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A tank is filled by three pipes with uniform flow. The first two pipes operating simultaneously fill the tank in the same time during which the tank is

filled by the third pipe alone. The second pipe fills the tank 5 hours faster than the first pipe and 4 hours slower than the third pipe. The time required by the first pipe is:



Let x hours be the time taken by ist pipe alone to fill the tank

then time taken by 2nd & 3rd pipe will be (x-5) & (x-9) hrs respectively

So, 1/x + 1/(x-5) = 1/(x-9)

x-5+x/x(x-5) = 1(x-9)

(2x-5)(x-9) = x(x-5)

x^2 - 18x + 45 = 0

(x-15) (x-3) = 0

(neglecting x=3) we have x =15

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