Taps A and B can fill an empty tank in 3 hours and 5 hours, respectively. Tap C can empty full tank in 15/2 hours. If all three taps are open at the same time, in how many hours will the tank be full?

Dear Student,

Please find below the solution to the asked query:

Given  : Time taken by tap A to fill the tank = 3 hours

Time taken by tap B to fill the tank = 5 hours

Time taken by tap C to empty the full tank = 7.5 hours (15/2)
So,

Tap A can fill tank in 1 hour   =$\frac{1}{3}$  of cistern   ,

Tap B can fill tank in 1 hour   = $\frac{1}{5}$ of cistern
And

Tap C can empty the tank in 1 hour   = $\frac{2}{15}$ of cistern
As tank being emptied by tap C .

Therefore,

( A + B + C ) ’s 1 hours net work =  $\frac{1}{3}$ + $\frac{1}{5}$ - $\frac{2}{15}$ =  $\frac{5+ 3 - 2}{15} = \frac{6}{15}= \frac{2}{5}$

So,

We can say that when all three taps open tank can fill in 2.5 hours .                                       ( Ans )


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