank is (a) = 15 minutes.

Pipe A can fill the water tank in 1 hour is equal to \(\frac{ 1 }{ a }\) = \(\frac{ 1 }{ 15 }\).

Pipe B can take the time to fill the water tank is (b) = 20 minutes.

Pipe B can fill the water tank in 1 hour is equal to \(\frac{ 1 }{ b }\) = \(\frac{ 1 }{ 20 }\).

Pipe A and Pipe B together can fill the water tank in 1 hour =\(\frac{ 1 }{ a }\) + \(\frac{ 1 }{ b }\).

By substituting the values in the above equation, then we will get as

\(\frac{ 1 }{ 15 }\) + \(\frac{ 1 }{ 20 }\).

L.C.M of 15 and 20 is equal to 60.

So,\(\frac{ 1 }{ 15 }\) + \(\frac{ 1 }{ 20 }\) = \(\frac{ 4 + 3 }{ 60 }\)= \(\frac{ 7 }{ 60 }\).

Total time to fill the water tank by two pipes A and B = \(\frac{ ab }{ a+b }\).

Therefore, the total time taken by two Pipes A and B to fill the water tank or cistern is equal to \(\frac{ 60 }{ 7 }\) minutes.

**Example 2.**

Pipe A can fill the cistern in 2 hours and Pipe B can empty the cistern in 4 hours. If two pipes are opened at a time, find the time to fill the cistern?

**Solution:**

As per the details, Time taken to fill the cistern by pipe A = a = 2 hours.

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

Time taken to empty the cistern by pipe B = b = 4 hours.

So, Pipe B can fill the cistern in 1 hour = \(\frac{ 1 }{ b }\) = \(\frac{ 1 }{ 4 }\).

Pipe A and Pipe B together can fill the cistern in 1 hour = \(\frac{ 1 }{ a }\) – \(\frac{ 1 }{ b }\).

Substitute the values in the above equation. Then, we will get like \(\frac{ 1 }{ a }\) – \(\frac{ 1 }{ b }\) = \(\frac{ 1 }{ 2 }\) – \(\frac{ 1 }{ 4 }\).

L.C.M of 2 and 4 is 4.

\(\frac{ b – a }{ ab }\)= \(\frac{ 2 – 1 }{ 4 }\)= \(\frac{ 1 }{ 4 }\).

Total time taken by two pipes A and B to fill the cistern is equal to\(\frac{ ab }{ b-a }\).

Therefore, Pipe A and Pipe B can fill the cistern in 4 hours.

**Example 3.**

Three pipes are connected to the cistern and three pipes are working as inlet pipes. Pipe A, Pipe B, and Pipe C can fill the water tank in 6 hours, 8 hours, and 10 hours respectively. Calculate the time to fill the water tank by three pipes together?

**Solution:**

As per the information, Pipe A can fill the water tank in 6 hours. That is a = 6 hours.

So, Pipe A can fill the water tank in 1 hour = \(\frac{ 1 }{ a }\)= \(\frac{ 1 }{ 6 }\)thpart of the water tank.

Pipe B can fill the water tank in 8 hours. That is b = 8 hours.

So, Pipe B can fill the water tank in 1 hour = \(\frac{ 1 }{ b }\) = \(\frac{ 1 }{ 8 }\)thpart of the water tank.

Pipe C can fill the water tank in 10 hours. That is c = 10 hours.

So, Pipe C can fill the water tank in 1 hour = \(\frac{ 1 }{ c }\)= \(\frac{ 1 }{ 10 }\)th part of the water tank.

Three pipes at a time can fill the water tank in 1 hour = \(\frac{ 1 }{ a }\) + \(\frac{ 1 }{ b }\) + \(\frac{ 1 }{ c }\)

= \(\frac{ 1 }{ 6 }\) + \(\frac{ 1 }{ 8 }\) + \(\frac{ 1 }{ 10 }\).

=\(\frac{ bc + ac + ab }{ abc }\).

= \(\frac{ (8) (10) + (6) (10) + (6) (8) }{ (6) (8) (10) }\).

= \(\frac{ 80 + 60 + 48}{ 480 }\) = \(\frac{ 188 }{ 480 }\).

Total time taken by three pipes to fill the water tank =\(\frac{ abc }{ bc + ac + ab}\).

= \(\frac{ 480 }{ 188 }\) hours= 2.5 hours

Therefore, Pipe A, Pipe B, and Pipe C together can fill the water tank in 2.5 hours.

**Example 4.**

Two Pipes are connected to the cistern and two pipes can fill the cistern in 10 minutes by working together. Pipe A can take 5 minutes less time than Pipe B to fill the cistern. Find the time to fill the cistern by Pipe B?

**Solution:**

As per the information, Pipe A and Pipe B together can fill the cistern in 10 minutes. That is

\(\frac{ ab }{ a+b }\)= 10 minutes.

Pipe A can fill the cistern in â€˜Xâ€™ minutes = a. (which is 5 minutes less than pipe B).

So, Pipe A can fill the cistern in 1 hour \(\frac{ 1 }{ a }\)= \(\frac{ 1 }{ x }\)th part of the cistern.

Pipe B can fill the cistern in â€˜X + 5â€™ minutes = b (which is 5 minutes faster than Pipe A).

So, Pipe B can fill the cistern in 1 hour \(\frac{ 1 }{ b }\)= \(\frac{ 1 }{ x + 5 }\)th part of the cistern.

Pipe A and Pipe B can fill the cistern in 1 hour = \(\frac{ 1 }{ a }\) + \(\frac{ 1 }{ b }\).

= \(\frac{ 1 }{ x }\) + latex]\frac{ 1 }{ x + 5 }[/latex]).

\(\frac{ a + b }{ ab }\)= \(\frac{ X + X + 5}{ (X) (X + 5) }\)= \(\frac{ 2X + 5 }{ (X) (X + 5) }\).

Total time to fill the cistern by Pipe A and Pipe B = \(\frac{ ab }{ a+b }\)= \(\fra{ (X) (X + 5) }{ 2X + 5 }\)= 10 minutes.

(X)(X+5) = 10(2X + 5).

X^2 + 5X = 20X + 50.

X^2 + 5X â€“ 20X â€“ 50 = 0.

X^2 â€“ 15X â€“ 50 = 0.

X^2 â€“ 10X â€“ 5X â€“ 50 = 0.

X(X â€“ 10) â€“ 5(X â€“ 10) = 0.

(X â€“ 10) (X â€“ 5) = 0.

X = 10 and 5.

If X = 10 (that is Pipe A can fill the cistern in 10 minutes), then Pipe B = X + 5 = 10 + 5 = 15 minutes.

So, Pipe B alone can fill the cistern in 15 minutes.

If X = 5 (that is Pipe A can fill the cistern in 5 minutes), then Pipe B = X + 5 = 5 + 5 = 10 minutes.

So, Pipe B alone can fill the cistern in 10 minutes.

**Example 5.**

Pipe A can fill the water tank 2 times faster than Pipe B. If we open two pipes at a time, then the water tank can be filled within 24 minutes. Find the Pipe A and Pipe B working time to fill the water tank alone?

**Solution:**

Let us consider the time taken by pipe B to fill the water tank as â€˜xâ€™ minutes.

So, Pipe B can fill the water tank in 1 minute = \(\frac{1}{a}\)= \(\frac{1}{X}\).

Pipe A can fill the water tank 2 times faster than Pipe B. That is, Time taken to fill the water tank by Pipe A = 2x.

Pipe A can fill the water tank in 1 minute = \(\frac{ 1 }{ b }\) = \(\frac{ 1 }{ 2X }\).

Pipe A and Pipe B together can fill the water tank in 1 minute = \(\frac{ 1 }{ a }\) + \(\frac{ 1 }{ b }\)

= \(\frac{ 1 }{ X }\) + latex]\frac{ 1 }{ 2X }[/latex].

=\(\frac{ a + b }{ ab }\)= \(\frac{ 1 }{ 3X }\).

Total time to fill the water tank by Pipe A and Pipe B = \(\frac{ ab }{ a+b }\)= 3x = 24 minutes.

X= \(\frac{ 24 }{ 3 }\)= 8 minutes.

So, Pipe A alone can fill the water tank in (2x) = (2) (8) = 16 minutes and Pipe B alone can fill the water tank in x = 8 minutes.

**Example 6.**

Two Pipes A and B are connected to the water tank. Pipe A and Pipe B can fill the water tank in 10 hours and 12 hours. After some time Pipe B is closed and the water tank is filled within 5 hours. Calculate how many hours, we need to close Pipe B to fill the water tank?

**Solution:**

As per the given details, Pipe A can fill the water tank = a = 10 hours.

Pipe b can fill the water tank = b = 12 hours.

After closing Pipe B, the time taken to fill the water tank = x = 5 hours.

{b[1 – x/a]} = {12[1 – 5/10]} = {12[5/10]} = {12[1/2]} = 6 hours.

Therefore, Pipe B is closed after 6 hours.

**Example 7.**

Three pipes are connected to the water tank. They are Pipe A, Pipe B, and Pipe C. Pipe A and Pipe B are working as inlet pipes. Pipe A and Pipe B can take 4 hours and 6 hours of time to fill the water tank. Pipe C is a working outlet pipe and it can empty the water tank in 5 hours. By opening the three pipes at a time, find the time to fill the water tank?

**Solution:**

As per the given information, three pipes are connected to the water tank.

Time taken by Pipe A to fill the water tank = a = 4 hours.

So, Pipe A can fill the water tank in 1 hour = \(\frac{ 1 }{ a }\)= \(\frac{ 1 }{ 4 }\).

Time taken by Pipe B to fill the water tank = b = 6 hours.

Therefore, Pipe B can fill the water tank in 1 hour = \(\frac{ 1 }{ b }\) = \(\frac{ 1 }{ 6 }\).

Time taken by Pipe C to empty the water tank = c = 5 hours.

So, Pipe C can empty the water tank in 1 hour = \(\frac{ 1 }{ c }\) = \(\frac{ 1 }{ 5 }\).

Water tank filled by three taps in 1 hour = \(\frac{ 1 }{ a }\) + \(\frac{ 1 }{ b }\) – \(\frac{ 1 }{ c }\).

Substitute the values in the above equation. That is,

\(\frac{ 1 }{ a }\) + \(\frac{ 1 }{ b }\) – \(\frac{ 1 }{ c }\) = \(\frac{ 1 }{ 4 }\) + \(\frac{ 1 }{ 6 }\) â€“ \(\frac{ 1 }{ 5 }\).

=\(\frac{ bc + ac – ab }{ abc }\).

=\(\frac{ (6) (5) + (4) (5) – (4) (6) }{ (4) (6) (5) }\).

= \(\frac{ 30 + 20 – 24 }{ 120 }\) = \(\frac{ 26 }{ 120 }\).

Total time taken by three pipes A,B,and C to fill the water tank = \(\frac{ abc }{ bc + ac – ab }\)= \(\frac{ 120 }{ 26 }\) hours.

Therefore, three Pipes can fill the water tank in \(\frac{ 120 }{ 26 }\) hours.