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

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==Solution 2==
 
==Solution 2==
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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>. ~Andrew_Lu with some small edits from: TabHawaii
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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==
 
==Video Solution==

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

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