A, B, C are pipes attached to a cistern. A an | vedclass

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(A) The work done by pipes $A, B, C$ in a cycle of $3$ minutes is:$\frac{1}{20} + \frac{1}{30} - \frac{1}{15} = \frac{3 + 2 - 4}{60} = \frac{1}{60}$ p

Vedclass · Accepted Answer

$167$

Vedclass · Answer

(A) The work done by pipes $A, B, C$ in a cycle of $3$ minutes is:$\frac{1}{20} + \frac{1}{30} - \frac{1}{15} = \frac{3 + 2 - 4}{60} = \frac{1}{60}$ part of the tank.To avoid overfilling,we consider the capacity filled before the last cycle. The tank capacity is $1$. Pipe $A$ fills $\frac{1}{20}$ and $B$ fills $\frac{1}{30}$.We need to reach a point where the remaining part is $\le \frac{1}{20} + \frac{1}{30} = \frac{5}{60} = \frac{1}{12}$.In $165$ minutes ($55$ cycles of $3$ minutes),the part filled is $55 	imes \frac{1}{60} = \frac{55}{60} = \frac{11}{12}$.Remaining part $= 1 - \frac{11}{12} = \frac{1}{12}$.In the $166$th minute,pipe $A$ fills $\frac{1}{20}$ part. Remaining $= \frac{1}{12} - \frac{1}{20} = \frac{5-3}{60} = \frac{2}{60} = \frac{1}{30}$.In the $167$th minute,pipe $B$ fills $\frac{1}{30}$ part. Thus,the tank is full.Total time $= 165 + 1 + 1 = 167$ minutes.

$A, B, C$ are pipes attached to a cistern. $A$ and $B$ can fill it in $20$ and $30$ minutes respectively,while $C$ can empty it in $15$ minutes. If $A, B, C$ are kept open successively for $1$ minute each,how soon will the cistern be filled? (in minutes)

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