AMC 8 · 2012 · #9

Grade 4 algebra

Archetype Arithmetic Expression Decomposition └ Sub-archetype Fraction / Expression Sum Evaluation

systems-of-equationslinear-equations-two-var convert-to-algebraidentify-subproblems ↑ Prerequisites: multi-digit-arithmeticlinear-equations-one-var

📏 Medium solution 💡 3 insights

📘 View easy version →

Problem

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?

Pick an answer.

(A)

$hspace{.05in}61$

(B)

$hspace{.05in}122$

(C)

$hspace{.05in}139$

(D)

$hspace{.05in}150$

(E)

$hspace{.05in}161$

AMC 8 2012 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

Try it yourself first — the explanation is most useful after you’ve attempted it.

View mode:

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?

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

#9 Solve an Easier Related Problem 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 $200 \times 2$.

$$200 \times 2 = 400 \text{ legs}$$

💡 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.

#9 Solve an Easier Related Problem 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.

$$522 - 400 = 122 \text{ extra legs}$$

💡 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.

#9 Solve an Easier Related Problem 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.

$$\text{mammals} = \dfrac{122}{2} = 61$$

💡 Each bird-to-mammal swap is worth $+2$ legs, so dividing the leg surplus by $2$ counts the mammals directly — no algebra needed.

#9 Solve an Easier Related Problem 4.OA.A.3 Step 4

Subtract the mammals from the total to get the birds, since heads $+$ heads $= 200$.

$$\text{birds} = 200 - 61 = 139 \;\Rightarrow\; \textbf{(C)}$$

💡 Heads are conserved at $200$, so once one species is counted, the other is just a subtraction.

[1] #9 3.OA.A.3 Solve the easier version first: what if all $200$ animals were two-legged birds?

[2] #9 4.OA.A.3 Compare the easy answer to the real count. The real zoo has $522$ legs, but the

[3] #9 4.OA.A.3 Figure out the cost of one swap. Turning a bird into a mammal keeps the head cou

[4] #9 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.3 Use 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.3 Solve 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!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 4)

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