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== | |
| − | |||
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. | ||
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 <math>400 + 1(2) = 402</math> legs. If we swapped two birds for two mammals, there would be <math>400 + 2(2) = 404</math> legs. If we swapped 50 birds for 50 mammals, there would be <math>400 + 50(2) = 500</math> 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 <math>400 + 1(2) = 402</math> legs. If we swapped two birds for two mammals, there would be <math>400 + 2(2) = 404</math> legs. If we swapped 50 birds for 50 mammals, there would be <math>400 + 50(2) = 500</math> legs. | ||
| − | + | ==Solution 3: Add two legs for each bird== | |
| + | Let's add 2 legs for each bird and assume each bird has four-legged as each mammal. So the total legs of these birds and mammals would be <math>4*200=800</math>. Actually, there were only <math>522</math> legs. The difference between these two numbers exactly gives us the number of all the legs we added for all birds: <math>800 - 522 = 278</math>. Because each bird was added by 2 legs, so the total number of birds would be <math> 278/2 = \boxed{\textbf{(C)}\ 139} </math> two-legged birds. ---LarryFlora | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/CsmpiJC-FBA ~David | ||
| + | |||
| + | https://youtu.be/fQzxnsnp40c ~savannahsolver | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2012|num-b=8|num-a=10}} | {{AMC8 box|year=2012|num-b=8|num-a=10}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 17:58, 15 April 2023
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 add 2 legs for each bird and 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 between these two numbers exactly gives us the number of all the legs we added for all birds: 
. Because each bird was added by 2 legs, so the total number of birds would be 
two-legged birds. ---LarryFlora
Video Solution
https://youtu.be/CsmpiJC-FBA ~David
https://youtu.be/fQzxnsnp40c ~savannahsolver
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. 
