Question 48

A tap can empty a tank in one hour. A second tap can empty it in 30 min If both the taps operate simultaneously, how much time is needed to empty the tank ?

Solution

1 minute work of tap A = $$\frac{1}{60}$$

1 minute work of tap B = $$\frac{1}{30}$$

=> (A + B)'s 1 minute work = $$\frac{1}{60} + \frac{1}{30}$$

= $$\frac{1 + 2}{60} = \frac{3}{60}$$

= $$\frac{1}{20}$$

$$\therefore$$ Time taken by (A + B) to empty the tank = 20 min

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