Three pipes can fill a water tank constantly.The time taken for the first two pipes working simultaneously to fill the tank is equivalent to time taken by the third pipe to fill it alone.If the second pipe fills the tank in 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:

Answer 1

T1*T2/(T1+T2) = T3.
T1*(T1-10)/(T1+(T1-10)) = T1-18.

(T1^2-10T1)/(2T1-10) = T1-18.
Multiply by 2T1-10:
T1^2-10T1 = 2T1^2-36T1-10T1+180
-T1^2 + 36T1 - 180 = 0
T1^2 - 36T1 + 180 = 0.
-6*-30 = 180, -6 + -30 = -36 = B.
(T1-6)(T1-30) = 0.

T1-6 = 0, T1 = 6 h.

T1-30 = 0, T1 = 30 Hours = Solution.

Thirty hours was selected because T2 cannot be 10 hours less than six hours.