Difference between revisions of "2023 AMC 12A Problems/Problem 2"
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==Solution 2== | ==Solution 2== | ||
| − | Let <math>p = </math> weight of a pizza | + | Let <math>p =</math> weight of a pizza. |
| − | <math>o = </math> weight of a cup of orange | + | <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 below: | Write the equation below: | ||
| − | < | + | <cmath>\frac{1}{3} p + \frac{7}{2}\cdot\frac{1}{4} = \frac{3}{4} p + \frac{1}{2}\cdot\frac{1}{4}</cmath> |
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> | + | <math>p = \frac{9}{5} = \boxed{\textbf{(A) }1\frac{4}{5}}.</math> |
~d_code | ~d_code | ||
Revision as of 19:49, 9 November 2023
Contents
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 ![\[\frac{1}{3}x+\frac{7}{2}y=\frac{3}{4}x+\frac{1}{2}y.\]](http://latex.artofproblemsolving.com/4/2/8/428df82b977e9784956fa0bfdc5b37f60f72762c.png)
Rearranging, we get 
Plugging in pounds for gives 
~ItsMeNoobieboy
Solution 2
Let 
weight of a pizza.
weight of a cup of orange.
weight of a cup of orange.
From the problem, we know that 
.
Write the equation below:
![\[\frac{1}{3} p + \frac{7}{2}\cdot\frac{1}{4} = \frac{3}{4} p + \frac{1}{2}\cdot\frac{1}{4}\]](http://latex.artofproblemsolving.com/f/9/0/f9081673d77e9ec514454909c293307aedf2bf6f.png)
Solving for 
:
~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. 
