Question

Pipes A and B running together can fill a cistern in 6 minutes. If B takes 5 minutes more than A to fill the cistern then, what will be the respective times in which A and B will fill the cistern separately.

Answer 1

Suppose pipe A fills the cistern in x min

Therefore pipe B will fill the cistern in (x + 5) min.

\(\therefore\) In one minute, pipe A and B can together fill \(\Big[\frac{1}{X}+\frac{1}{X+5}\Big]\) part of the cistern

Given \(\frac{1}{X}+\frac{1}{X+5}\) = \(\frac{1}{6}\)

\(\Rightarrow\) \(\frac{X+5+X}{X(X+5)}\)= \(\frac{1}{6}\) \(\Rightarrow\) \(\frac{2X+5}{X^2+5}\) = \(\frac{1}{6}\)

\(\Rightarrow\) 12x + 30 = x² + 5

\(\Rightarrow\) x²– 7x – 30 = 0

\(\Rightarrow\) x² – 10x + 3x – 30 = 0

\(\Rightarrow\) x(x – 10) + 3(x – 10) = 0

\(\Rightarrow\) (x + 3) (x – 10) = 0

\(\Rightarrow\) x = –3 or 10

Neglecting negative values x = 10

\(\therefore\) A and B will the cistern separately in 10 min and 15 min.