A cistern has two pipes. One can fill it with water in
8 hours and other can empty it in 5 hours. In how
many hours will the cistern be emptied if both the
pipes are opened together when 3/4 of the cistern is
already full of water?
Answers
Answer :
In 10 hours the cistern be emptied if both the pipes are opened together when 3/4 of the cistern is already full of water .
Explanation :
We are given a cistern having 2 pipes.
1st cistern can fill the water in 8 hours while, other can empty in 5 hours.
So,
★ Part of the tank filled by the 1st pipe in 1 hour is 1/8 hours and part of the cistern emptied by the 2nd pipe is 1/5 hours.
When both the pipes are opened together then, empty part of the tank is,
⇛ 1/5 - 1/8
⇛ 8 - 5/40
⇛ 3/40
As we have got that, 3/40 th part of the tank is emptied in 1 hour , Then,
★ 3/4th part of the tank will be emptied in,
➟ 40/3 × 3/4 [As 3/40 part is emptied in 1 hour.]
➟ 40/4
➟ 10 hours.
Therefore, in 10 hours the cistern be emptied if both the pipes are opened together when 3/4 of the cistern is already full of water .
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Additional information :
★ In Cistern related questions, we are supposed to find out, what portion of the tank each of the pipes fill or get empty and then we'll form an equation on this value, that's all .
★ If pipe can fill a tank in ' x hr ', then part filled in 1 hr = 1/x hrs .
★ If the same pipe can empty a tank in ' y hr ', then part emptied in 1 hr = 1/y hrs .
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Answer:
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