Question 136

Two pipes can fill a tank in 10 hours and 16 hours respectively. A third pipe can empty the tank in 40 hours. If all the three pipes are opened and they function simultaneously then in how much time the tank will be full ? (in hours)

Solution

If all the three pipes are opened simultaneously, part of the tank filled in 1 hour

= $$\frac{1}{10} + \frac{1}{16} - \frac{1}{40}$$

= $$\frac{8 + 5 - 2}{80} = \frac{11}{80}$$

$$\therefore$$ Required time = $$\frac{1}{\frac{11}{80}}$$

= $$\frac{80}{11} = 7\frac{3}{11}$$ hours

=> Ans - (D)

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