Difficulty: Medium
Correct Answer: 55 km/h
Explanation:
Introduction / Context:This is a classic “time saved with higher speed” equation. Over a fixed distance, increasing speed reduces time according to t = D/v. Setting up the difference in times equal to the saved amount yields a quadratic in v that can be solved exactly.
Given Data / Assumptions:
Concept / Approach:Equation: 715/v − 715/(v+10) = 2. Simplify to solve for v. Positive, realistic highway values are expected.
Step-by-Step Solution:
715[(1/v) − (1/(v+10))] = 2.715 * (10) / (v(v+10)) = 2 ⇒ 7150 = 2v(v+10).v^2 + 10v = 3575 ⇒ v^2 + 10v − 3575 = 0.Discriminant = 100 + 14300 = 14400; sqrt = 120.v = (−10 + 120)/2 = 55 km/h (ignoring negative root).Verification / Alternative check:Times: at 55 km/h → 13 h; at 65 km/h → 11 h; saving = 2 h as required.
Why Other Options Are Wrong:60 or 65 km/h do not yield a 2-hour difference for 715 km; 36 km/h is too low and unrealistic here.
Common Pitfalls:Arithmetic slips in clearing denominators; accepting the negative quadratic root.
Final Answer:55 km/h
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