Difference between revisions of "2023 AMC 12A Problems/Problem 2"
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<math>\textbf{(A) }1\frac{4}{5}\qquad\textbf{(B) }2\qquad\textbf{(C) }2\frac{2}{5}\qquad\textbf{(D) }3\qquad\textbf{(E) }3\frac{3}{5}</math> | <math>\textbf{(A) }1\frac{4}{5}\qquad\textbf{(B) }2\qquad\textbf{(C) }2\frac{2}{5}\qquad\textbf{(D) }3\qquad\textbf{(E) }3\frac{3}{5}</math> | ||
| − | ==Solution 1== | + | ==Solution 1 (Substitution)== |
Use a system of equations. Let <math>x</math> be the weight of a pizza and <math>y</math> be the weight of a cup of orange slices. | Use a system of equations. Let <math>x</math> be the weight of a pizza and <math>y</math> be the weight of a cup of orange slices. | ||
We have <cmath>\frac{1}{3}x+\frac{7}{2}y=\frac{3}{4}x+\frac{1}{2}y.</cmath> | We have <cmath>\frac{1}{3}x+\frac{7}{2}y=\frac{3}{4}x+\frac{1}{2}y.</cmath> | ||
Revision as of 08:20, 10 November 2023
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 (Substitution)
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:
be the weight of a pizza.
be the 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
Solution 3
P/3 + 7/2 R = 3/4 P + R/2 where P = pizza weight and R = weight of cup of oranges
Since oranges weigh 1/4 pound per cup, the oranges on the LHS weigh 7/2 cups x 1/4 pounds/cup = 7/8 pound, and those on the RHS weigh 1/2 cup x 1/4 pounds/cup = 1/8 pound.
So P/3 + 7/8 pound = 3/4 P + 1/8 pound; P/3 + 3/4 pound = 3/4 P.
Multiplying both sides by LCM (3,4) = 12, we have
4P + 9# = 9P; 5P = 9#; P = weight of a large pizza = 9/5 pounds = 
pounds.
~Dilip
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. 
