A man travels 28 km downstream in a motor boat and immediately returns upstream to his starting point. The upstream journey takes twice as long as the downstream journey. If the speed of the river flow were doubled, the total time for going 28 km downstream and back would be 672 minutes. What are the speeds of the boat in still water and of the river flow (in km/h)?

Introduction / Context:
This is a multi-step boats and streams problem involving both time ratios and a hypothetical change in the speed of the current. You must use relative speed concepts downstream and upstream, together with the given conditions, to determine both the speed of the boat in still water and the speed of the river flow.

Given Data / Assumptions:

Distance downstream = 28 km; same distance upstream back to start = 28 km.
Time upstream is twice the time downstream.
If the speed of the river flow were doubled, the total time for 28 km downstream and back would be 672 minutes = 11.2 hours.
Let the speed of the boat in still water be b km/h.
Let the speed of the river flow be c km/h.
Downstream speed = b + c; upstream speed = b - c (with c < b so that upstream is possible).

Concept / Approach:
We first convert the time ratio into an algebraic relation using times for a fixed distance: time = distance / speed. This gives b in terms of c. Then, for the hypothetical scenario with doubled current speed (2c), we express the total round-trip time in terms of c only and set it equal to 11.2 hours. Solving this equation gives c, and substituting back yields b.

Step-by-Step Solution:
Step 1: Use the given time ratio condition. Downstream time t₁ = 28 / (b + c). Upstream time t₂ = 28 / (b - c). Given t₂ = 2t₁, so 28 / (b - c) = 2 * [28 / (b + c)]. Cancel 28: 1 / (b - c) = 2 / (b + c). Cross-multiply: b + c = 2(b - c) ⇒ b + c = 2b - 2c ⇒ 3c = b. So b = 3c. Step 2: Use the hypothetical doubled current scenario. New current speed = 2c, still-water speed remains b = 3c. New downstream speed = 3c + 2c = 5c. New upstream speed = 3c - 2c = c. Total time T = 28 / (5c) + 28 / c. Given T = 672 minutes = 11.2 hours. Compute T in terms of c: T = 28(1/(5c) + 1/c) = 28[(1 + 5)/(5c)] = 28 * 6/(5c) = 168/(5c). Set 168/(5c) = 11.2 ⇒ 168 = 11.2 * 5c = 56c ⇒ c = 3 km/h. Then b = 3c = 9 km/h.

Verification / Alternative check:
With b = 9 km/h and c = 3 km/h: original downstream speed = 12 km/h, upstream speed = 6 km/h. Downstream time = 28/12 ≈ 2.333 h, upstream time = 28/6 ≈ 4.667 h, and t₂ is twice t₁, as required. With doubled current 6 km/h, downstream speed = 15 km/h, upstream speed = 3 km/h. Total time = 28/15 + 28/3 ≈ 1.867 + 9.333 ≈ 11.2 h (672 minutes), confirming both conditions.

Why Other Options Are Wrong:
12 km/h, 3 km/h and 8 km/h, 2 km/h do not satisfy both the original time ratio and the 672-minute condition simultaneously.
9 km/h, 6 km/h makes the current too fast; with b = 9 and c = 6, upstream speed would be only 3 km/h and the original time ratio would not match the given 2:1 relation correctly for 28 km each way.

Common Pitfalls:
Many students incorrectly add or average speeds instead of using the formula time = distance / speed in each phase. Another common mistake is forgetting to convert minutes to hours when inserting the 672-minute condition. Keeping symbols for b and c, solving the time-ratio equation first, and only then applying the second condition keeps the solution clear and manageable.

Final Answer:
The speed of the boat in still water is 9 km/h, and the speed of the river flow is 3 km/h.