Difference between revisions of "2012 AMC 8 Problems/Problem 9"
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==Solution 3: Add two legs for each bird== | ==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 | 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 | ||
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+ | ==Video Solution== | ||
+ | https://youtu.be/CsmpiJC-FBA | ||
==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 22:17, 7 October 2021
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
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 |
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