In a 200 m race, B can give A a start of 10 m (so A runs 190 m when B runs 200 m). In the same 200 m race, C can give B a start of 20 m (so B runs 180 m when C runs 200 m). How much start can C give A in the same race?
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A30 m
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B29 m
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C27 m
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D25 m
Answer
Correct Answer: 29 m
Explanation
Introduction / Context:Head-starts yield speed ratios. Chain them to compare two competitors who have not raced directly. Then, compute the distance A would cover when C finishes the 200 m, and infer the start C can give A.
Given Data / Assumptions:
- B vs A: vB/vA = 200/190 = 20/19.
- C vs B: vC/vB = 200/180 = 10/9.
- Target race = 200 m with C finishing the full 200 m.
Concept / Approach:vC/vA = (vC/vB)*(vB/vA) = (10/9)*(20/19) = 200/171. When C runs 200 m, A covers 200*(vA/vC) = 200*(171/200) = 171 m, so the start equals 200 − 171.
Step-by-Step Solution:
A’s distance at C’s finish = 171 m.Hence C can give A = 200 − 171 = 29 m.Verification / Alternative check:Time equality check confirms the same value since distance is proportional to speed at a common finish time.
Why Other Options Are Wrong:30, 27, 25 m do not match the chained ratio result 200/171.
Common Pitfalls:Adding start values directly (10 + 20) instead of chaining speed ratios, which is incorrect.
Final Answer:29 m