Two taps running together can fill a cistern in minutes. If one tap takes 1 minute more than the other to fill the cistern, find the time in which each tap separately can fill the cistern.

Let the first tap take x minutes to fill the cistern.

So, let the second tap take (x + 1) minutes to fill the cistern.

And let the volume of cistern be V.

Volume of cistern filled by first tap in 1 minute = V/x

Volume of cistern filled by second tap in 1 minute = V/(x + 1)

Volume of cistern filled by both the taps in 1 minute = V/x + V/(x + 1)

Volume of cistern filled by both the taps in minutes i.e. 30/11 minutes =

∴ V =

⇒ = 1

⇒ 30(2x + 1) = 11(x² + x)

⇒ 60x + 30 = 11x² + 11x

⇒ 11x² + 11x – 60x – 30 = 0

⇒ 11x² – 49x – 30 = 0

By factorisation method,

⇒ 11x² – 55x + 6x – 30 = 0

⇒ 11x(x – 5) + 6(x – 5) = 0

⇒ (x – 5) (11x + 6) = 0

⇒ (x – 5) = 0 (or) (11x + 6) = 0

⇒ x = 5 (or) x = -6/11

Since time cannot be negative, x = 5

So, the time taken by the first tap to fill the cistern = 5 minutes.

And the time taken by the second tap to fill the cistern = 5 + 1 = 6 minutes

Ans. The time in which the first pipe and the second tap separately can fill the cistern is 5 and 6 minutes respectively.