Relative speed with head start: A runs 1 3⁄8 times as fast as B. If A gives B a start of 120 m and they finish together, find the race distance (the winning post distance from A’s start).

Difficulty: Medium

Correct Answer: 440 m

Explanation:


Introduction / Context:
This problem is about equalizing finish times with a head start when one runner is faster by a known factor.


Given Data / Assumptions:

  • Speed ratio v_A : v_B = 11 : 8
  • Head start for B = 120 m
  • They finish together


Concept / Approach:
Set equal times: distance/v_A = (distance − headstart)/v_B. Substitute the known ratio and solve for the distance.


Step-by-Step Solution:

Let D be race distance from A's start.D / (11k) = (D − 120) / (8k) ⇒ (8/11)D = D − 120D − (8/11)D = 120 ⇒ (3/11)D = 120 ⇒ D = 440 m


Verification / Alternative check:
Plug D = 440 to confirm both times equal: 440/(11k) vs. 320/(8k) = 40/k each.


Why Other Options Are Wrong:
Other distances do not satisfy the time equality for the given ratio.


Common Pitfalls:
Adding/subtracting speeds incorrectly or forgetting to subtract the head start from B’s running distance.


Final Answer:
440 m

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