Problem 4 · AMC 8 Stretch
Core
Algebra & Patterns
Arithmetic & Operations
work-backwardtranslate-text-into-mathematics
Forty-two birds sit on three trees. Then 3 birds fly from tree 1 to tree 2, and 7 birds fly from tree 2 to tree 3. Now there are twice as many birds on tree 2 as on tree 1, and twice as many on tree 3 as on tree 2. How many birds were on each tree at the start?
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Answer: 9, 16, and 17 birds on trees 1, 2, 3
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Hint 1 of 4
The nice moment is AFTER the birds move, when the counts come in the ratio 1 : 2 : 4.
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Hint 2 of 4
Call the number on tree 1 (after moving) one 'part'. Then tree 2 has 2 parts and tree 3 has 4 parts. How many parts total, and what do they add to?
Still stuck? Show hint 3 →
Hint 3 of 4
\(1 + 2 + 4 = 7\) parts make 42 birds, so one part is \(42 \div 7 = 6\). Now you know all three after-counts.
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Approach: Solve the easy after-state ratio, then work backward
- After the move the three trees are in the ratio 1 : 2 : 4. Think of tree 1's count as one part, so the trees hold \(1 + 2 + 4 = 7\) parts totaling 42 birds: 1 part \(= 42 \div 7 = 6\). So after the move the trees hold 6, 12, and 24 birds.
- Rewind: tree 1 lost 3 birds, so it started with \(6 + 3 = 9\). Tree 3 gained 7, so it started with \(24 - 7 = 17\). Tree 2 is the rest: \(42 - 9 - 17 = 16\).
- Check: tree 2 received 3 and lost 7, a change of \(-4\), and \(16 - 4 = 12\). Correct!
- Originally there were 9, 16, and 17 birds on trees 1, 2, and 3.
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