CAT 2023 Slot 2 · Quant · Question 11

All three must be at the same spot at the same instant, so start with the pair whose data is complete — Ravi and Ashok.

They approach head-on, so their speeds add: the closing speed is 40+50=90 km/h.

Converting, 90 km/h is 90\times \frac{5}{18}=25 metres per second.

They are 225 metres apart, so they meet after \frac{225}{25}=9 seconds.

That fixes the meeting instant, and Vijay must arrive at the same point at the same 9-second mark.

Vijay is 54 metres behind Ravi and travelling the same way, so he must close that 54-metre gap in 9 seconds.

His speed advantage over Ravi must therefore be \frac{54}{9}=6 metres per second.

Converting back, 6 m/s is 6\times \frac{18}{5}=21.6 km/h.

So Vijay's speed is 40+21.6=61.6 km/h — the fourth option.

Shortcut. The whole question is two relative-speed steps, and the second never needs absolute positions: the 54-metre gap and the 9-second window give the extra speed directly. Working in metres per second throughout and converting once at the end avoids every awkward fraction.