Management Aptitude Test Sep 2013 Paper

(b) Let the pipe B is closed x min before, it get completely filled.
Now, pipe A is open for 20 min
Similarly, pipe B is closed after (20 - x) min.
Then, by given condition,
$\frac{1}{30}$ x 20 + $\frac{1}{40}$ (20 - x) = 1
$\frac{2}{3}+\frac{20-\mathit{x}}{40}$ = 1
$\frac{80+60-3\mathit{x}}{120}=1$
140 -3x = 120
140 -120 = 3x ⇒ 3x = 20 ⇒ x = $\frac{20}{3}$
∴ Pipe B is closed after (20 - x) min
= 20 - $\frac{20}{3}$ = $\frac{60-30}{3}=\frac{40}{3}$ = 13 $\frac{1}{3}$ min