CAT 2022 Slot 3 · Quant · Question 21

Both ships move straight towards the port, so the angle at the port stays {60}^{∘} throughout.

Initially each ship is 24 km from the port.

After the slower ship covers 8 km, it is 24-8=16 km from the port.

Let the faster ship then be d km from the port, so the triangle has sides 16 and d meeting at {60}^{∘}.

The right angle cannot be at the port, since that angle is fixed at {60}^{∘}.

Suppose the right angle is at the slower ship. Then the port-to-faster side is the hypotenuse, and \cos{60}^{∘}=\frac{16}{d}, giving d=32.

That is impossible, since the faster ship started only 24 km out and has been moving closer.

So the right angle is at the faster ship, making the port-to-slower side the hypotenuse.

Then \cos{60}^{∘}=\frac{d}{16}, so d=16\times \frac{1}{2}=8 km.

The faster ship has therefore covered 24-8=16 km while the slower covered 8 km.

Their speeds are in the ratio 16:8=2:1.

The faster ship still has 8 km to go, and in that time the slower covers half as much, that is 4 km.

The slower ship's remaining distance is 16-4=12 km — the second option.

Shortcut. The angle at the port never changes, so the right angle must sit at one of the ships, and only the faster one gives a distance under 24. Once the speed ratio is seen to be 2:1, the final step is simply halving the faster ship's remaining 8 km and subtracting.