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?
Aptitude
Races and Games
Difficulty: Medium
Choose an option
-
A52 m
-
B75 m
-
C69 m
-
D70 m
Answer
Correct Answer: 75 m
Explanation
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 mVerification / 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