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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> cups of orange slices. A cup of orange slices weight <math>\frac{1}{4}</math> of a pound. What is the weight, in pounds, of a large pizza?
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==Problem==
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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 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?
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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>
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==Solution 1==
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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.
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We have <cmath>\frac{1}{3}x+\frac{7}{2}y=\frac{3}{4}x+\frac{1}{2}y.</cmath>
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Rearranging, we get <cmath>\begin{align*}
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\frac{5}{12}x&=3y, \\
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x&=\frac{36}{5}y.
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\end{align*}</cmath>
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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>
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~ItsMeNoobieboy

Revision as of 20:33, 9 November 2023

Problem

The weight of $\frac{1}{3}$ of a large pizza together with $3 \frac{1}{2}$ cups of orange slices is the same as the weight of $\frac{3}{4}$ of a large pizza together with $\frac{1}{2}$ cup of orange slices. A cup of orange slices weighs $\frac{1}{4}$ of a pound. What is the weight, in pounds, of a large pizza? $\textbf{(A) }1\frac{4}{5}\qquad\textbf{(B) }2\qquad\textbf{(C) }2\frac{2}{5}\qquad\textbf{(D) }3\qquad\textbf{(E) }3\frac{3}{5}$

Solution 1

Use a system of equations. Let $x$ be the weight of a pizza and $y$ be the weight of a cup of orange slices. We have \[\frac{1}{3}x+\frac{7}{2}y=\frac{3}{4}x+\frac{1}{2}y.\] Rearranging, we get \begin{align*} \frac{5}{12}x&=3y, \\ x&=\frac{36}{5}y. \end{align*} Plugging in $\frac{1}{4}$ pounds for $y$ gives $\frac{9}{5}=\boxed{\textbf{(A) }1\frac{4}{5}}.$

~ItsMeNoobieboy

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