In billiards, A can give B 6 points in a game to 50, and A can give C 13 points in a game to 65. In a game to 55, how many points can B give C (so that the game is fair)?

Introduction / Context:
Handicap statements like “A gives B 6 in 50” imply proportional scoring ability: when A scores 50, B scores 44. Converting these into ability ratios allows fair handicaps to be computed for new target totals.

Given Data / Assumptions:

• A vs B to 50: B scores 44 when A scores 50 ⇒ A:B = 50:44 = 25:22.
• A vs C to 65: C scores 52 when A scores 65 ⇒ A:C = 65:52 = 5:4.

Concept / Approach:
From A:B = 25:22 and A:C = 5:4 ⇒ take A = 25 units; then B = 22, C = (4/5)×25 = 20. Hence B:C = 22:20 = 11:10. In a fair game to 55 for B, C should be proportionally at 55×(10/11) = 50.

Step-by-Step Solution:

Verification / Alternative check:
Scaling any common multiple maintains the 11:10 ratio.

Why Other Options Are Wrong:
7, 8, or 10 would distort the B:C proportionality implied by the original handicaps.

Common Pitfalls:
Adding or averaging handicaps directly without converting to ability ratios.

Final Answer:
5