CAT 2025 Slot2 Quant: Rita and Sneha can row a boat at 5 km/h and 6 km/h in still water,

Step 1: Find the River Velocity ($v$)
Sneha's speed ($u_s$) = 6 km/h. Distance ($d$) = 14 km.
Time difference = 48 mins = $\frac{48}{60} = \frac{4}{5}$ hours.
Equation: $\frac{14}{6-v} - \frac{14}{6+v} = \frac{4}{5}$
$14 \left[ \frac{(6+v)-(6-v)}{36-v^2} \right] = \frac{4}{5} \implies \frac{14(2v)}{36-v^2} = \frac{4}{5}$
$70v = 36 - v^2 \implies v^2 + 70v - 36 = 0$. (By observation $v=1$: $1+70-36 \neq 0$, let's recheck calculation).
$14 \times 2v \times 5 = 4(36-v^2) \implies 35v = 36-v^2 \implies v^2 + 35v - 36 = 0$
$(v+36)(v-1) = 0 \implies \mathbf{v = 1 \text{ km/h}}$.

Step 2: Calculate Rita's One-Way Distance ($D$)
Rita's speed ($u_r$) = 5 km/h. Total time = 100 mins = $\frac{100}{60} = \frac{5}{3}$ hours.
$\frac{D}{5-1} + \frac{D}{5+1} = \frac{5}{3} \implies \frac{D}{4} + \frac{D}{6} = \frac{5}{3}$
$\frac{3D + 2D}{12} = \frac{5}{3} \implies \frac{5D}{12} = \frac{5}{3} \implies \mathbf{D = 4 \text{ km}}$.

Step 3: Total Distance Covered
Rita goes from point A to B and returns to A.
Total Distance = $D + D = 4 + 4 = \mathbf{8 \text{ km}}$.