Difference between revisions of "2023 AMC 12A Problems/Problem 2"
(fixed typos in problem, added answer choices, added a solution) |
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− | The weight of <math>\frac{1}{3}</math> of a large pizza together with <math>3 \frac{1}{2}</math> cups of orange slices is the same weight of <math>\frac{3}{4}</math> of a large pizza together with <math>\frac{1}{2}</math> | + | ==Problem== |
+ | The weight of <math>\frac{1}{3}</math> of a large pizza together with <math>3 \frac{1}{2}</math> cups of orange slices is the same as the weight of <math>\frac{3}{4}</math> of a large pizza together with <math>\frac{1}{2}</math> cup of orange slices. A cup of orange slices weighs <math>\frac{1}{4}</math> of a pound. What is the weight, in pounds, of a large pizza? | ||
+ | <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== | ||
+ | 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> | ||
+ | Rearranging, we get <cmath>\begin{align*} | ||
+ | \frac{5}{12}x&=3y, \ | ||
+ | x&=\frac{36}{5}y. | ||
+ | \end{align*}</cmath> | ||
+ | Plugging in <math>\frac{1}{4}</math> pounds for <math>y</math> gives <math>\frac{9}{5}=\boxed{\textbf{(A) }1\frac{4}{5}}.</math> | ||
+ | |||
+ | ~ItsMeNoobieboy |
Revision as of 19:33, 9 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
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