Boat A travels downstream from Point X to Point Y in $3$ hours less than the time taken by Boat B to travel upstream from Point Y to Point Z. The distance between X and Y is $20$ km, which is half of the distance between Y and Z. The speed of Boat B in still water is $10$ km/h and the speed of Boat A in still water is equal to the speed of Boat B upstream. What is the speed of Boat A in still water? (Consider the speed of the current to be the same.) [RBI Gr. 'B' (Phase I) Exam, 2015]

### Concept & Algebraic Formulation Break complex relational problems down by expressing all unknown speeds in terms of a single variable, typically the speed of the current. $$ \text{Time} = \frac{\text{Distance}}{\text{Speed}} $$ ### Step-by-Step Solution * Let the speed of the current be $c$. * Analyze Distances: Distance X to Y = $20$ km. Distance Y to Z = $20 \times 2 = 40$ km. * Analyze Boat B: Speed in still water = $10$ km/h. Upstream speed = $10 - c$. Time for B to travel Y to Z upstream = $\frac{40}{10 - c}$. * Analyze Boat A: Speed in still water equals Boat B's upstream speed, so it is $10 - c$. Downstream speed of A = $(10 - c) + c = 10$ km/h. Time for A to travel X to Y downstream = $\frac{20}{10} = 2$ hours. * Set up the equation based on the time relationship: Time of A = Time of B - $3$. $2 = \frac{40}{10 - c} - 3$ * Solve for $c$: $5 = \frac{40}{10 - c}$ $10 - c = 8$ $c = 2$ km/h. * Calculate Boat A's still water speed: $10 - c = 10 - 2 = 8$ km/h. ### Exam Strategy & Shortcut Observe that Boat A's downstream speed perfectly cancels out the current variable $(10 - c + c = 10)$, immediately giving Boat A's time as $2$ hours. This drastically simplifies the equation setup right from the start. ### Common Pitfall Losing track of the variables and forgetting that Boat A's still water speed already incorporates a subtraction of $c$, leading to double-counting the current when calculating downstream speed. ### Final Answer Therefore, the correct answer is **(d) 8 km/h**.