Boats and Streams – Two conditions on upstream vs downstream times: Kapil rows in a river where his usual still-water speed is u mile/h and the current is c mile/h. For 12 miles, he needs 6 hours less upstream than downstream. If he could double his rowing speed to 2u, then for a 24-mile round trip the downstream 12 miles would take only 1 hour less than the upstream 12 miles. What is the current speed c?

Introduction / Context:
This problem gives two time-gap conditions across upstream and downstream legs for the same distances under two different rowing capabilities. The unknowns are the usual still-water speed u and the current c. We translate the time gaps into equations in u and c and solve simultaneously.

Given Data / Assumptions:

• Distance per leg = 12 miles.
• Usual speeds: downstream u + c and upstream u − c.
• Condition 1: 12/(u − c) − 12/(u + c) = 6 hours.
• Condition 2 (with 2u): 12/(2u − c) − 12/(2u + c) = 1 hour.

Concept / Approach:
For fixed distances, time differences convert to rational expressions in u and c. Multiply through to clear denominators to form polynomial equations, then eliminate u to solve for c.

Step-by-Step Solution:
From Condition 1: 12[(u + c) − (u − c)]/(u^2 − c^2) = 6 ⇒ 24c = 6(u^2 − c^2) ⇒ u^2 − c^2 − 4c = 0.From Condition 2: 12[(2u + c) − (2u − c)]/(4u^2 − c^2) = 1 ⇒ 24c = 4u^2 − c^2 ⇒ 4u^2 − c^2 − 24c = 0.Eliminate u by subtracting 4*(first) from second: 3c^2 − 8c = 0 ⇒ c(3c − 8) = 0 ⇒ c = 8/3 mile/h (positive solution).

Verification / Alternative check:
Back-substitute c = 8/3 into either equation to find u and verify both conditions hold numerically.

Why Other Options Are Wrong:
2, 3, 5/2, and 10/3 mile/h do not satisfy both time-gap equations simultaneously.

Common Pitfalls:
Algebraic slips when clearing denominators, or incorrectly setting the direction of the time differences.

Final Answer:
8/3 mile/h