Difference between revisions of "2012 AMC 8 Problems/Problem 9"

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==Solution 3: Cheating the System==
 
==Solution 3: Cheating the System==
Let's 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>. It exactly gives us the assumed legs of birds by doing this subtraction: <math>800 - 522 = 278</math>. Each bird has 2 legs, the number of birds would be <math> 278/2 = \boxed{\textbf{(C)}\ 139} </math> two-legged birds$.  ---LarryFlora
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Let's assume each bird has four-legged as each mammal. So the total legs of these birds and mammals would be <math>4*200=800</math>. Actually, there were only <math>522</math> legs. The difference of these two numbers of legs exactly gives us the number of those assumed legs of birds: <math>800 - 522 = 278</math>. Because each bird only has 2 legs, the number of birds would be <math> 278/2 = \boxed{\textbf{(C)}\ 139} </math> two-legged birds$.  ---LarryFlora
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==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}}

Revision as of 13:19, 16 August 2021

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.

Solution 3: Cheating the System

Let's assume each bird has four-legged as each mammal. So the total legs of these birds and mammals would be $4*200=800$. Actually, there were only $522$ legs. The difference of these two numbers of legs exactly gives us the number of those assumed legs of birds: $800 - 522 = 278$. Because each bird only has 2 legs, the number of birds would be $278/2 = \boxed{\textbf{(C)}\ 139}$ two-legged birds$. ---LarryFlora

See Also

2012 AMC 8 (ProblemsAnswer KeyResources)
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

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