Math, asked by Anonymous, 6 months ago

Tap A can fill a tank in 10 hours, while tap B can empty the tank in 12 hours, If bon
the taps are opened together in the empty tank, in how many hours will the tank you
filled?

Answers

Answered by Anonymous
1

Answer:

60 hour's

Step-by-step explanation:

time taken by tap a to fill the tank = 10 hrs

work done by tap a in 1 hour =1/10

time taken by tap b to empty the tank =12 hours

work done by tap b in 1 hour = 1/12

work done by tap b in 1 hour = 1/10 -1/12= 6-5/60 = 1/60

therefore the time taken by the two taps to fill the tank 60 hours .

Answered by svaishnaviprakash
4

Answer:

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)

Step-by-step explanation:

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