Problem 9 · 2012 AMC 8
Easy
Algebra & Patterns
system-of-equationshead-leg-trick
The Fort Worth Zoo has a number of two-legged birds and a number of four-legged mammals. On one visit to the zoo, Margie counted 200 heads and 522 legs. How many of the animals that Margie counted were two-legged birds?
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Answer: C — 139 birds.
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Hint 1 of 2
Before reaching for two equations, try a what-if: pretend every animal is a 4-legged mammal. You'd over-count the legs — and the over-count is caused entirely by the birds.
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Hint 2 of 2
This is the assume-then-fix trick (a.k.a. the chicken-and-rabbit method): assume the extreme case, see the gap from reality, and let each swapped animal explain a fixed chunk of the gap.
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Approach: assume all 4-legged, then account for the shortage
- Imagine all 200 animals are 4-legged: that would be 200 × 4 = 800 legs — one easy multiplication instead of solving a system.
- Reality shows only 522 legs, a shortage of 800 − 522 = 278.
- Every bird counted as a mammal hides 2 legs (it has 2, not 4), so each bird explains exactly 2 of the missing legs: birds = 278 / 2 = 139.
- Sanity check: 139 birds + 61 mammals = 200 heads ✓, and 139×2 + 61×4 = 278 + 244 = 522 legs ✓.
Another way — two equations (heads and legs):
- Let t = birds, f = mammals. Heads: t + f = 200. Legs: 2t + 4f = 522.
- Double the head equation: 2t + 2f = 400, and subtract it from the leg equation to clear t... or subtract from the legs to clear f: 2f = 122, so f = 61.
- Then t = 200 − 61 = 139 birds. The shortcut above is just this subtraction done in your head.
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