CAT 2022 Slot 1 · Quant · Question 6

Let the two trains meet t minutes after they start.

After meeting, train A still has to cover the stretch train B has already covered, and train B still has to cover the stretch train A has already covered.

Train A's whole journey takes 10 minutes, so after the meeting it needs 10-t minutes.

Train B needs 9 minutes after the meeting, as given.

Before the meeting A covers t minutes' worth at its own speed; after it, B covers that same distance in 9 minutes.

So the ratio of speeds is \frac{\text{speed of A}}{\text{speed of B}}=\frac{9}{t}.

Applying the same reasoning the other way, \frac{\text{speed of A}}{\text{speed of B}}=\frac{t}{10-t}.

Equating the two, \frac{9}{t}=\frac{t}{10-t}, so {t}^{2}=9\left( 10-t \right).

Rearranging, {t}^{2}+9t-90=0, which factors as \left( t+15 \right)\left( t-6 \right)=0.

Time cannot be negative, so t=6 minutes.

Train B's total journey is the time before the meeting plus the time after: 6+9=15 minutes — the first option.

Shortcut. The meeting time is the geometric mean of the two post-meeting times, so t=\sqrt{\left( 10-t \right)\times 9}. Alternatively, since A's full trip is 10 minutes and it takes 10-t after meeting, trying t=6 gives \sqrt{4\times 9}=6 — a hit on the first guess.