Difference between revisions of "2012 AMC 8 Problems/Problem 9"
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<math> \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 </math> | <math> \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 </math> | ||
− | + | ==Solution 1: Algebra== | |
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Let the number of two-legged birds be <math>x</math> and the number of four-legged mammals be <math>y</math>. We can now use systems of equations to solve this problem. | Let the number of two-legged birds be <math>x</math> and the number of four-legged mammals be <math>y</math>. We can now use systems of equations to solve this problem. | ||
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By subtracting the second equation from the first equation, we find that <math> 2y = 122 \implies y = 61 </math>. Since there were <math> 200 </math> heads, meaning that there were <math> 200 </math> animals, there were <math> 200 - 61 = \boxed{\textbf{(C)}\ 139} </math> two-legged birds. | By subtracting the second equation from the first equation, we find that <math> 2y = 122 \implies y = 61 </math>. Since there were <math> 200 </math> heads, meaning that there were <math> 200 </math> animals, there were <math> 200 - 61 = \boxed{\textbf{(C)}\ 139} </math> 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 <math>200\cdot2=400</math> legs. | 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 <math>200\cdot2=400</math> legs. | ||
Latest revision as of 17:03, 28 March 2018
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.
Notice that we must gain legs. This means we must swap out birds. Therefore, there must be birds. Checking our work, we find that , and we are correct.
See Also
2012 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.