Three pipes can fill a water tank constant flow. The time taken by the first two pipes working simultaneously to fill the tank is equivalent to the time taken by the third pipe to fill it alone . If the second pipe fills the tank 10 hours faster than the first pipe and 8 hours slower than the third pipe then find out the time required by the first pipe is :
Solution: Suppose, first pipe alone takes x hours to fill the tank. Then, second and third pipes will take (x - 10) and (x - 18) hours respectively to fill the tank.
.'. 1/x + 1/(x - 10) = 1/(x - 18)
=> (x -10 + x)/(x(x -10)) = 1(x -18)
=> (2x-10) (x-18) = x(x-10)
=> x^2 - 36x + 180 = 0.
=> (x - 30) (x - 6) = 0.
=> x =30, x =6 (neglecting x=6)
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