The ratio of motorboat speed to current speed is 35 : 5 (that is, v : c = 7 : 1). The motorboat goes with the current in 5 h 10 min. How long will it take to come back against the current (choose the nearest option provided)?

Introduction / Context:
Given the ratio of still water speed to current, we can treat both as multiples of a common unit k. The downstream time gives the downstream distance, which we then use to compute the upstream time over the same distance. Since the options are in coarse 10-minute increments, select the closest.

Given Data / Assumptions:

• v : c = 35 : 5 ⇒ v = 7k and c = k.
• Downstream time t_down = 5 h 10 min = 31/6 h.
• Distance downstream equals distance upstream.

Concept / Approach:
Downstream speed = v + c = 8k; upstream speed = v − c = 6k. Let the one-way distance be D. Then D = (downstream speed) * (downstream time) = 8k * (31/6). Upstream time = D / (6k). Cancel k and compute the value numerically.

Step-by-Step Solution:

Verification / Alternative check:
Using exact fractions maintains independence from k; only the ratio matters. The computed upstream time is approximately 6 h 53 min.

Why Other Options Are Wrong:
5 h 50 min and 6 h are too short; 12 h 10 min is far too long. The nearest provided option to approximately 6 h 53 min is 6 h 50 min.

Common Pitfalls:
Assuming symmetric times or averaging speeds. Upstream is significantly slower, so time is longer.

Final Answer:
6 h 50 min