A water tank is filled by two pipes in 48 minutes. If the larger pipe alone can fill it in 40 minutes less time than the smaller pipe, find the time it takes for the smaller pipe to fill the tank alone.
+128406 Let the time for one pipe to fill the tank be = T (in minutes)
Let the time for the other pipe to fill the T + 40 (in minutes)
So....in one minute the first pipe fills 1/T of the tank and the second fills 1 / (T + 40) of the tank
And both pipes fill1/48 of the tank in one minute
So....we have this equation
1 /T + 1/ (T + 40) = 1/48 simplify
(T + 40 + T ) / [ (T) (T + 40) ] = 1/48 cross-multiply
48 (2T + 40) = T(T + 40)
96T + 1920 = T^2 + 40T rearrange as
T^2 - 56T - 1920 = 0 factor as
(T - 80) (T + 24) = 0
The first factor set to 0 and solved produces a positive solution for T = 80 (min)
The time for the smaller pipe to fill the tank = T + 40 = 120 min
