2012 AMC 8 Problems/Problem 9
Contents
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?
Solution 1: Algebra
Let the number of two-legged birds be and the number of four-legged mammals be . We can now use systems of equations to solve this problem.
Write two equations:
Now multiply the latter equation by .
By subtracting the second equation from the first equation, we find that . Since there were heads, meaning that there were animals, there were 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 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 legs. If we swapped two birds for two mammals, there would be legs. If we swapped 50 birds for 50 mammals, there would be legs.
Solution 3: Add two legs for each bird
Let's assume each bird has four-legged as each mammal. So the total legs of these birds and mammals would be . Actually, there were only legs. The difference of these two numbers of legs exactly gives us the number of those added-legs of birds: . Because each bird only has 2 legs, the number of birds would be two-legged birds$. ---LarryFlora
See Also
2012 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
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