There are two taps A and B to fill up a water | vedclass

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(D) Let the total capacity of the tank be $1$ unit.Rate of tap $A$ and tap $B$ working together $= \frac{1}{40}$ tank per minute.Rate of tap $A$ worki

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$120$

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(D) Let the total capacity of the tank be $1$ unit.Rate of tap $A$ and tap $B$ working together $= \frac{1}{40}$ tank per minute.Rate of tap $A$ working alone $= \frac{1}{60}$ tank per minute.Let the time taken by tap $B$ alone to fill the tank be $x$ minutes.Therefore,the rate of tap $B$ is $\frac{1}{x}$ tank per minute.We know that: $(	ext{Rate of } A) + (	ext{Rate of } B) = 	ext{Combined Rate}$.$\frac{1}{60} + \frac{1}{x} = \frac{1}{40}$.$\frac{1}{x} = \frac{1}{40} - \frac{1}{60}$.Taking the least common multiple $(LCM)$ of $40$ and $60,$ which is $120$:$\frac{1}{x} = \frac{3 - 2}{120} = \frac{1}{120}$.Thus,$x = 120$ minutes.

There are two taps $A$ and $B$ to fill up a water tank. The tank can be filled in $40$ $min,$ if both taps are on. The same tank can be filled in $60$ $min,$ if tap $A$ alone is on. How much time (in $minutes$) will tap $B$ alone take to fill up the same tank?

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