#### Question

One pipe can fill a cistern in 3 hours less than the other. The two pipes together can fill the cistern in 6 hours 40 minutes. Find the time that each pipe will take to fill the cistern.

#### Solution

Let one pipe fill the cistern in x hours and the other fills it in (x – 3) hours.

Given that the two pipes together can fill the cistern in 6 hours 40 minutes, i.e.,

`6 40/60 "hours" = 6 2/3 "hour" = 20/3 "hours"`

`1/x + 1/(x - 3) = 3/20`

`(x - 3 + x)/(x(x - 3)) = 3/20`

`(2x - 3)/(x^2 - 3x) = 3/20`

`40x - 60 = 3x^2 - 9x`

`3x^2 - 49x + 60 = 0`

`3x^2 - 45x - 4x + 60 = 0`

3x(x - 15) - 4(x - 15) = 0

(x - 15)(3x - 4) = 0

`x = 15, 4/3`

if x = 4/3 then x - 3 = `4/3 - 3= (4- 9)/3 = (-5)/3` which is not possible.

So x = 15

Thus oen pipe fill the cistern in 15 hours and other fills in (x - 3) =- 15 - 3 = 12 hours