Two taps A and B can fill a tank in 20 text{ | vedclass

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

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$167$

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(C) The work done by pipes $A, B$,and $C$ in a $3 	ext{ min}$ cycle is:$\frac{1}{20} + \frac{1}{30} - \frac{1}{15} = \frac{3 + 2 - 4}{60} = \frac{1}{60}$ part of the tank.To avoid emptying the tank at the end,we calculate the time taken to reach near the capacity. Let $n$ be the number of cycles.In $3n$ minutes,the tank filled is $\frac{n}{60}$.We need to ensure that after $A$ and $B$ work,the tank is full. The capacity to be filled before the last two steps is $1 - (\frac{1}{20} + \frac{1}{30}) = 1 - \frac{5}{60} = 1 - \frac{1}{12} = \frac{11}{12}$.$\frac{n}{60} \approx \frac{11}{12} \implies n = 55$.In $55 	imes 3 = 165 	ext{ min}$,the tank filled is $\frac{55}{60} = \frac{11}{12}$ part.Remaining part $= 1 - \frac{11}{12} = \frac{1}{12}$.In the $166^{	ext{th}} 	ext{ min}$,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 $167^{	ext{th}} 	ext{ min}$,pipe $B$ fills the remaining $\frac{1}{30}$ part.Total time $= 165 + 1 + 1 = 167 	ext{ min}$.

Two taps $A$ and $B$ can fill a tank in $20 \text{ min}$ and $30 \text{ min}$,respectively. An outlet pipe $C$ can empty the full tank in $15 \text{ min}$. If $A, B$,and $C$ are opened alternatively,each for $1 \text{ min}$,how long (in $\text{min}$) will the tank take to be filled?

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