Difficulty: Medium
Correct Answer: 210 km
Explanation:
Introduction / Context:This is a speed-exchange puzzle. Two trains start together over the same origin–destination span. After the faster train has covered D km, both instantly swap speeds and continue, arriving simultaneously. We form total-time expressions before and after the swap and equate them.
Given Data / Assumptions:
Concept / Approach:Let t1 = time until swap. Then D = 7u * t1 (distance Taj covers), while Ajanta covers 6u * t1. After swapping, Taj's remaining distance = L − D at speed 6u; Ajanta's remaining = L − 6u t1 at speed 7u. Equal arrival times ⇒ post-swap travel times are equal.
Step-by-Step Solution:
t1 = D / (7u); Ajanta has covered 6D/7 by then.Set (L − D)/(6u) = (L − 6D/7)/(7u).Multiply by 42u: 7(L − D) = 6(L − 6D/7) ⇒ 7L − 7D = 6L − 36D/7.Rearrange: L = (13/7)D ⇒ D = (7/13)L = (7/13)*390 = 210 km.Verification / Alternative check:Plugging D = 210 satisfies symmetry of post-swap times; any other D breaks equality.
Why Other Options Are Wrong:150 km and 190 km do not make the post-swap times equal; “can’t be determined” is incorrect because algebra yields a unique D.
Common Pitfalls:Assuming they must meet at the swap point; the swap is instantaneous and simultaneous, not collocated.
Final Answer:210 km
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