Pipe A Can Fill A Cistern In 6 Hours Less Than Pipe B. Both The Pipes

1.	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?
Answer» 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. 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.

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?

Answer»

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.

Let's assume time required by Pipe A to fill the cistern = X hours