Two water taps can fill a tank in 9 3/8 hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately . Find the time in which each tap can separately fill the tank?

### Asked by Aryan.sharma30j | 10th Feb, 2018, 01:19: PM

Let the larger diameter tap fills the tank alone in (*x* – 10) hours.

In 1 hour, the smaller diameter tap can fill 1/*x* part of the tank.

In 1 hour, the larger diameter tap can fill 1/(*x* – 10) part of the tank.

Two water taps together can fill a tank in 75 / 8 hours.

But in 1 hour the taps fill 8/75 part of the tank.

1 / x + 1 / (x – 10) = 8 / 75.

( x – 10 + x ) / x ( x – 10) = 8 / 75.

2( x – 5) / ( x^{2} – 10 x) = 8 / 75.

4x^{2} – 40x = 75x – 375.

4x^{2} – 115x + 375 = 0

4x^{2} – 100x – 15x + 375 = 0

4x ( x – 25) – 15( x – 25) = 0

( 4x -15)( x – 25) = 0.

x = 25, 15/ 4.

But x = 15 / 4 then x – 10 = -25 /4 which is not possible since time

But x = 25 then x – 10 = 15.

Larger diameter of the tap can the tank 15 hours and smaller diameter of the tank can fill

the tank in 25 hours.

### Answered by | 10th Feb, 2018, 04:11: PM

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