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, find the time in which each pipe would/fill the tank.

Answer 1

Let the time taken by one pipe = x minutes

Then time taken by second pipe = (x + 5) minutes

Time taken by both pipes = `11(1)/(9)`  minutes

Now according to the condition.

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

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

⇒ `(x + 5 + x)/(x^2 + 5x) = (9)/(100)`

⇒ `(2x + 5)/(x^2 + 5x) = (9)/(100)`

⇒ 9x² + 45x = 200x + 500

⇒ 9x² + 45x - 200x - 500 = 0

⇒ 9x² - 155x - 500 = 0

⇒ 9x² - 180x + 25x - 500 = 0

⇒ 9x(x - 20) + 25(x - 20) = 0

⇒ (x - 20)(9x + 25) = 0

Either x - 20 = 0,

then x = 20.

or

9x + 25 = 0,

then 9x = -25

⇒ x = `(-25)/(9)`

but is not possible as it is in negative.

x = 20

Hence, the first pipe can fill the tank in 20 minutes

and second pipe can do the same in 20 + 5 = 25 minutes.