Question

A tap can fill an empty tank in 12 Hours and another tap can empty half the tank in 10 hours l. it both the taps are opened simpultaneously how long would it take for the empty tank to be filled to to half its capacity.​

Answer 1

Let, the total volume of the tank be x litres.

In case of first tap,

in 12 hrs, it supplies x litres of water

So, in 1 hr it supplies x/12 litres of water

And, in case of 2nd tap,

in 10 hrs, it removes x/2 litres of water

In 1 hr, it removes x/20 litres of water

Therefore after 1 hr the ultimate amount of water remaining in the tank, when both the taps are on..

=X/12-X/20.

=x/30

Let the required number of hours be Y

Then (x/30)*Y=X

Y=30 hrs (ans)

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Answer 2

Let, the total volume of the tank be x litres.

In case of first tap,

in 12 hrs, it supplies x litres of water

So, in 1 hr it supplies x/12 litres of water

And, in case of 2nd tap,

in 10 hrs, it removes x/2 litres of water

In 1 hr, it removes x/20 litres of water

Therefore after 1 hr the ultimate amount of water remaining in the tank, when both the taps are on..

=X/12-X/20.

=x/30

Let the required number of hours be Y

Then (x/30)*Y=X

Y=30 hrs (ans)

I hope my answer is helpful for you ✌️☺️.