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Question:
Two pipers running together can fill a cistern in 3/40 minutes. If one pipe takes 3 minutes more than the other to fill it, find the time in which each pipe would fill the cistern.
Answer:

Let one pipe take x minutes to fill the cistern.

So, in 1 minute, it will fill 1/x part of the cirstern.

Then other pipe take x + 3 minutes to fill the cistern.

So, in 1 minute, it will fill 1/(x + 3) part of the cirstern.

Now, two pipes together fill cistern in 40/13 minutes

So, two pipes together fill cistern in 1 minutes = 1/(40/13) = 13/40

Now, 1/x + 1/(x + 3) = 13/40

=> (x + 3 + x)/{x(x + 3)} = 13/40

=> (2x + 3)/(x2 + 3x) = 13/40

=> 40(2x + 3) = 13(x2 + 3x)

=> 13x2 + 39x = 80x + 120

=> 13x2 + 39x - 80x - 120 = 0

=> 13x2 - 41x - 120 = 0

=> 13x2 - 65x + 24x - 120 = 0

=> 13x(x - 5) + 24(x - 5) = 0

=> (x - 5)*(13x + 24) = 0

=> x = 5, -24/13

Since x can not be negative,

So, x = 5

So, the first pipe take 5 minutes and second pipe take 8 minutes to fill the cistern.

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