Difference between revisions of "2012 AMC 8 Problems/Problem 1"
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<math>\textbf{(A)}\hspace{.05in}6 \qquad \textbf{(B)}\hspace{.05in}6\dfrac23 \qquad \textbf{(C)}\hspace{.05in}7\dfrac12 \qquad \textbf{(D)}\hspace{.05in}8 \qquad \textbf{(E)}\hspace{.05in}9 </math> | <math>\textbf{(A)}\hspace{.05in}6 \qquad \textbf{(B)}\hspace{.05in}6\dfrac23 \qquad \textbf{(C)}\hspace{.05in}7\dfrac12 \qquad \textbf{(D)}\hspace{.05in}8 \qquad \textbf{(E)}\hspace{.05in}9 </math> | ||
| − | ==Solution== | + | ==Solution 1== |
Since Rachelle uses <math>3</math> pounds of meat to make <math>8</math> hamburgers, she uses <math>\frac{3}{8}</math> pounds of meat to make one hamburger. She'll need 24 times that amount of meat for 24 hamburgers, or <math>\frac{3}{8} \cdot 24 = \boxed{\textbf{(E)}\ 9}</math>. | Since Rachelle uses <math>3</math> pounds of meat to make <math>8</math> hamburgers, she uses <math>\frac{3}{8}</math> pounds of meat to make one hamburger. She'll need 24 times that amount of meat for 24 hamburgers, or <math>\frac{3}{8} \cdot 24 = \boxed{\textbf{(E)}\ 9}</math>. | ||
| + | |||
| + | ==Solution 2== | ||
| + | So we have the proportion <math>\frac{3}{8}</math>=<math>\frac{x}{24}, to solve for x, multiply the left side of the equation by </math>\frac{3}{3}$. We find the x=9. | ||
| + | ~Andrew_Lu | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2012|before=First Problem|num-a=2}} | {{AMC8 box|year=2012|before=First Problem|num-a=2}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 19:59, 23 October 2020
Contents
Problem
Rachelle uses 3 pounds of meat to make 8 hamburgers for her family. How many pounds of meat does she need to make 24 hamburgers for a neighborhood picnic?
Solution 1
Since Rachelle uses 
pounds of meat to make 
hamburgers, she uses 
pounds of meat to make one hamburger. She'll need 24 times that amount of meat for 24 hamburgers, or 
.
Solution 2
So we have the proportion =
\frac{3}{3}$. We find the x=9.
~Andrew_Lu
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. 
