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2023 AMC 12A Problems/Problem 2: Difference between revisions

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~ItsMeNoobieboy
~ItsMeNoobieboy
==Solution 1==
Let <math>p = weight of a pizza</math>
    <math>o = weight of an orange</math>
From the problem, we know that <math>o = \frac{1}{4}</math>.
Write the equation as below:
<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>:
<math>\frac{5}{12} p = \frac{3}{4}</math>
<math>p = \frac{9}{5} = \boxed{\textbf{(A) }1\frac{4}{5}.</math>
~d_code


==See Also==
==See Also==

Revision as of 19:46, 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

Solution 1

Let $p = weight of a pizza$ 

   $o = weight of an orange$

From the problem, we know that $o = \frac{1}{4}$.

Write the equation as below: $\frac{1}{3} p + \frac{7}{2}\cdot\frac{1}{4} = \frac{3}{4} + \frac{1}{2}\cdot\frac{1}{4}$ Solving for $p$: $\frac{5}{12} p = \frac{3}{4}$ $p = \frac{9}{5} = \boxed{\textbf{(A) }1\frac{4}{5}.$ (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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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