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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Solve this multiple-choice question and choose the correct option.

Accepted Answer

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. Final Answer:50

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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