Answer :

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.


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