Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2012 AMC 8 Problems/Problem 1"

(→‎Solution)
m (→‎Solution 2)
 
(24 intermediate revisions by 10 users not shown)
Line 1: Line 1:
 
==Problem==
 
==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?
+
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?
  
 
<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==
−
===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===
+
==Solution 2==
−
Rachelle uses meat for hamburgers in the ratio <math>\frac{3\text{ lb}}{8\text{ hamburgers}}</math>. If we multiply this ratio on top and bottom by 3 to get 24 hamburgers on the bottom, we get <math>\frac{3\text{ lb}\cdot 3}{8\text{ hamburgers}\cdot 3} = \frac{9\text{ lb}}{24\text{ hamburgers}}</math>. Therefore, she uses 9 pounds of meat for 24 hamburgers, making the answer <math>\boxed{\textbf{(E)}\ 9}</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>. This gives us <math>\frac{9}{24}=\frac{x}{24}</math>. For <math>\frac{9}{24}</math> to equal <math>\frac{x}{24}</math>, we find that <math>x=9</math>. The answer is then <math>\boxed{(E)}</math>. ~Andrew_Lu with some small edits from: TabHawaii
 +
 
 +
==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 12:32, 19 November 2024

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?

$\textbf{(A)}\hspace{.05in}6 \qquad \textbf{(B)}\hspace{.05in}6\dfrac23 \qquad \textbf{(C)}\hspace{.05in}7\dfrac12 \qquad \textbf{(D)}\hspace{.05in}8 \qquad \textbf{(E)}\hspace{.05in}9$

Solution 1

Since Rachelle uses $3$ pounds of meat to make $8$ hamburgers, she uses $\frac{3}{8}$ pounds of meat to make one hamburger. She'll need 24 times that amount of meat for 24 hamburgers, or $\frac{3}{8} \cdot 24 = \boxed{\textbf{(E)}\ 9}$.

Solution 2

We have the proportion $\frac{3}{8}$ = $\frac{x}{24}$. To solve for $x$, multiply the left side of the equation by $\frac{3}{3}$. This gives us $\frac{9}{24}=\frac{x}{24}$. For $\frac{9}{24}$ to equal $\frac{x}{24}$, we find that $x=9$. The answer is then $\boxed{(E)}$. ~Andrew_Lu with some small edits from: TabHawaii

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_8_Problems/Problem_1&oldid=235040"