#### Question

Two pipes running together can fill a tank in `11 1/9` minutes. If one pipe takes 5 minutes more than the other to fill the tank separately, find the time in which each pipe would fill the tank separately.

#### Solution

Let the first pipe takes *x* minutes to fill the tank. Then the second pipe will takes (x + 5)minutes to fill the tank.

Since, the first pipe takes *x* minutes to fill the tank.

Therefore, portion of the tank filled by the first pipe in one minutes = 1/x

So, portion of the tank filled by the first pipe in `11 1/9` minutes `=100/(9x)`

Similarly,

Portion of the tank filled by the second pipe in `11 1/9` minutes `=100/(9(x+5))`

It is given that the tank is filled in `11 1/9`minutes.

So,

`100/(9x)+100(9(x+5))=1`

`(100(x+5)+100x)/(9x(x+5))=1`

100x + 500 + 100x = 9x^{2} - 45x

9x^{2} + 45x - 200x - 500 = 0

9x^{2} - 155x - 500 = 0

9x^{2} - 180x + 25x - 500 = 0

9x(x - 20) + 25(x - 20) = 0

(x - 20)(9x + 25) = 0

x - 20 = 0

x = 20

Or

9x + 25 = 0

9x = -25

x = -25/9

But, x cannot be negative.

Therefore, when *x* = 20 then

x + 5 = 20 + 5 = 25

Hence, the first water tape will takes 20 min to fill the tank, and the second water tape will take 25 min to fill the tank.