CAT 2024 Slot 1 · Quant · Question 15

Let Anil's speed be p and Sunil's be s, both in km/h.

They approach at p+s and meet after 1.5 hours, so \left( p+s \right)\times 1.5=45 and p+s=30.

At the meeting point Anil has covered 1.5p and Sunil has covered 1.5s.

After meeting, Anil still has Sunil's stretch to cover, taking \frac{1.5s}{p} hours.

Likewise Sunil still has Anil's stretch, taking \frac{1.5p}{s} hours.

Anil finishes 1 hour 15 minutes later, so \frac{1.5s}{p}-\frac{1.5p}{s}=1.25.

Testing p=12 with s=30-12=18: the two times are \frac{27}{12}=2.25 and \frac{18}{18}=1 hours.

Their difference is 2.25-1=1.25 hours, which matches, so Anil's speed is 12 km/h.

Shortcut. For two travellers starting together, the speeds are in the inverse ratio of the square roots of their post-meeting times: \frac{p}{s}=\sqrt{\frac{{t}_{s}}{{t}_{p}}}=\sqrt{\frac{1}{2.25}}=\frac{2}{3}. Combined with p+s=30 that gives p=12 at once.