Boats and Streams – Round trip time with a known current: The stream current is 1 km/h. A motorboat goes 35 km upstream and returns 35 km downstream to the start in a total of 12 hours. What is the speed of the motorboat in still water?

Introduction / Context:This is a symmetric up-and-down trip with a fixed current. Total time is the sum of upstream and downstream legs. The unknown still-water speed u is found by solving a rational equation built from the two legs.

Given Data / Assumptions:

• Current v = 1 km/h.
• Upstream distance = downstream distance = 35 km.
• Total time upstream + downstream = 12 h.

Concept / Approach:Time upstream = 35/(u − 1); time downstream = 35/(u + 1). Set their sum to 12 and solve for u by clearing denominators.

Step-by-Step Solution:35/(u − 1) + 35/(u + 1) = 12.35[(u + 1) + (u − 1)]/(u^2 − 1) = 12 ⇒ 70u/(u^2 − 1) = 12.12u^2 − 12 − 70u = 0 ⇒ 12u^2 − 70u − 12 = 0.Discriminant = 5476 = 74^2 ⇒ u = (70 + 74)/24 = 144/24 = 6 km/h.

Verification / Alternative check:Times: 35/5 = 7 h upstream and 35/7 = 5 h downstream sum to 12 h.

Why Other Options Are Wrong:Values other than 6 km/h do not satisfy the time equation when substituted.

Common Pitfalls:Forgetting to use the same distance for both legs or miscomputing the discriminant when solving the quadratic.

Final Answer:6 km/h