Difference between revisions of "2012 AMC 8 Problems/Problem 1"
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==Solution 2== | ==Solution 2== | ||
| − | We have the proportion <math>\frac{3}{8}</math> = <math>\frac{x}{24}</math> | + | We have the proportion <math>\frac{3}{8}</math> = <math>\frac{x}{24}</math>. To solve for <math>x</math>, multiply the left side of the equation by <math>\frac{3}{3}</math>. We find that <math>x=9</math>. ~Andrew_Lu |
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| + | ==Video Solution== | ||
| + | https://youtu.be/GG61DsRqoo4 ~savannahsolver | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2012|before=First Problem|num-a=2}} | {{AMC8 box|year=2012|before=First Problem|num-a=2}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 20:22, 29 June 2022
Problem
Rachelle uses 3 pounds of meat to make 8 hamburgers for her family. How many pounds of meat does she need to make 24 hamburgers for a neighbourhood picnic?
Solution 1
Since Rachelle uses 
pounds of meat to make 
hamburgers, she uses 
pounds of meat to make one hamburger. She'll need 24 times that amount of meat for 24 hamburgers, or 
.
Solution 2
We have the proportion = 
. To solve for 
, multiply the left side of the equation by 
. We find that 
. ~Andrew_Lu
Video Solution
https://youtu.be/GG61DsRqoo4 ~savannahsolver
See Also
| 2012 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by First Problem |
Followed by Problem 2 | |
| 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. 
