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?

Question

Solve this multiple-choice question and choose the correct option.

Accepted Answer

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: D / (5k) = (D − 30) / (3k) ⇒ (3/5)D = D − 30D − (3/5)D = 30 ⇒ (2/5)D = 30 ⇒ D = 75 m 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

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?

More Questions from Races and Games

Discussion & Comments