Two water taps together can fill a tank in `9 3/8`. 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.

#### Solution

Let the first water tape takes *x* hours to fill the tank. Then the second water tape will takes (x + 10) hours to fill the tank.

Since, the faster water tape takes *x* hours to fill the tank.

Therefore, portion of the tank filled by the faster water tape in one hour = 1/x

So, portion of the tank filled by the faster water tape in `9 3/8` hours `= 75/(8x)`

Similarly,

Portion of the tank filled by the slower water tape in `9 3/8` hours `=75/(8(x+10))`

It is given that the tank is filled in `9 3/8` hours.

So,

`75/(8x)+75/(8(x + 10))=1`

`(75(x+10)+75x)/(8x(x+10))=1`

75x + 750 + 75x = 8x2 + 80x

8x2 + 80x - 150x - 750 = 0

8x2 - 70x - 750 = 0

2(4x2 - 35x - 375) = 0

4x2 - 35x - 375 = 0

4x2 - 60x + 25x - 375 = 0

4x(x - 15) + 25(x - 15) = 0

(4x + 25)(x - 15) = 0

4x + 25 = 0

4x = -25

x = -25/4

Or

x - 15 = 0

x = 15

But, x cannot be negative.

Therefore, when x = 15 then

x + 10 = 15 + 10 = 25

Hence, the first water tape will takes 15 hours to fill the tank, and the second water tape will takes 25 hours to fill the tank.