2012 AMC 8 Problems/Problem 9

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?

$\textbf{(A)}\hspace{.05in}61\qquad\textbf{(B)}\hspace{.05in}122\qquad\textbf{(C)}\hspace{.05in}139\qquad\textbf{(D)}\hspace{.05in}150\qquad\textbf{(E)}\hspace{.05in}161$

Solution 1: Algebra

Let the number of two-legged birds be $x$ and the number of four-legged mammals be $y$. We can now use systems of equations to solve this problem.

Write two equations:

$2x + 4y = 522$

$x + y = 200$

Now multiply the latter equation by $2$.

$2x + 4y = 522$

$2x + 2y = 400$

By subtracting the second equation from the first equation, we find that $2y = 122 \implies y = 61$. Since there were $200$ heads, meaning that there were $200$ animals, there were $200 - 61 =  \boxed{\textbf{(C)}\ 139}$ two-legged birds.

Solution 2: Cheating the System

First, we "assume" there are 200 two-legged birds only, and 0 four-legged mammals. Of course, this poses a problem, as then there would only be $200\cdot2=400$ legs.

Now we have to do some swapping--for every two-legged bird we swap for a four-legged mammal, we gain 2 legs. For example, if we swapped one bird for one mammal, giving 199 birds and 1 mammal, there would be $400 + 1(2) = 402$ legs. If we swapped two birds for two mammals, there would be $400 + 2(2) = 404$ legs. If we swapped 50 birds for 50 mammals, there would be $400 + 50(2) = 500$ legs.

Notice that we must gain $522-400 = 122$ legs. This means we must swap out $122\div2 = 61$ birds. Therefore, there must be $200-61 = \boxed{\textbf{(C)}\ 139}$ birds. Checking our work, we find that $139\cdot2 + 61 \cdot 4 = 522$, and we are correct.

See Also

2012 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
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All AJHSME/AMC 8 Problems and Solutions

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