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>
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==Solution==
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Since Rachelle uses 3 pounds of meat to make 8 hamburgers, she uses <math>\frac{3}{8}</math> pounds of meat to make one hamburger. To find out the pounds of meat she needs for 24 hamburgers, multiply <math>\frac{3}{8}</math> and 24 to obtain the answer, <math>\boxed{\textbf{(E)}\ 9}</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2012|before=First Problem|num-a=2}}
 
{{AMC8 box|year=2012|before=First Problem|num-a=2}}

Revision as of 10:38, 24 November 2012

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?

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

Solution

Since Rachelle uses 3 pounds of meat to make 8 hamburgers, she uses $\frac{3}{8}$ pounds of meat to make one hamburger. To find out the pounds of meat she needs for 24 hamburgers, multiply $\frac{3}{8}$ and 24 to obtain the answer, $\boxed{\textbf{(E)}\ 9}$.

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