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Harddeduction

~135s

Which box holds the prize

Three boxes are labelled A, B and C, and exactly one holds a prize. Box A says 'the prize is not here', Box B says 'the prize is here', and Box C says 'the prize is not in B'. If exactly one statement is true, where is the prize?

Show the answer and the worked solution

Correct answer: A.

Box A

Worked solution

  1. Prize in A: A false, B false, C true → 1 true. ✓
  2. Prize in B: A true, B true, C false → 2 true. ✗
  3. Prize in C: A true, B false, C true → 2 true. ✗
  4. Only the first case gives exactly one true statement, so the prize is in Box A.

Answer: Box A.

The idea

Test each case exhaustively against the constraint. Counting the true statements per case is faster than reasoning about the statements directly.

Shortcut. B ('here') and C ('not in B') contradict each other, so exactly one of them is true — that uses up the single truth. A's statement must be false, so the prize is in A.

Common trap. "Box C" — taking A's statement as the lone truth, ruling out A and B, and not noticing that C's 'not in B' is then true as well.

Mathematical Reasoning

27 of 35

Level

Hard

Target time

~135s

Which box holds the prize — Aptitude Question

Which box holds the prize: Three boxes are labelled A, B and C, and exactly one holds a prize. Answer and worked solution on CodeKairo.

Question

Three boxes are labelled A, B and C, and exactly one holds a prize. Box A says 'the prize is not here', Box B says 'the prize is here', and Box C says 'the prize is not in B'. If exactly one statement is true, where is the prize?

Options

  1. Box A
  2. Box B
  3. Box C
  4. Cannot be determined
Show the answer and the worked solution

Correct answer: A. Box A

Worked solution

  1. Prize in A: A false, B false, C true → 1 true. ✓
  2. Prize in B: A true, B true, C false → 2 true. ✗
  3. Prize in C: A true, B false, C true → 2 true. ✗
  4. Only the first case gives exactly one true statement, so the prize is in Box A.

Answer: Box A.

The idea

Test each case exhaustively against the constraint. Counting the true statements per case is faster than reasoning about the statements directly.

Shortcut. B ('here') and C ('not in B') contradict each other, so exactly one of them is true — that uses up the single truth. A's statement must be false, so the prize is in A.

Common trap. "Box C" — taking A's statement as the lone truth, ruling out A and B, and not noticing that C's 'not in B' is then true as well.

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