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

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==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?
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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>
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==Solution 2==
 
==Solution 2==
So we have the proportion <math>\frac{3}{8}</math> = <math>\frac{x}{24}</math> to solve for x, multiply the left side of the equation by <math>\frac{3}{3}</math>. We find that x=9.
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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>. We find that <math>x=9</math>. ~Andrew_Lu
~Andrew_Lu
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==Video Solution==
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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 21:22, 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?

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

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 $\frac{3}{3}$. We find that $x=9$. ~Andrew_Lu

Video Solution

https://youtu.be/GG61DsRqoo4 ~savannahsolver

See Also

2012 AMC 8 (ProblemsAnswer KeyResources)
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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