Pipes and Cistern-RRB

Pipes And Cistern

Pipes and cisterns problems are almost the same as those of Time and work problems. Thus, if a pipe fills a tank in 6 hrs, then the pipe fills $\frac{1}{6}$ th of the tank in 1 hr. the only difference with Pipes and Cisterns problems is that there are outlets as well as inlets.

Inlet

An inlet pipe is connected with a tank and it fills tank. So a pipe which fills up the tank is known as inlet.

Outlet

A outlet pipe is connected with a tank and it empties tank. So a pipe which empties the tank is known as outlet.

If a pipe can fill a tank in x hours, then: part filled in 1 hour = $\frac{1}{x}$ .

If a pipe can empty a tank in y hours, then:part emptied in 1 hour = $\frac{1}{y}$

If a pipe can fill a tank in x hours and another pipe can empty the full tank in y hours (where y > x), then on opening both the pipes, then the net part filled in 1 hour =( $\frac{1}{x}$ – $\frac{1}{y}$ )

If a pipe can fill a tank in x hours and another pipe can empty the full tank in y hours (where x > y), then the net part emptied in 1 hour =( $\frac{1}{y}$ – $\frac{1}{x}$ )

Question 1:

Two pipes A and B can fill a tank in 20 and 30 minutes respectively. If both the pipes are used together, then how long will it take to fill the tank?

Solution

Pipe A can fill a tank in 20 minutes, hence in 1 minute it fills $\frac{1}{20}$ th.

Pipe B can fill a tank in 30 minutes, hence in 1 minute it fills $\frac{1}{30}$ th

When both the pipes are used together part filled by A + B together in 1 min.

= $\frac{1}{20}$ + $\frac{1}{30}$ = $\frac{5}{60}$ = $\frac{1}{12}$

Hence both pipes can fill the tank in 12 minutes.

Question 2:

Two pipes A and B can fill a cistern in 37 minutes and 45 minutes respectively. Both pipes are opened. The cistern will be filled in just half an hour, if the B is turned off after.

Solution

Let B be turned off after x minutes.

Then, part filled by (A + B) in x min. + Part filled by A in (30 –x) min. = 1

Hence, x( $\frac{2}{75}$ . + $\frac{1}{45}$ ) + (30 –x) $\frac{2}{75}$ =1

( $\frac{11}{225}$ )x + (60 – 2x) $\frac{1}{75}$ =1

11x+ 180 – 6x = 225.

x= 9

Hence the answer.

Question 3:

If two pipes function simultaneously, the reservoir is filled in 12 hrs. One pipe fills the reservoir 10 hrs faster than the other. How many hrs does the faster pipe take to fill the reservoir?

Solution

Let the faster pipe fills the tank in x hrs.

Then the slower pipe fills it x + 10 hrs.

When both are opened the reservoir will be filled in x(x+10) $\frac{1}{x}$ + (x+10) = 12

Hence x² + 10x = 12(2x + 10)

x² + 10x -24x -120 =0

x² -14x -120 = 0

x² – 20x + 6x -120 = 0

x(x – 20) + 6 (x – 20) = 0

(x – 20 )(x +6) = 0

x = 20, x = – 6(Impossible)

Hence the faster pipe can fill in 20 hrs.

Question 4:

Two pipes A and B can fill a cistern in 1.5 hour and 100 minutes respectively. There is also an outlet C. If all the three pipes are opened together, the tank is full in 60 minutes. How much time will be taken by C to empty the full tank?

Solution

Two pipes A and B can fill a cistern in 1.5 hour i.e 90 and 100 minutes.

Work done by C in 1 minute = $\frac{1}{90}$ + $\frac{1}{100}$ – $\frac{1}{60}$

= $\frac{4}{900}$ = $\frac{1}{225}$

Hence C can empty the tank in 225 minutes.

Question 5:

A pipe can fill a tank in 25 hrs. Due to a leakage in the bottom, it is filled in 50 hrs. If the tank is full, how much time will the leak take to empty it?

Solution

A pipe can fill in 1 hour $\frac{1}{25}$ th of tank.

So work done due to leak in $\frac{1}{25}$ – $\frac{1}{50}$ = $\frac{1}{50}$ hr.

Hence the leak will empty the tank in 50 hrs.

Question 6:

Pipe P can fill a tank in 40 hours while Pipe Q alone can fill it in 50 hours and Pipe R can empty the full tank in 60 hours. If all the pipes are opened together, how much time will be needed to make the tank full?

Solution

Pipe P can fill a tank in 40 hrs.

In 1 hr it can fill $\frac{1}{40}$ .

Similarly B can fill in 1 hr $\frac{1}{50}$ .

But pipe C can empty in 60 hrs mean in one hour $\frac{1}{60}$

Hence if the all the pipes are open then in 1 hour $\frac{1}{40}$ + $\frac{1}{50}$ – $\frac{1}{60}$

= $\frac{17}{600}$

Hence the tank will be full in $\frac{600}{17}$ hrs.

Question 7:

Two pipes P and Q can fill a tank in 30 hours and 45 hours respectively. If both the pipes are opened together, how much time will be taken to fill the tank?

Solution

Part filled by P alone in one hour = $\frac{1}{30}$

Part filled by Q alone in one hour = $\frac{1}{45}$

Part filled by P + Q together in one hour

= $\frac{1}{30}$ + $\frac{1}{45}$ = $\frac{5}{90}$ = $\frac{1}{18}$

Hence both the pipes can fill the tank in 18 hours

Question 8:

Two pipes A and B can fill a tank in 16 minutes and 24 minutes respectively. If both the pipes are opened simultaneously, after how much time should B be closed so that that the tank is full in 12 minutes?

Solution

Let B be closed after x minutes.

Then part filled by A+B in x minutes + part filled by A in (12 –x) minutes = 1 Hence x( $\frac{1}{16}$ + $\frac{1}{24}$ ) + (12 – x) $\frac{1}{16}$ = 1

=>x( $\frac{5}{48}$ ) + $\frac{3}{4}$ – $\frac{x}{16}$ = 1

=> $\frac{5x}{48}$ – $\frac{3x}{48}$ = 1 – $\frac{3}{4}$

=> $\frac{2x}{48}$ = $\frac{1}{4}$

=> $\frac{x}{24}$ = $\frac{1}{4}$

=> x =6

So B should be closed after 6 min.

Question 9:

A tank has a leak which would empty it in 12 hrs. A tap is turned on which admits 6 liters a minutes into the tank, and it is now emptied in 16 hrs. How many liters does the tank hold?

Solution

A tank has a leak which would empty it in 12 hrs and emptied in 16 hrs

The filler tap can fill the tank in $\frac{16.12}{(16 - 12)}$ = $\frac{192}{4}$ =48 hrs.

A tap admits 6 liters a minutes into the tank

Hence capacity of a tank = 48

Question 10:

A tank is filled in 5 hours by three pipes A, B and C. The pipe C is twice as fast as B and B is twice as fast as A. How much time will pipe A alone take to fill the tank?

Solution

Suppose pipe A alone takes x hours to fill the tank. Then, pipes B and C will take x/2 and x/4, hours respectively to fill the tank.

Hence $\frac{1}{x}$ + $\frac{2}{x}$ + $\frac{4}{x}$ = $\frac{1}{5}$

⇒ $\frac{7}{x}$ = $\frac{1}{5}$

⇒x =35 hrs

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