Question

. A pipe can fill a cistern in 4 min and another pipe can
fill it in 5 min, but a third pipe can empty it in 2 min.
The first two pipes are kept open for 2 min in the
beginning and then the third pipe is also opened.
Time taken to empty the cistern is?​

Answer 1

Given :- A pipe can fill a cistern in 4 min and another pipe can fill it in 5 min, but a third pipe can empty it in 2 min.

The first two pipes are kept open for 2 min in the

beginning and then the third pipe is also opened.

Time taken to empty the cistern is ?

Answer :-

→ LCM of 5 , 4 and 2 = 20 units = Let total capacity of cistern .

so,

→ Efficiency of pipe A filling = Total work / Time taken = 20/4 = 5 units / min.

→ Efficiency of pipe B filling = Total work / Time taken = 20/5 = 4 units / min.

→ Efficiency of pipe C empty = Total work / Time taken = 20/2 = 10 units / min.

now,

→ In first 2 min. both A + B fills = 2 * (5 + 4) = 18 units .

and then,

→ Efficiency of all empty = (5 + 4) - 10 = 9 - 10 = (-1) units / min.

therefore,

→ Time taken to empty the cistern = 18/1 = 18 minutes (Ans.)

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