Two ladies start simultaneously from Cities A and B, 24 km apart, walking toward each other. The first lady walks 2 km/h faster than the second and reaches B exactly 1 hour before the second reaches A. What are their speeds (km per hour)?
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A5 km,3 km
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B7 km, 5 km
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C8 km, 6 km
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D16 km, 14 km
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ENone of these
Answer
Correct Answer: 8 km, 6 km
Explanation
Introduction / Context:Both travellers start together and traverse the entire 24 km, one going from A to B and the other from B to A. Let the slower speed be v km/h and the faster be v + 2. The faster arrives 1 hour earlier, giving a time-difference equation that can be solved for v directly.
Given Data / Assumptions:
- Total one-way distance for each = 24 km.
- Faster speed = v + 2; slower speed = v.
- Time difference between arrivals = 1 h.
Concept / Approach:Arrival times: T_fast = 24/(v+2), T_slow = 24/v. Since the faster arrives earlier by 1 h: T_slow − T_fast = 1. Solve for v by clearing denominators.
Step-by-Step Solution:
24/v − 24/(v+2) = 1.24[(v+2 − v)/(v(v+2))] = 1 ⇒ 24 * 2 / (v(v+2)) = 1.v(v+2) = 48 ⇒ v^2 + 2v − 48 = 0 ⇒ (v − 6)(v + 8) = 0.v = 6 km/h (positive root) ⇒ faster = 8 km/h.Verification / Alternative check:Times: 24/6 = 4 h; 24/8 = 3 h; difference = 1 h as required.
Why Other Options Are Wrong:Other pairs do not differ by exactly 2 km/h or do not yield a 1-hour arrival gap over 24 km.
Common Pitfalls:Using relative speed meeting-time equations instead of whole-journey arrival times; mixing roles of the two speeds.
Final Answer:8 km, 6 km