CAT 2022 Slot 3 · Quant · Question 10

Let the faster car travel at u km/h and the slower at v=60 km/h, with the gap between them being D km.

Travelling towards each other, they close the gap at u+v, taking 1.5 hours: D=1.5\left( u+v \right).

Travelling the same way, the faster closes at u-v, taking 10.5 hours: D=10.5\left( u-v \right).

Setting the two expressions equal, 1.5\left( u+v \right)=10.5\left( u-v \right).

Dividing both sides by 1.5, u+v=7\left( u-v \right).

Expanding, u+v=7u-7v, so 8v=6u.

That gives u=\frac{4v}{3}, confirming u>v as expected.

With v=60, the faster car travels at \frac{4\times 60}{3}=80 km/h.

The question asks only how far the slower car goes before they meet head-on, which takes 1.5 hours.

That distance is 60\times 1.5=90 km — the third option.

Shortcut. The final step needs only the slower car's own speed and the 1.5-hour meeting time, so 60\times 1.5=90 can be written down immediately. The 10.5-hour condition is there to tempt you into solving for the gap, which is never required.