Head start with speed factor (corrected statement): A runs 1 2⁄3 times as fast as B. If A gives B a start of 30 m and they finish together, how far from the start must the winning post be?

Introduction / Context:
The original text had a likely typo (“1% times”). By the Recovery-First Policy we minimally repair to a standard form: A runs 1 2/3 times as fast as B. With a 30 m head start to B, find the race distance so they tie.

Given Data / Assumptions (after minimal repair):

• v_A : v_B = 5 : 3
• Head start to B = 30 m
• Finish together

Concept / Approach:
Equalize times: D / v_A = (D − 30) / v_B. Solve for D with v_A/v_B = 5/3.

Step-by-Step Solution:

Verification / Alternative check:
Times equal: 75/(5k) = 45/(3k) = 15/k.

Why Other Options Are Wrong:
They do not satisfy the time equality for ratio 5:3 with a 30 m head start.

Common Pitfalls:
Using speed difference instead of ratio, or forgetting to subtract the head start from B’s distance.

Final Answer:
75 m