Difficulty: Medium
Correct Answer: 2 km/hr
Explanation:
Introduction / Context:When two travel-time equations are provided for different upstream/downstream mixes, we can solve the system for still-water speed b and current c.
Given Data / Assumptions:
Concept / Approach:Form two equations in b and c using time = distance/speed for each leg and add them. Solve the linear-fractional system.
Step-by-Step Solution:
24/(b − c) + 36/(b + c) = 6 … (1)36/(b − c) + 24/(b + c) = 6.5 … (2)Subtract (2) − (1): 12/(b − c) − 12/(b + c) = 0.512 * ( (b + c) − (b − c) ) / ( (b − c)(b + c) ) = 0.5 ⇒ 12 * (2c) / (b^2 − c^2) = 0.524c = 0.5 (b^2 − c^2) ⇒ b^2 − c^2 = 48cAdd (1) and (2): 60/(b − c) + 60/(b + c) = 12.5 ⇒ 60 * 2b / (b^2 − c^2) = 12.5120b / (b^2 − c^2) = 12.5 ⇒ b^2 − c^2 = 120b / 12.5 = 9.6bEquate: 9.6b = 48c ⇒ b = 5cUse b = 5c in b^2 − c^2 = 48c ⇒ 25c^2 − c^2 = 48c ⇒ 24c^2 = 48c ⇒ c = 2 km/h (c ≠ 0)Verification / Alternative check:With c = 2 and b = 10, upstream = 8, downstream = 12. Trip 1 time = 24/8 + 36/12 = 3 + 3 = 6 h; Trip 2 time = 36/8 + 24/12 = 4.5 + 2 = 6.5 h.
Why Other Options Are Wrong:1, 1.5, 2.5 do not satisfy both equations concurrently.
Common Pitfalls:Attempting to average distances or speeds directly; the harmonic nature of times requires careful algebra, not naive averaging.
Final Answer:2 km/h.
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