Problem 3 · AMC 8 Stretch
Core
Algebra & Patterns
Number Theory
symmetryreduce-and-expandlogical-reasoning
Two friends want to buy a horse, but neither has enough alone. The first says, 'If you gave me half of your money, I'd have exactly enough to buy the horse.' The second says, 'And if you gave me a third of your money, I'd have exactly enough to buy the horse.' Find the smallest whole-number amounts each friend could have, and the price of the horse. (Give the horse's price.)
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Answer: First has 3, second has 4, horse costs 5
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Hint 1 of 4
Turn the sentences into equations. Let \(A\) and \(B\) be the friends' money and \(H\) the horse's price. 'I plus half of yours' gives \(A + \tfrac{B}{2} = H\); the second gives \(B + \tfrac{A}{3} = H\).
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Hint 2 of 4
Multiply the first equation by 2 and the second by 3 to clear fractions: \(2A + B = 2H\) and \(A + 3B = 3H\).
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Hint 3 of 4
Substitute to compare \(A\) and \(B\).
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Approach: Set up two equations, clear fractions, solve the ratio
- Let \(A\) = first friend's money, \(B\) = second's, \(H\) = horse price. Then \(A + \tfrac{1}{2}B = H\) and \(B + \tfrac{1}{3}A = H\).
- Clear fractions: \(2A + B = 2H\) and \(A + 3B = 3H\). From the first, \(B = 2H - 2A\).
- Substitute: \(A + 3(2H - 2A) = 3H \Rightarrow A + 6H - 6A = 3H \Rightarrow -5A = -3H \Rightarrow A = \tfrac{3}{5}H\), and \(B = 2H - \tfrac{6}{5}H = \tfrac{4}{5}H\). So \(A : B : H = 3 : 4 : 5\).
- Smallest whole numbers: \(A = 3, B = 4, H = 5\). Check: \(3 + \tfrac12(4) = 5\) and \(4 + \tfrac13(3) = 5\). The horse costs 5.
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