A medium question on logic puzzles for P3–P4, with a step-by-step solution, diagrams and the answer. The Hua Xia Cup's P4 qualifying round and finals have true-or-false statements and match-scoring questions. A favourite thinking workout, too.
The question
4 pupils hold a chess tournament: every two of them play each other exactly once.
A win scores 2 points and a loss 0; in a draw each player scores 1.
When it's all over, Yan has 4, Lok has 3 and Keung has 2.
How many points does Ka-yan have?
Show the hints (2)
💡Hint 1: With every two of the 4 playing once, how many games are there in all?
💡Hint 2: In each game, whoever wins or if it's a draw, how many points do the two players get between them?
Worked solution
Have a go first, then read the solution step by step.
1.Each plays the other 3: 4 × 3 = 12, but that counts every game twice, so there are 12 ÷ 2 = 6 games.
2.Every game hands out 2 points in all (2 + 0, or 1 + 1 for a draw), so everyone's points add up to 6 × 2 = 12.
3.Ka-yan: 12 − (4 + 3 + 2) = 12 − 9 = 3.
⚠️Common mistake: counting 4 × 3 = 12 games. A against B and B against A are the same game, counted twice.
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