2023 AMC 12A Problems/Problem 2

The following problem is from both the 2023 AMC 10A #2 and 2023 AMC 12A #2, so both problems redirect to this page.

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 (Substitution)

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 2

Let: $p$ be the weight of a pizza. $o$ be the weight of a cup of orange.

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

Write the equation below: $\frac{1}{3} p + \frac{7}{2}\cdot\frac{1}{4} = \frac{3}{4} p + \frac{1}{2}\cdot\frac{1}{4}$

Solving for $p$ : $\frac{5}{12} p = \frac{3}{4}$

$p = \frac{9}{5} = \boxed{\textbf{(A) }1\frac{4}{5}}.$

~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 = $\boxed{\textbf{(A) 1 4/5}}$ pounds.

~Dilip

Video Solution by Math-X (First understand the problem!!!)

https://youtu.be/cMgngeSmFCY?si=ULlMU09VdlpsRW3n&t=205

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.

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