A cistern has three pipes A, B and C. Pipes A | vedclass

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(B) Pipe $A$ fills $1/3$ of the cistern in $1 \, h$. Pipe $B$ fills $1/4$ of the cistern in $1 \, h$. Pipe $C$ empties $1$ full cistern in $1 \, h$.At

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$7:12 \, pm$

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(B) Pipe $A$ fills $1/3$ of the cistern in $1 \, h$. Pipe $B$ fills $1/4$ of the cistern in $1 \, h$. Pipe $C$ empties $1$ full cistern in $1 \, h$.At $4:00 \, pm$, pipe $A$ has worked for $1 \, h$, filling $1/3$ of the cistern.At $5:00 \, pm$, pipe $A$ has worked for $2 \, h$ and pipe $B$ has worked for $1 \, h$. Total filled part $= (2 	imes 1/3) + (1 	imes 1/4) = 2/3 + 1/4 = 11/12$.From $5:00 \, pm$ onwards, all three pipes $A, B,$ and $C$ are open. The net work done per hour $= 1/3 + 1/4 - 1 = (4+3-12)/12 = -5/12$. The negative sign indicates the cistern is being emptied at a rate of $5/12$ per hour.Time taken to empty the $11/12$ part $= (11/12) / (5/12) = 11/5 \, h = 2.2 \, h$.$2.2 \, h = 2 \, h$ and $0.2 	imes 60 \, min = 2 \, h \, 12 \, min$.Starting from $5:00 \, pm$, adding $2 \, h \, 12 \, min$ gives $7:12 \, pm$.

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