Question

Two pipes running together can fill an empty cistern in `4 20/21` min. If one pipe takes 5 minutes more than the other takes to fill the empty cistern. Find the time in which each pipe would fill the cistern.

Answer 1

Let the time taken by each pipe be x minutes and y minutes.

It is given that one pipe takes 5 minutes more than the other pipe to fill the empty cistern.

⇒ x = y + 5

So, one minute work of the pipe = `1/x = 1/(y+5)`

Both pipes running together can fill the empty cistern in `4 20/21 = 104/21` minutes.

⇒ `1/x + 1/y = 21/104`

⇒ `1/(y + 5) = 1/y = 21/104`

⇒ `(1 xx y)/(y(y + 5)) + (1 xx (y + 5))/(y(y + 5)) = 21/104`

⇒ `(y + y + 5)/(y(y + 5) = 21/104`

⇒ `(2y + 5)/(y^2 + 5y) = 21/40`

⇒ 104(2y + 5) = 21(y² + 5y)

⇒ 208y + 520 = 21y² + 105y

⇒ 21y² + 105y − 208y − 520 = 0

⇒ 21y² − 103y − 520 = 0

⇒ 21y² + 65y − 168y − 105 = 0

⇒ y(21y + 65) − 8(21y + 65) = 0

⇒ (21y + 65)(y − 8) = 0

⇒ (21y + 65) = 0 or (y − 8) = 0

⇒ y = `−65/21`​ or y = 8

As minutes cannot be negative.

So, one pipe takes 8 minutes and the other pipe takes 8 + 5 = 13 minutes.

∴ The time taken by pipes = 8 minutes and 13 minutes.