CAT 2025 Slot 2 · Quant · Question 19

Let the river flow at v km/h, so Sneha rows at 6-v upstream and 6+v downstream.

48 minutes is \frac{48}{60}=0.8 hours.

So \frac{14}{6-v}-\frac{14}{6+v}=0.8.

Combining the fractions, \frac{14\left[ \left( 6+v \right)-\left( 6-v \right) \right]}{36-{v}^{2}}=\frac{28v}{36-{v}^{2}}=0.8.

Clearing, 28v=28.8-0.8{v}^{2}, so 0.8{v}^{2}+28v-28.8=0, that is {v}^{2}+35v-36=0.

Factorising, \left( v+36 \right)\left( v-1 \right)=0, and a speed is positive, so v=1 km/h.

Rita therefore rows at 5-1=4 km/h upstream and 5+1=6 km/h downstream.

Covering the same distance d each way in \frac{100}{60}=\frac{5}{3} hours: \frac{d}{4}+\frac{d}{6}=\frac{5}{3}.

So \frac{5d}{12}=\frac{5}{3} and d=4 km.

The question asks for the total distance, which is both legs: 2\times 4=8 km.

Shortcut. For a round trip the average speed is the harmonic mean of 4 and 6, namely \frac{2\times 4\times 6}{4+6}=4.8 km/h. In \frac{5}{3} hours that covers 4.8\times \frac{5}{3}=8 km — the total distance, in one multiplication, with no need to find d first.