A cistern is filled by two taps A and B. Tap | vedclass

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Question

(A) The rate of filling by tap $A$ is $\frac{1}{20}$ of the cistern per minute.The rate of filling by tap $B$ is $\frac{1}{25}$ of the cistern per min

Vedclass · Accepted Answer

$11$ $minutes$

Vedclass · Answer

(A) The rate of filling by tap $A$ is $\frac{1}{20}$ of the cistern per minute.The rate of filling by tap $B$ is $\frac{1}{25}$ of the cistern per minute.Combined rate of filling when both taps are open is $\frac{1}{20} + \frac{1}{25} = \frac{5+4}{100} = \frac{9}{100}$ of the cistern per minute.In $5$ $minutes$,the part filled is $\frac{9}{100} 	imes 5 = \frac{9}{20}$.The remaining part to be filled is $1 - \frac{9}{20} = \frac{11}{20}$.Since tap $B$ is turned off,tap $A$ fills the remaining part.Time taken by tap $A$ to fill $\frac{11}{20}$ part is $\frac{11}{20} \div \frac{1}{20} = \frac{11}{20} 	imes 20 = 11$ $minutes$.

$A$ cistern is filled by two taps $A$ and $B$. Tap $A$ can fill it in $20$ $minutes$ and tap $B$ in $25$ $minutes$. Both the taps are kept open for $5$ $minutes$ and then the second tap $(B)$ is turned off. The cistern will be completely filled in another:

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