CAT 2023 Slot 3 · Quant · Question 13

Let the distance between the ports be D, the first boat's speed in still water be u, and the river's speed be w.

Downstream in 2 hours gives u+w=\frac{D}{2}.

Upstream in 3 hours gives u-w=\frac{D}{3}.

Adding, 2u=\frac{5D}{6}, so u=\frac{5D}{12}.

Subtracting, 2w=\frac{D}{6}, so w=\frac{D}{12}.

Scale the distance to D=12 for convenience, making u=5 and w=1.

Let the second boat's speed be v; its round trip takes 6 hours, so \frac{12}{v-1}+\frac{12}{v+1}=6.

Combining the left side, \frac{12\left( v+1 \right)+12\left( v-1 \right)}{{v}^{2}-1}=\frac{24v}{{v}^{2}-1}.

So 24v=6\left({v}^{2}-1 \right), which simplifies to {v}^{2}-4v-1=0.

The positive root is v=\frac{4+\sqrt{20}}{2}=2+\sqrt{5}\approx 4.24.

Since 4.24<5, the second boat is the slower one.

Its downstream time from A to B is \frac{12}{v+1}=\frac{12}{3+\sqrt{5}}.

Rationalising by multiplying top and bottom by 3-\sqrt{5}, \frac{12\left( 3-\sqrt{5}\right)}{9-5}=3\left( 3-\sqrt{5}\right).

That is the first option, about 2.29 hours.

Shortcut. The first boat's two times, 2 and 3 hours, fix the stream at one-fifth of that boat's speed without any algebra: the speeds are in the ratio 3:2, so still-water to stream is 5:1. Scaling D=12 then keeps every number whole until the final surd.