Difference between revisions of "2012 AMC 8 Problems/Problem 1"
(Undo revision 137200 by Srijankundu (talk)) (Tag: Undo) |
(→Problem) |
||
(One intermediate revision by the same user 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 | + | 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> |
Latest revision as of 15:17, 23 November 2020
Contents
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?
Solution 1
Since Rachelle uses pounds of meat to make hamburgers, she uses pounds of meat to make one hamburger. She'll need 24 times that amount of meat for 24 hamburgers, or .
Solution 2
So we have the proportion = to solve for x, multiply the left side of the equation by . We find that x=9. ~Andrew_Lu
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.