CAT 2023 Slot 2 · Quant · Question 10

Let Pipe A take a hours to fill the tank, so Pipe B empties it in a-1 hours, and let Pipe C take c hours.

All three together fill the tank in 2 hours, so \frac{1}{a}-\frac{1}{a-1}+\frac{1}{c}=\frac{1}{2}, the minus sign marking B as a drain.

In the second scenario, B and C run together for one hour, filling \frac{1}{c}-\frac{1}{a-1} of the tank.

Then C alone runs for one hour fifteen minutes, that is \frac{5}{4} hours, filling \frac{5}{4c}.

Together these complete the tank: \frac{1}{c}-\frac{1}{a-1}+\frac{5}{4c}=1, that is \frac{9}{4c}-\frac{1}{a-1}=1.

Subtracting the first equation from this one, the \frac{1}{a-1} terms cancel: \frac{9}{4c}-\frac{1}{c}-\frac{1}{a}=1-\frac{1}{2}.

So \frac{5}{4c}-\frac{1}{a}=\frac{1}{2}.

Test the options, remembering a<5. Take c=\frac{3}{2} hours, that is 90 minutes.

Then \frac{5}{4c}=\frac{5}{6}, so \frac{1}{a}=\frac{5}{6}-\frac{1}{2}=\frac{1}{3} and a=3 hours, comfortably under five.

Verifying the first condition with a=3 and c=\frac{3}{2}: \frac{1}{3}-\frac{1}{2}+\frac{2}{3}=\frac{1}{2}, which holds.

Verifying the second: \frac{9}{4}\times \frac{2}{3}-\frac{1}{2}=1.5-0.5=1, which also holds.

So Pipe C takes 90 minutes — the first option.

Shortcut. Subtracting the two scenarios cancels the awkward drain term \frac{1}{a-1} at once, leaving the single clean relation \frac{5}{4c}=\frac{1}{2}+\frac{1}{a}. From there the options can simply be tried, and the condition a<5 confirms the one that fits.