Pipe A can fill a tank in 12 hours and pipe B can fill it in 15 hours seperately. A third pipe C can empty in 20 hours, Intially pipe A was opened, after one hour pipe B was opened and then after 1 hour when pipe B, was opened pipe C was also opened. In how many total hours the tank will be full?

9 \frac{2}{3}

hours

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6 \frac{2}{3}

hours

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10 hours

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none of these

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Solution

The correct option is A $9 \frac{2}{3}$ hours

Efficiency of pipe A = 8.33%

Efficiency of pipe B = 6.67%

Efficiency of pipe C = 5%

When tap C was opened,pipe A had filled 16.66% capacity

When tap C was opened,pipe B had filled 6.67% capacity

Therefore rest capacity of the tank to be filled = 100 - 23.34 = 76.66%

Now, the net efficiency of A, B and C = 10%

Hence, pipes A, B and C will take = $\frac{76.66}{10}$ = 7.66 = $7 \frac{2}{3}$

Total time = 2hours + $7 \frac{2}{3}$ hours = $9 \frac{2}{3}$ hours.

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