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(B) Let the work done by taps $A, B,$ and $C$ in $1$ $min$ be $\frac{1}{a}, \frac{1}{b},$ and $\frac{1}{c}$ respectively.Given that the three taps tog

Vedclass · Accepted Answer

$24$

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(B) Let the work done by taps $A, B,$ and $C$ in $1$ $min$ be $\frac{1}{a}, \frac{1}{b},$ and $\frac{1}{c}$ respectively.Given that the three taps together fill the cistern in $10$ $min$,so $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{10}$.Given that tap $A$ alone fills it in $30$ $min$,so $\frac{1}{a} = \frac{1}{30}$.Given that tap $B$ alone fills it in $40$ $min$,so $\frac{1}{b} = \frac{1}{40}$.Substituting these values into the first equation:$\frac{1}{30} + \frac{1}{40} + \frac{1}{c} = \frac{1}{10}$.$\frac{1}{c} = \frac{1}{10} - \frac{1}{30} - \frac{1}{40}$.Taking the least common multiple $(LCM)$ of $10, 30,$ and $40$,which is $120$:$\frac{1}{c} = \frac{12 - 4 - 3}{120} = \frac{5}{120} = \frac{1}{24}$.Therefore,tap $C$ alone will take $24$ $min$ to fill the cistern.

Three taps $A, B$ and $C$ together can fill an empty cistern in $10$ $min.$ The tap $A$ alone can fill it in $30$ $min$ and the tap $B$ alone can fill it in $40$ $min.$ How long will the tap $C$ alone take to fill it? (in $minutes$)

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