In a 1000 m race, A can give B a 50 m start and C an 80 m start. In a 1000 m race between B and C, how many metres start should B give C?

Introduction / Context:Head starts relative to a common runner (A) let us infer speed ratios. From A:B and A:C, derive B:C and convert that to the start B should give C over the same official distance.

Given Data / Assumptions:

• In 1000 m for A, B covers 950 m ⇒ A:B = 1000:950 = 20:19.
• In 1000 m for A, C covers 920 m ⇒ A:C = 1000:920 = 25:23.
• Uniform speeds.

Concept / Approach:Use B/A = 19/20 and C/A = 23/25 to get B/C = (19/20)/(23/25) = (19*25)/(20*23) = 95/92. When B runs 1000 m, C runs 1000*(92/95), so the start is the difference to 1000.

Step-by-Step Solution: C's distance when B finishes = 1000 * (92/95) = 968 8/19 m Required start = 1000 − 1000*(92/95) = 1000*(3/95) = 31 11/19 m

Verification / Alternative check:As a decimal, 31 11/19 m ≈ 31.5789 m; B:C ≈ 1.0326, consistent with B being slightly faster.

Why Other Options Are Wrong:31 m and 312⁄19 m (≈ 16.42 m) do not match the exact fractional start; 3011⁄19 m is far too large.

Common Pitfalls:Subtracting 50 and 80 directly; forgetting to convert starts into speed ratios via a common reference.

Final Answer:3111⁄19 m