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Difference between revisions of "2023 AMC 12A Problems/Problem 2"
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{{duplicate|[[2023 AMC 10A Problems/Problem 2|2023 AMC 10A #2]] and [[2023 AMC 12A Problems/Problem 2|2023 AMC 12A #2]]}}
 
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==Problem==
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==Problem 2==
 
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?
 
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>
 
<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>

Revision as of 06:36, 26 August 2024

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

Problem 2

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 $y$ by the given gives $\frac{9}{5}=\boxed{\textbf{(A) }1\frac{4}{5}}.$

~ItsMeNoobieboy ~walmartbrian

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

$\frac{P}{3} + \frac{7}{2} R = \frac{3}{4} P + \frac{R}{2}$ where $P$ is the pizza weight and $R$ is the weight of cup of oranges Since oranges weigh $\frac{1}{4}$ pound per cup, the oranges on the LHS weigh $\frac{7}{2}$ cups x $\frac{1}{4}$ pounds/cup = $\frac{7}{8}$ pound, and those on the RHS weigh $\frac{1}{2}$ cup x $\frac{1}{4}$ pounds/cup = $\frac{1}{8}$ pound.

So $\frac{P}{3}$ + $\frac{7}{8}$ pound = $\frac{3}{4} P$ + $\frac{1}{8}$ pound; $\frac{P}{3}$ + $\frac{3}{4}$ pound = $\frac{3}{4} P$.

Multiplying both sides by $\text{lcm}(3,4) = 12$, we have $4P + 9 = 9P$; $5P = 9$; $P$ = weight of a large pizza = $\frac{9}{5}$ pounds = $\boxed{\textbf{(A)}1 \frac{4}{5}}$ pounds.

~Dilip ~$\LaTeX$ by A_MatheMagician

Video Solution (easy to digest) by Power Solve

https://www.youtube.com/watch?v=8huvzWTtgaU

Video Solution (🚀 Just 1 min 🚀)

https://youtu.be/7ADHHpSNMsE

~Education, the Study of Everything

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

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

Video Solution by CosineMethod [🔥Fast and Easy🔥]

https://www.youtube.com/watch?v=CpboCxGBcWY

Video Solution

https://youtu.be/k8hnq3pPpc0

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)


Video Solution (Easy to Understand) by DR.GOOGLE (YT: Pablo's Math)

https://youtu.be/N5cHw8ODT_I

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

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