Starts into effective speeds: In a 1 km race A gives B a start of 50 m, and B gives C a start of 40 m. In a 2 km race, what start (in metres) can A give C so that they finish together?
-
A222 m
-
B220 m
-
C167 m
-
D176 m
-
ENone of these
Answer
Correct Answer: 176 m
Explanation
Introduction / Context:Starts translate to speed ratios: if A gives B 50 m in 1000 m, A is proportionally faster. Chaining two such relations lets us compare A directly to C and scale to a 2000 m race.
Given Data / Assumptions:
- 1 km race: A vs B → B runs 950 m when A runs 1000 m.
- 1 km race: B vs C → C runs 960 m when B runs 1000 m.
- Uniform speeds; proportional times.
Concept / Approach:Speed ratios: vA/vB = 1000/950 = 20/19; vB/vC = 1000/960 = 25/24. Then vA/vC = (20/19)*(25/24) = 125/114.
Step-by-Step Solution:
For a 2000 m race: distance that C covers when A runs 2000 m is 2000 * (vC/vA)vC/vA = 114/125C's distance = 2000 * 114/125 = 1824 mRequired start to C = 2000 - 1824 = 176 mVerification / Alternative check:Scale both speeds by a common time; the same 176 m emerges from proportional distances.
Why Other Options Are Wrong:220/222/167 are mismatched scalings; only 176 m satisfies the chained ratios precisely.
Common Pitfalls:Adding starts directly instead of converting to speed ratios; not scaling to the new race length.
Final Answer:176 m