Two taps A and B can fill a tank in 25 min an | vedclass

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(D) Let the capacity of the tank be $1$ unit.Normal rate of tap $A = \frac{1}{25}$ tank per minute.Normal rate of tap $B = \frac{1}{20}$ tank per minu

Vedclass · Accepted Answer

$15$

Vedclass · Answer

(D) Let the capacity of the tank be $1$ unit.Normal rate of tap $A = \frac{1}{25}$ tank per minute.Normal rate of tap $B = \frac{1}{20}$ tank per minute.Due to improper opening,the actual rate of tap $A = \frac{1}{25} 	imes \frac{5}{6} = \frac{1}{30}$ tank per minute.Actual rate of tap $B = \frac{1}{20} 	imes \frac{2}{3} = \frac{1}{30}$ tank per minute.Combined rate of both taps $= \frac{1}{30} + \frac{1}{30} = \frac{2}{30} = \frac{1}{15}$ tank per minute.Therefore,the time taken to fill the tank $= \frac{1}{1/15} = 15$ $min$.

Two taps $A$ and $B$ can fill a tank in $25$ $min$ and $20$ $min$ respectively. However,the taps are not opened properly,so taps $A$ and $B$ allow only $\frac{5}{6}$ and $\frac{2}{3}$ of their normal water flow,respectively. How long will they take to fill the tank (in $min$)?

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