A pipe can fill a cistern in 12 minutes and another can fill it in 15 minutes, but a third pipe can empty it in 6 minutes. The first two are kept open for 6 minutes in the beginning and then the third pipe is also opened, in what time will the cistern be emptied?
Answers
Answer:
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Step-by-step explanation:
=> Let the capacity of the tank = LCM of (12,15, 6) = 60 litres.
=> Rate of work done by the first pipe = 60/12
⇒ 5 L/min
=> Rate of work done by the second pipe = 60/15
⇒ 4L/min
=> Rate of work done by the draining pipe = 60/6
⇒10L/min
=> If all the pipes are opened simultaneously, (5+4-10) litres will be filled in a minute.
=> Since this comes out to be negative, no water will be filled at any point of time.
=> For the 6 minutes only filling pipes are opened,
=> Rate of both filling pipes = 9L/min, (5+4).
=> So in 6 minutes, 9 × 6 = 54 litres water is filled.
=> When the third pipe is opened, all three pipes will work simultaneously, at a rate of (5+4 – 10) = - 1 ltr per min.
=> This implies that in a minute, 1 litre water will be drained.
=> So, 54 litres water will be drained in 54 minutes. So the tank will be emptied in 54 minutes.
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Answer:
Let the number of minutes taken to empty the cistern be x min.
According to the question,
x6−x+512−x+512=0
x6−x12−512−x15−515=0
x6−x12−x15=512+515
10x−5x−4x60=25+2060
x60=4560=x=45min.