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?

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:

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