AMC 8 · 2012 · #9
Easy mode Grade 4Problem
A zoo has only two kinds of animals: birds with legs each, and mammals with legs each.
Margie visits the zoo and counts all the animals. She counts heads in total, and legs in total.
How many of the animals are birds?
Pick an answer.
Toolkit + CCSS Solution
Understand
Restated: At the zoo, Margie counts $200$ heads and $522$ legs across two-legged birds and four-legged mammals. How many of those $200$ animals are birds?
Givens: Every animal has exactly $1$ head, so total animals $= 200$; Birds have $2$ legs, mammals have $4$ legs; Total legs counted $= 522$; Answer choices: (A) $61$, (B) $122$, (C) $139$, (D) $150$, (E) $161$
Unknowns: The number of two-legged birds among the $200$ animals
Understand
Restated: At the zoo, Margie counts $200$ heads and $522$ legs across two-legged birds and four-legged mammals. How many of those $200$ animals are birds?
Givens: Every animal has exactly $1$ head, so total animals $= 200$; Birds have $2$ legs, mammals have $4$ legs; Total legs counted $= 522$; Answer choices: (A) $61$, (B) $122$, (C) $139$, (D) $150$, (E) $161$
Plan
Primary tool: #9 Solve an Easier Related Problem
Secondary: #6 Guess and Check
The reference solution uses algebra (Tool #13), but a 4th grader can crack this without variables. Tool #9 (Easier Related Problem): pretend every animal is a bird and count legs — that gives $400$. The real count is $522$, so the extra legs must come from swapping some birds for mammals. Each swap adds $2$ legs, so dividing the leg shortage by $2$ instantly gives the mammal count. Tool #6 (Guess and Check) is a clean backup: plug each answer choice into $\text{birds} \cdot 2 + (200 - \text{birds}) \cdot 4$ and see which gives $522$.
Execute — Answer: C
3.OA.A.3 Step 1 - Solve the easier version first: what if all $200$ animals were two-legged birds?
- Then the leg count would simply be $200 \times 2$.
💡 Replacing the mixed flock with an all-bird flock turns a 2-unknown puzzle into a single multiplication — exactly the Grade 3 "multiply within $100$ to solve word problems" move.
4.OA.A.3 Step 2 - Compare the easy answer to the real count.
- The real zoo has $522$ legs, but the all-bird version only has $400$.
- The difference is the "extra" legs that mammals contribute.
💡 The gap between the easy version and the real version measures exactly how much "mammal-ness" is in the flock — a Grade 4 multi-step word-problem reasoning move.
4.OA.A.3 Step 3 - Figure out the cost of one swap.
- Turning a bird into a mammal keeps the head count the same but raises the leg count by $4 - 2 = 2$.
- So every $2$ extra legs corresponds to one mammal.
💡 Each bird-to-mammal swap is worth $+2$ legs, so dividing the leg surplus by $2$ counts the mammals directly — no algebra needed.
4.OA.A.3 Step 4 Subtract the mammals from the total to get the birds, since heads $+$ heads $= 200$.
💡 Heads are conserved at $200$, so once one species is counted, the other is just a subtraction.
3.OA.A.3 Solve the easier version first: what if all $200$ animals were two-legged birds? 4.OA.A.3 Compare the easy answer to the real count. The real zoo has $522$ legs, but the 4.OA.A.3 Figure out the cost of one swap. Turning a bird into a mammal keeps the head cou 4.OA.A.3 Subtract the mammals from the total to get the birds, since heads $+$ heads $= 2 Review
Reasonableness: Plug the answer back in: $139$ birds give $139 \times 2 = 278$ legs, $61$ mammals give $61 \times 4 = 244$ legs, and $278 + 244 = 522$. Heads: $139 + 61 = 200$. Both totals match exactly. Also a sanity check — birds far outnumber mammals because $522$ is much closer to $400$ ($\text{all-bird}$) than to $800$ ($\text{all-mammal}$), so the answer should be near the bird end, and $139$ is indeed in that range.
Alternative: Tool #6 (Guess and Check) on the choices: for each candidate bird count $b$, compute $2b + 4(200 - b) = 800 - 2b$ and see if it equals $522$. That forces $2b = 278$, so $b = 139$. Testing the choices: (A) $b=61 \Rightarrow 678$ legs, (B) $b=122 \Rightarrow 556$, (C) $b=139 \Rightarrow 522$ ✓, (D) $b=150 \Rightarrow 500$, (E) $b=161 \Rightarrow 478$. Only (C) lands on $522$.
CCSS standards used (min grade 4)
3.OA.A.3Use multiplication and division within 100 to solve word problems (Computing the all-bird leg total $200 \times 2 = 400$ from the easier-problem setup.)4.OA.A.3Solve multi-step word problems using the four operations (Chaining $522 - 400 = 122$, $122 \div 2 = 61$, and $200 - 61 = 139$ to step from the easy version to the real answer.)
⭐ Pretend every animal is a bird first — then each extra pair of legs you're missing is one mammal hiding in the flock. Pure Grade 4 arithmetic, no algebra needed!
⭐ Pretend every animal is a bird first — then each extra pair of legs you're missing is one mammal hiding in the flock. Pure Grade 4 arithmetic, no algebra needed!
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