
Two pipes A and B can separately fill a cistern in 60 minutes and 75 minutes respectively. There is a third pipe at the bottom of the cistern to empty it, if all the three pipes are simultaneously opened, then the cistern is full in 50minutes, In how much time, the third pipe alone can empty the cistern?
$
(a){\text{ 110 minutes}} \\
(b){\text{ 100 minutes}} \\
(c){\text{ 120 minutes}} \\
(d){\text{ 90 minutes}} \\
$
Answer
598.5k+ views
Hint: In this question let the time taken by piston C to empty the cistern by x minutes. Use the unitary method to calculate the work done by each pipe in one minute, the total work done by all three pipes will eventually be the work done by pipe A, pipe B and pipe C as they are opened simultaneously.
Complete Step-by-Step solution:
Pipe A fill a cistern in 60 minutes.
Pipe B fills a cistern in 75 minutes.
Let the third pipe be C which empty the cistern alone in x minutes.
Now it is given that all the three pipes simultaneously opened, then the cistern is full in 50 minutes.
So pipe A one minute work = $\dfrac{1}{{60}}$.
Pipe B one minute work = $\dfrac{1}{{75}}$.
And the third pipe C one minute work = $ - \dfrac{1}{x}$ (‘-’ sign is because it empty the cistern).
All the pipes simultaneously do one minute work = $\dfrac{1}{{50}}$.
So pipe A one minute work + Pipe B one minute work + pipe C one minute work = All the pipes simultaneously one minute work.
$ \Rightarrow \dfrac{1}{{60}} + \dfrac{1}{{75}} - \dfrac{1}{x} = \dfrac{1}{{50}}$
Now simplify the above equation we have,
$ \Rightarrow \dfrac{1}{{60}} + \dfrac{1}{{75}} - \dfrac{1}{{50}} = \dfrac{1}{x}$
$ \Rightarrow \dfrac{{75 \times 50 + 60 \times 50 - 60 \times 75}}{{60 \times 75 \times 50}} = \dfrac{1}{x}$
$ \Rightarrow \dfrac{{2250}}{{60 \times 75 \times 50}} = \dfrac{1}{x}$
$ \Rightarrow x = \dfrac{{60 \times 75 \times 50}}{{2250}} = 100$
So the third pipe alone can empty the cistern in 100 minutes.
Hence option (B) is correct.
Note: In this question two pipes were filling the cistern and one pipe was emptying it hence the work done by the pipe which is emptying the cistern is taken with a negative sign. There is a misconception that work done can’t be negative but if the work done is relatively opposite to what others are doing than it is negative.
Complete Step-by-Step solution:
Pipe A fill a cistern in 60 minutes.
Pipe B fills a cistern in 75 minutes.
Let the third pipe be C which empty the cistern alone in x minutes.
Now it is given that all the three pipes simultaneously opened, then the cistern is full in 50 minutes.
So pipe A one minute work = $\dfrac{1}{{60}}$.
Pipe B one minute work = $\dfrac{1}{{75}}$.
And the third pipe C one minute work = $ - \dfrac{1}{x}$ (‘-’ sign is because it empty the cistern).
All the pipes simultaneously do one minute work = $\dfrac{1}{{50}}$.
So pipe A one minute work + Pipe B one minute work + pipe C one minute work = All the pipes simultaneously one minute work.
$ \Rightarrow \dfrac{1}{{60}} + \dfrac{1}{{75}} - \dfrac{1}{x} = \dfrac{1}{{50}}$
Now simplify the above equation we have,
$ \Rightarrow \dfrac{1}{{60}} + \dfrac{1}{{75}} - \dfrac{1}{{50}} = \dfrac{1}{x}$
$ \Rightarrow \dfrac{{75 \times 50 + 60 \times 50 - 60 \times 75}}{{60 \times 75 \times 50}} = \dfrac{1}{x}$
$ \Rightarrow \dfrac{{2250}}{{60 \times 75 \times 50}} = \dfrac{1}{x}$
$ \Rightarrow x = \dfrac{{60 \times 75 \times 50}}{{2250}} = 100$
So the third pipe alone can empty the cistern in 100 minutes.
Hence option (B) is correct.
Note: In this question two pipes were filling the cistern and one pipe was emptying it hence the work done by the pipe which is emptying the cistern is taken with a negative sign. There is a misconception that work done can’t be negative but if the work done is relatively opposite to what others are doing than it is negative.
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