Arithmetic Aptitude :: Pipes & Cistern
6. Pipe A can fill a cistern in 6 hours less than Pipe B. Both the pipes together can fill the cistern in 4 hours. How much time would A take to fill the cistern all by itself?
Let's assume time required by Pipe A to fill the cistern = X hours
So Time required by Pipe B to fill the cistern = (X + 6) hours
? Both Pipes (A+B) can fill cistern in 1 hour = [1/X + 1/(X + 6)]
Given Both pipe fill the cistern in 4 hours
=> [1/X + 1/(X + 6)] = 1/4 => [(X+6) + X]/(X+6)*x = 1/4
4X + 24 + 4X = X2 + 6x
X2 - 2X - 24 = 0
(X-6)(X+4) = 0
=> A can fill cistern in 6 hours.