Problem 5 · AMC 8 Stretch
Core
Algebra & Patterns
Number Theory
symmetryreduce-and-expand
Three children pool their pocket money. You are told only the pair totals: Anya and Ben together have 20 dollars; Ben and Carla together have 30 dollars; Carla and Anya together have 40 dollars. How much does Carla have?
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Answer: Carla has 25 dollars (Anya 15, Ben 5)
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Hint 1 of 4
Write the three facts as equations: \(A + B = 20\), \(B + C = 30\), \(C + A = 40\). Each child shows up more than once — a hint there's a slicker move than one variable at a time.
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Hint 2 of 4
Try ADDING all three pair-totals. Every child appears in exactly two of the pairs, so \(20 + 30 + 40\) counts everyone's money TWICE.
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Hint 3 of 4
\(20 + 30 + 40 = 90 = 2 \times (A + B + C)\), so all three together have \(90 \div 2 = 45\) dollars.
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Approach: Add all the equations to get the total, then subtract
- Let \(A, B, C\) be Anya, Ben, Carla's money: \(A + B = 20\), \(B + C = 30\), \(C + A = 40\).
- Add all three: \((A+B) + (B+C) + (C+A) = 90\). Each child appears in two pairs, so the left side is \(2(A+B+C)\), giving \(A + B + C = 45\).
- Peel off each child: \(C = 45 - (A+B) = 45 - 20 = 25\); \(A = 45 - (B+C) = 45 - 30 = 15\); \(B = 45 - (C+A) = 45 - 40 = 5\).
- Check: \(15 + 5 = 20\), \(5 + 25 = 30\), \(25 + 15 = 40\). So Carla has 25 dollars (Anya 15, Ben 5).
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