Difference between revisions of "2023 AMC 12A Problems/Problem 2"
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~ItsMeNoobieboy | ~ItsMeNoobieboy | ||
− | ==Solution | + | ==Solution 2== |
− | Let <math>p = weight of a pizza | + | Let <math>p = </math> weight of a pizza |
− | <math>o = | + | <math>o = </math> weight of a cup of orange |
From the problem, we know that <math>o = \frac{1}{4}</math>. | From the problem, we know that <math>o = \frac{1}{4}</math>. | ||
− | Write the equation | + | Write the equation below: |
<math>\frac{1}{3} p + \frac{7}{2}\cdot\frac{1}{4} = \frac{3}{4} + \frac{1}{2}\cdot\frac{1}{4}</math> | <math>\frac{1}{3} p + \frac{7}{2}\cdot\frac{1}{4} = \frac{3}{4} + \frac{1}{2}\cdot\frac{1}{4}</math> | ||
+ | |||
Solving for <math>p</math>: | Solving for <math>p</math>: | ||
<math>\frac{5}{12} p = \frac{3}{4}</math> | <math>\frac{5}{12} p = \frac{3}{4}</math> | ||
− | <math>p = \frac{9}{5} = \boxed{\textbf{(A) }1\frac{4}{5} | + | |
+ | <math>p = \frac{9}{5} = \boxed{\textbf{(A) }1\frac{4}{5}</math>. | ||
~d_code | ~d_code |
Revision as of 19:47, 9 November 2023
Contents
[hide]Problem
The weight of of a large pizza together with cups of orange slices is the same as the weight of of a large pizza together with cup of orange slices. A cup of orange slices weighs of a pound. What is the weight, in pounds, of a large pizza?
Solution 1
Use a system of equations. Let be the weight of a pizza and be the weight of a cup of orange slices. We have Rearranging, we get Plugging in pounds for gives
~ItsMeNoobieboy
Solution 2
Let weight of a pizza
weight of a cup of orange
From the problem, we know that .
Write the equation below:
Solving for :
$p = \frac{9}{5} = \boxed{\textbf{(A) }1\frac{4}{5}$ (Error compiling LaTeX. Unknown error_msg).
~d_code
See Also
2023 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
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 AMC 10 Problems and Solutions |
2023 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 1 |
Followed by Problem 3 |
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 AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.