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

Revision as of 10:21, 23 November 2013

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?

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

Solution

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

Rachelle uses meat for hamburgers in the ratio $\frac{3\text{ lb}}{8\text{ hamburgers}}$ . If we multiply this ratio on top and bottom by 3 to get 24 hamburgers on the bottom, we get $\frac{3\text{ lb}\cdot 3}{8\text{ hamburgers}\cdot 3} = \frac{9\text{ lb}}{24\text{ hamburgers}}$ . Therefore, she uses 9 pounds of meat for 24 hamburgers, making the answer $\boxed{\textbf{(E)}\ 9}$ .

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.

@@ Line 5: / Line 5: @@
 ==Solution==
-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. To find out the pounds of meat she needs for <math>24</math> hamburgers, multiply <math>\frac{3}{8}</math> and <math>24</math> to obtain the answer, <math>\boxed{\textbf{(E)}\ 9}</math>.
+===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>.
+===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>.
 ==See Also==
 {{AMC8 box|year=2012|before=First Problem|num-a=2}}
 {{MAA Notice}}

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