Two pipes,A and B, can separately fill a cist | vedclass

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Question

(A) Work done by pipe $A$ in $1$ minute $= \frac{1}{15}$.Work done by pipe $B$ in $1$ minute $= \frac{1}{18}$.Work done by pipes $(A+B)$ in $6$ minute

Vedclass · Accepted Answer

$16 \frac{1}{2} 	ext{ minutes}$

Vedclass · Answer

(A) Work done by pipe $A$ in $1$ minute $= \frac{1}{15}$.Work done by pipe $B$ in $1$ minute $= \frac{1}{18}$.Work done by pipes $(A+B)$ in $6$ minutes $= 6 	imes (\frac{1}{15} + \frac{1}{18}) = 6 	imes (\frac{6+5}{90}) = 6 	imes \frac{11}{90} = \frac{11}{15}$.So,$\frac{11}{15}$ of the cistern is filled in $6$ minutes.Now,all three pipes $(A+B+C)$ are opened. The net work done in $1$ minute $= \frac{1}{15} + \frac{1}{18} - \frac{1}{6} = \frac{6+5-15}{90} = \frac{-4}{90} = -\frac{2}{45}$.The negative sign indicates that the cistern is being emptied at a rate of $\frac{2}{45}$ per minute.Time taken to empty $\frac{11}{15}$ of the cistern $= \frac{11/15}{2/45} = \frac{11}{15} 	imes \frac{45}{2} = \frac{33}{2} = 16 \frac{1}{2}$ minutes.

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