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

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==Problem==
{{duplicate|[[2023 AMC 10A Problems/Problem 2|2023 AMC 10A #2]] and [[2023 AMC 12A Problems/Problem 2|2023 AMC 12A #2]]}}
 
==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>


==Solution 1==
==Solution 1 (Substitution)==
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.
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>
We have <cmath>\frac{1}{3}x+\frac{7}{2}y=\frac{3}{4}x+\frac{1}{2}y.</cmath>
Line 10: Line 13:
x&=\frac{36}{5}y.
x&=\frac{36}{5}y.
\end{align*}</cmath>
\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>
Plugging in <math>\frac{1}{4}</math> pounds for <math>y</math> by the given gives <math>\frac{9}{5}=\boxed{\textbf{(A) }1\frac{4}{5}}.</math>


~ItsMeNoobieboy
~ItsMeNoobieboy
~walmartbrian


==Solution 2==
==Solution 2==
Let <math>p</math> be the weight of a pizza.
Let:
    <math>o</math> be the weight of a cup of orange.
<math>p</math> be the weight of a pizza.
<math>o</math> be the weight of a cup of orange.


From the problem, we know that <math>o = \frac{1}{4}</math>.
From the problem, we know that <math>o = \frac{1}{4}</math>.
Line 29: Line 34:


~d_code
~d_code
==Solution 3==
<math>\frac{P}{3} + \frac{7}{2} R = \frac{3}{4} P + \frac{R}{2}</math> where <math>P</math> is the pizza weight and <math>R</math> is the weight of cup of oranges
Since oranges weigh <math>\frac{1}{4}</math> pound per cup, the oranges on the LHS weigh <math>\frac{7}{2}</math> cups x <math>\frac{1}{4}</math> pounds/cup = <math>\frac{7}{8}</math> pound,
and those on the RHS weigh <math>\frac{1}{2}</math> cup x <math>\frac{1}{4}</math> pounds/cup = <math>\frac{1}{8}</math> pound.
So <math>\frac{P}{3}</math> + <math>\frac{7}{8}</math> pound = <math>\frac{3}{4} P</math> + <math>\frac{1}{8}</math> pound; <math>\frac{P}{3}</math> + <math>\frac{3}{4}</math> pound = <math>\frac{3}{4} P</math>.
Multiplying both sides by <math>\text{lcm}(3,4) = 12</math>, we have
<math>4P + 9 = 9P</math>; <math>5P = 9</math>; <math>P</math> = weight of a large pizza = <math>\frac{9}{5}</math> pounds = <math>\boxed{\textbf{(A)}1 \frac{4}{5}}</math> pounds.
~Dilip
~<math>\LaTeX</math> by A_MatheMagician
==Video Solution by Little Fermat==
https://youtu.be/h2Pf2hvF1wE?si=IoOvSsibLfs3rusB&t=280
~little-fermat
==Video Solution by Math-X (First understand the problem!!!)==
https://youtu.be/GP-DYudh5qU?si=7af0rivWz7M-Pc71&t=285
==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 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==
==See Also==

Latest revision as of 20:12, 27 October 2025

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 by Little Fermat

https://youtu.be/h2Pf2hvF1wE?si=IoOvSsibLfs3rusB&t=280 ~little-fermat

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

https://youtu.be/GP-DYudh5qU?si=7af0rivWz7M-Pc71&t=285

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 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

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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