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Question
Two pipes flowing together can fill a cistern in 6 minutes. If one pipe takes 5 minutes more than the other to fill the cistern, find the time in which each pipe would fill the cistern.
Solution
Let the time taken by the two pipes to fill the cistern be x and x + 5 min. respectively.
In 1 min., the first pipe can fill `(1)/x` of the cistern. In 1 min., the second pipe can fill `(1)/(x + 5)` of the cistern then
`(1)/x + (1)/(x + 5) = (1)/(6)`
⇒ `(x + 5 + x)/(x(x + 5)) = (1)/(6)`
⇒ `(2x + 5)/(x^2 + 5x) = (1)/(6)`
⇒ x2 + 5x = 12x + 30
⇒ x2 - 7x - 30 = 0
⇒ x2 - 10x + 3x - 30 = 0
⇒ x(x - 10) + 3(x - 10) = 0
⇒ (x - 10)(x + 3) = 0
⇒ x - 10 = 0 or x = -3
⇒ x = 10 or x = -3
Since, time cannot be negative.
So, x = 10 and x + 5 = 10 + 5 = 15.
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