AMC8 2012

AMC8 2012 · Q9

AMC8 2012 · Q9. It mainly tests Systems of equations, Word problems (algebra).

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?

沃思堡动物园有一些两腿鸟类和一些四腿哺乳动物。玛吉一次参观时数到200个头和522条腿。玛吉数到的动物中有多少是两腿鸟类？

(A) 61 61

(B) 122 122

(D) 150 150

(E) 161 161

Answer

Correct choice: (C)

正确答案：（C）

Solution

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.

设两腿鸟类数量为$x$，四腿哺乳动物数量为$y$。我们用方程组来解这个问题。列两个方程： $2x + 4y = 522$ $x + y = 200$ 将第二个方程乘以2： $2x + 4y = 522$ $2x + 2y = 400$ 用第一个方程减去第二个方程，得$2y = 122 \implies y = 61$。共有200个头，即200只动物，因此两腿鸟类数量为$200 - 61 = \boxed{\textbf{(C)}\ 139}$。

Topics