Point handicap consistency: A can give B 15 points; A can give C 22 points; and B can give C 10 points (all in the same “N-point” game). How many points make the game?

Difficulty: Medium

Correct Answer: 50

Explanation:

Introduction / Context:
In point handicaps, “A gives B 15” in an N-point game means when A scores N, B would score N − 15 in the same time, implying a speed ratio. Consistency across three pairwise handicaps pins down N.

Given Data / Assumptions:

A gives B 15 → vA/vB = N/(N − 15).
A gives C 22 → vA/vC = N/(N − 22).
B gives C 10 → vB/vC = N/(N − 10).

Concept / Approach:
Ratios must satisfy (vA/vB) * (vB/vC) = vA/vC. Substitute the expressions and solve for N.

Step-by-Step Solution:

(N/(N − 15)) * (N/(N − 10)) = N/(N − 22)Cancel one N: N/(N − 15) * 1/(N − 10) = 1/(N − 22)Cross-multiply: (N − 22) * N = (N − 15)(N − 10)Expand: N^2 − 22N = N^2 − 25N + 150 ⇒ 3N = 150 ⇒ N = 50

Verification / Alternative check:
Plug N = 50 back to get ratios 50/35, 50/28, and 50/40; they satisfy the identity.

Why Other Options Are Wrong:
60/70/80 do not satisfy the consistency equation for all three pairwise handicaps.

Common Pitfalls:
Adding or averaging handicaps; correct method uses multiplicative speed ratios.

Point handicap consistency: A can give B 15 points; A can give C 22 points; and B can give C 10 points (all in the same “N-point” game). How many points make the game?

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