2025 AMC 12A Problems/Problem 2: Difference between revisions

Latest revision as of 17:58, 10 November 2025

The following problem is from both the 2025 AMC 10A #2 and 2025 AMC 12A #2, so both problems redirect to this page.

Problem

A box contains $10$ pounds of a nut mix that is $50$ percent peanuts, $20$ percent cashews, and $30$ percent almonds. A second nut mix containing $20$ percent peanuts, $40$ percent cashews, and $40$ percent almonds is added to the box resulting in a new nut mix that is $40$ percent peanuts. How many pounds of cashews are now in the box?

$\textbf{(A) } 3.5 \qquad\textbf{(B) } 4 \qquad\textbf{(C) } 4.5 \qquad\textbf{(D) } 5 \qquad\textbf{(E) } 6$

Solution 1

We are given $0.2(10) = 2$ pounds of cashews in the first box.

Denote the pounds of nuts in the second nut mix as $x.$

$5 + 0.2x = 0.4(10 + x)$ $0.2x = 1$ $x = 5$

Thus, we have $5$ pounds of the second mix.

$0.4(5) + 2 = 2 + 2 = \boxed{\text{(B) }4}$

~pigwash

~yuvaG (Formatting)

~LucasW (Minor LaTeX)

Solution 2

Let the number of pounds of nuts in the second nut mix be $x$ . Therefore, we get the equation $0.5 \cdot 10 + 0.2 \cdot x = 0.4(x+10)$ . Solving it, we get $x=5$ . Therefore the amount of cashews in the two bags is $0.2 \cdot 10 + 0.4 \cdot 5 = 4$ , so our answer choice is $\boxed{\textbf{(B)} 4}$ .

~iiiiiizh

~yuvaG - $\LaTeX$ Formatting ;)

~Amon26(really minor edits)

Solution 3

The percent of peanuts in the first mix is $10\%$ away from the total percentage of peanuts, and the percent of peanuts in the second mix is $20\%$ away from the total percentage. This means the first mix has twice as many nuts as the second mix, so the second mix has $5$ pounds. $0.20 \cdot 10 + 0.40 \cdot 5 = 4$ pounds of cashews. So our answer is, $\boxed{\textbf{(B)}4}$ ~LUCKYOKXIAO

~LEONG2023-Latex

Solution 4

Note that we can set the information given in the problem into a table shown below:

$\renewcommand{\arraystretch}{1.5} \begin{centering} \begin{array}{| c | c | c |} \hline \text{Peanuts} & \text{Cashews} & \text{Almonds}\\ \hline 5 & 2 & 3\\ \hline \frac{2}{10}x & \frac{4}{10}x & \frac{4}{10}x\\ \hline \end{array} \end{centering}$

We are given that the new nut mix will contain $40\%$ peanuts. Hence, $5 + \frac{2}{10}x$ is $40\%$ of the total mix which is $10 + x$ . Solving the equation $5 + \frac{2}{10}x = \frac{2}{5} \cdot (10 + x)$ yields $x=5.$ Therefore, the number of cashews in the new mix is equal to $2 + \frac{2}{5} \cdot 5 = \boxed{\textbf{(B)} 4}$ .

~Moonlight11

~TehSovietOnion (LaTeX)

Solution 5(extremely long, overcomplicated, never use on the test)

Note: This got messed up when putting into the wiki and it has been re-interpreted by AI. Please review this solution carefully and correct any AI errors.

1️⃣ Measure-Theoretic Setup

Let (Ω, F, μ) be a finite measure space, where Ω = {peanuts, cashews, almonds}.

Define a density function f_i : Ω → [0,1] representing the probability distribution (composition) of each mix i:

- f₁(peanuts) = 0.5, f₁(cashews) = 0.2, f₁(almonds) = 0.3 - f₂(peanuts) = 0.2, f₂(cashews) = 0.4, f₂(almonds) = 0.4

Each mix corresponds to a measure ν_i = m_i f_i μ, where m_i is the total mass (10 lb for i=1, unknown x lb for i=2).

The combined measure is: ν = ν₁ + ν₂ = (m₁f₁ + m₂f₂)μ

The normalized mixture (probability measure for composition) is: f = (m₁f₁ + m₂f₂) / (m₁ + m₂)

We are told that f(peanuts) = 0.4.

2️⃣ Functional Equation in Measure Form

This is equivalent to: [m₁f₁(peanuts) + m₂f₂(peanuts)] / (m₁ + m₂) = 0.4

Substitute m₁ = 10: [10(0.5) + x(0.2)] / (10 + x) = 0.4

Same as before — but this time we view x as a scalar measure parameter in the space of signed measures.

Solving yields: x = 5

3️⃣ Abstract Affine Geometry View

Let Δ₂ = {(p,c,a) ∈ ℝ³ : p+c+a=1, p,c,a≥0}, the 2-simplex representing all possible nut compositions.

Each mix is a point in this simplex: - v₁ = (0.5, 0.2, 0.3) - v₂ = (0.2, 0.4, 0.4)

The combined mix lies on the affine line joining them: v = (10v₁ + 5v₂) / 15

The map Φ: (ℝ₊)² → Δ₂, (m₁,m₂) ↦ (m₁v₁ + m₂v₂)/(m₁ + m₂) is an affine morphism of positive cones that collapses scalar measures to compositions.

The constraint π_p(v) = 0.4 defines a hyperplane section of the simplex, and the intersection with the line segment joining v₁, v₂ defines a unique barycentric coordinate λ = 1/3.

This corresponds to an affine convex combination: v = (1-λ)v₁ + λv₂, λ = 1/3

4️⃣ Categorical Abstract Algebra Interpretation

We can view the mixing process as a functor: Mix: (FinMeas, +) → (Δ₂, convex combinations)

where each object is a measure with labeled components (mass and composition), and morphisms are scalar additions of measures.

The condition "final mix has 40% peanuts" is a natural transformation constraint between two functors: Φ, Ψ: FinMeas → ℝ

where: - Φ(ν) = total mass of peanuts - Ψ(ν) = total mass

We require Φ(ν)/Ψ(ν) = 0.4.

This induces a categorical equation that forces the unique morphism ratio ν₂:ν₁ = 1:2.

Hence x = 5.

5️⃣ Differential-Geometric / Tangent-Space Insight

On the manifold M = Δ₂, the line of mixtures parameterized by x is a 1D affine submanifold: γ(x) = (10v₁ + xv₂)/(10 + x)

The constraint surface S = {v ∈ Δ₂ : p = 0.4} is a codimension-1 affine submanifold (a plane slice).

The intersection S ∩ Im(γ) is transversal because the derivative dπ_p(γ'(x)) ≠ 0.

Hence there exists a unique transverse intersection point x = 5.

That transversality guarantees that the equilibrium composition is structurally stable under small perturbations of the parameters — i.e., you could wiggle the percentages slightly and the solution still exists and varies smoothly.

6️⃣ Return to measurable quantity

Total cashew mass: M_cashew = 10(0.20) + 5(0.40) = 2 + 2 = 4 pounds

Video Solution by Power Solve

https://youtu.be/QBn439idcPo?si=jrzzKE72p29BIDQZ&t=102

Chinese Video Solution

https://www.bilibili.com/video/BV1S52uBoE8d/

~metrixgo

Video Solution (Intuitive, Quick Explanation!)

https://youtu.be/Qb-9KDYDDX8

~ Education, the Study of Everything

Video Solution (Fast and Easy)

https://youtu.be/YpJ3QZTmDuw?si=ucvH15JKX2tw4SKZ ~ Pi Academy

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=dAeyV60Hu5c

Video Solution by Daily Dose of Math

https://youtu.be/LN5ofIcs1kY

~Thesmartgreekmathdude

Video Solution

https://youtu.be/gWSZeCKrOfU

~MK

Video Solution

https://youtu.be/l1RY_C20Q2M

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2025 AMC 12A Problems/Problem 2: Difference between revisions

Latest revision as of 17:58, 10 November 2025

Contents

Problem

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5(extremely long, overcomplicated, never use on the test)

Video Solution by Power Solve

Chinese Video Solution

Video Solution (Intuitive, Quick Explanation!)

Video Solution (Fast and Easy)

Video Solution by SpreadTheMathLove

Video Solution by Daily Dose of Math

Video Solution

Video Solution

See Also

@@ Line 1: / Line 1: @@
+{{duplicate|[[2025 AMC 10A Problems/Problem 2|2025 AMC 10A #2]] and [[2025 AMC 12A Problems/Problem 2|2025 AMC 12A #2]]}}
 ==Problem==
+A box contains <imath>10</imath> pounds of a nut mix that is <imath>50</imath> percent peanuts, <imath>20</imath> percent cashews, and <imath>30</imath> percent almonds. A second nut mix containing <imath>20</imath> percent peanuts, <imath>40</imath> percent cashews, and <imath>40</imath> percent almonds is added to the box resulting in a new nut mix that is <imath>40</imath> percent peanuts. How many pounds of cashews are now in the box?
+<imath>\textbf{(A) } 3.5 \qquad\textbf{(B) } 4 \qquad\textbf{(C) } 4.5 \qquad\textbf{(D) } 5 \qquad\textbf{(E) } 6</imath>
+==Solution 1==
+We are given <imath>0.2(10) = 2</imath> pounds of cashews in the first box.
+Denote the pounds of nuts in the second nut mix as <imath>x.</imath>
+<cmath>5 + 0.2x = 0.4(10 + x)</cmath>
+<cmath>0.2x = 1</cmath>
+<cmath>x = 5</cmath>
+Thus, we have <imath>5</imath> pounds of the second mix.
+<cmath>0.4(5) + 2 = 2 + 2 = \boxed{\text{(B) }4}</cmath>
+~pigwash
+~yuvaG (Formatting)
+~LucasW (Minor LaTeX)
+==Solution 2==
+Let the number of pounds of nuts in the second nut mix be <imath>x</imath>. Therefore, we get the equation <imath>0.5 \cdot 10 + 0.2 \cdot x = 0.4(x+10)</imath>. Solving it, we get <imath>x=5</imath>. Therefore the amount of cashews in the two bags is <imath>0.2 \cdot 10 + 0.4
+\cdot 5 = 4</imath>, so our answer choice is <imath>\boxed{\textbf{(B)} 4}</imath>.
+~iiiiiizh
+~yuvaG - <imath>\LaTeX</imath> Formatting ;)
+~Amon26(really minor edits)
+==Solution 3==
+The percent of peanuts in the first mix is <imath>10\%</imath> away from the total percentage of peanuts, and the percent of peanuts in the second mix is <imath>20\%</imath> away from the total percentage. This means the first mix has twice as many nuts as the second mix, so the second mix has <imath>5</imath> pounds.
+<imath>0.20 \cdot 10 + 0.40 \cdot 5 = 4</imath> pounds of cashews. So our answer is, <imath>\boxed{\textbf{(B)}4}</imath>
+~LUCKYOKXIAO
+~LEONG2023-Latex
+==Solution 4==
+Note that we can set the information given in the problem into a table shown below:
+<cmath>\renewcommand{\arraystretch}{1.5}
+\begin{centering}
+\begin{array}{| c | c | c |}
+\hline
+\text{Peanuts} & \text{Cashews} & \text{Almonds}\\
+\hline
+& 2 & 3\\
+\hline
+\frac{2}{10}x & \frac{4}{10}x & \frac{4}{10}x\\
+\hline
+\end{array}
+\end{centering}
+</cmath>
+We are given that the new nut mix will contain <imath>40\%</imath> peanuts. Hence, <imath>5 + \frac{2}{10}x</imath> is <imath>40\%</imath> of the total mix which is <imath>10 + x</imath>.
+Solving the equation <imath>5 + \frac{2}{10}x = \frac{2}{5} \cdot (10 + x)</imath> yields <imath>x=5.</imath> Therefore, the number of cashews in the new mix is equal to <imath>2 + \frac{2}{5} \cdot 5 = \boxed{\textbf{(B)}  4}</imath>.
+~Moonlight11
+~TehSovietOnion (LaTeX)
+==Solution 5(extremely long, overcomplicated, never use on the test)==
+Note: This got messed up when putting into the wiki and it has been re-interpreted by AI. Please review this solution carefully and correct any AI errors.
+️⃣ Measure-Theoretic Setup
+Let (Ω, F, μ) be a finite measure space, where Ω = {peanuts, cashews, almonds}.
+Define a density function f_i : Ω → [0,1] representing the probability distribution (composition) of each mix i:
+- f₁(peanuts) = 0.5, f₁(cashews) = 0.2, f₁(almonds) = 0.3
+- f₂(peanuts) = 0.2, f₂(cashews) = 0.4, f₂(almonds) = 0.4
+Each mix corresponds to a measure ν_i = m_i f_i μ, where m_i is the total mass (10 lb for i=1, unknown x lb for i=2).
+The combined measure is:
+ν = ν₁ + ν₂ = (m₁f₁ + m₂f₂)μ
+The normalized mixture (probability measure for composition) is:
+f = (m₁f₁ + m₂f₂) / (m₁ + m₂)
+We are told that f(peanuts) = 0.4.
+️⃣ Functional Equation in Measure Form
+This is equivalent to:
+[m₁f₁(peanuts) + m₂f₂(peanuts)] / (m₁ + m₂) = 0.4
+Substitute m₁ = 10:
+[10(0.5) + x(0.2)] / (10 + x) = 0.4
+Same as before — but this time we view x as a scalar measure parameter in the space of signed measures.
+Solving yields: '''x = 5'''
+️⃣ Abstract Affine Geometry View
+Let Δ₂ = {(p,c,a) ∈ ℝ³ : p+c+a=1, p,c,a≥0}, the 2-simplex representing all possible nut compositions.
+Each mix is a point in this simplex:
+- v₁ = (0.5, 0.2, 0.3)
+- v₂ = (0.2, 0.4, 0.4)
+The combined mix lies on the affine line joining them:
+v = (10v₁ + 5v₂) / 15
+The map Φ: (ℝ₊)² → Δ₂, (m₁,m₂) ↦ (m₁v₁ + m₂v₂)/(m₁ + m₂) is an affine morphism of positive cones that collapses scalar measures to compositions.
+The constraint π_p(v) = 0.4 defines a hyperplane section of the simplex, and the intersection with the line segment joining v₁, v₂ defines a unique barycentric coordinate λ = 1/3.
+This corresponds to an affine convex combination:
+v = (1-λ)v₁ + λv₂, λ = 1/3
+️⃣ Categorical Abstract Algebra Interpretation
+We can view the mixing process as a functor:
+Mix: (FinMeas, +) → (Δ₂, convex combinations)
+where each object is a measure with labeled components (mass and composition), and morphisms are scalar additions of measures.
+The condition "final mix has 40% peanuts" is a natural transformation constraint between two functors:
+Φ, Ψ: FinMeas → ℝ
+where:
+- Φ(ν) = total mass of peanuts
+- Ψ(ν) = total mass
+We require Φ(ν)/Ψ(ν) = 0.4.
+This induces a categorical equation that forces the unique morphism ratio ν₂:ν₁ = 1:2.
+Hence '''x = 5'''.
+️⃣ Differential-Geometric / Tangent-Space Insight
+On the manifold M = Δ₂, the line of mixtures parameterized by x is a 1D affine submanifold:
+γ(x) = (10v₁ + xv₂)/(10 + x)
+The constraint surface S = {v ∈ Δ₂ : p = 0.4} is a codimension-1 affine submanifold (a plane slice).
+The intersection S ∩ Im(γ) is transversal because the derivative dπ_p(γ'(x)) ≠ 0.
+Hence there exists a unique transverse intersection point '''x = 5'''.
+That transversality guarantees that the equilibrium composition is structurally stable under small perturbations of the parameters — i.e., you could wiggle the percentages slightly and the solution still exists and varies smoothly.
+️⃣ Return to measurable quantity
+Total cashew mass:
+M_cashew = 10(0.20) + 5(0.40) = 2 + 2 = '''4 pounds'''
+==Video Solution by Power Solve==
+https://youtu.be/QBn439idcPo?si=jrzzKE72p29BIDQZ&t=102
+==Chinese Video Solution==
+https://www.bilibili.com/video/BV1S52uBoE8d/
+~metrixgo
+== Video Solution (Intuitive, Quick Explanation!) ==
+https://youtu.be/Qb-9KDYDDX8
+~ Education, the Study of Everything
+== Video Solution (Fast and Easy) ==
+https://youtu.be/YpJ3QZTmDuw?si=ucvH15JKX2tw4SKZ ~ Pi Academy
+==Video Solution by SpreadTheMathLove==
+https://www.youtube.com/watch?v=dAeyV60Hu5c
+==Video Solution by Daily Dose of Math==
+https://youtu.be/LN5ofIcs1kY
-== Solution 1==
+~Thesmartgreekmathdude
+==Video Solution==
+https://youtu.be/gWSZeCKrOfU
-The initial box has 10 pounds. With <imath>50</imath> percent of it being peanuts, there are <imath>0.5\cdot10 = 5</imath> pounds of peanuts.
+~MK
-We then add <imath>x</imath> pounds of a second mix, which is <imath>20</imath> percent peanuts, causing the peanuts to now be <imath>40</imath> percent of the total. We write the equation
+==Video Solution==
-<cmath>\frac{5+0.2x}{10+x} = 0.4</cmath>
+https://youtu.be/l1RY_C20Q2M
-<cmath>5+0.2x = 4+x</cmath>
-<cmath>x=5.</cmath>
-This means the second mix was a total of <imath>5</imath> pounds. Because <imath>40</imath> percent of that is cashews, there are <imath>0.4\cdot 5 = 2</imath> cashews in the second mix. The original mixture was <imath>20</imath> percent cashews, so there were <imath>0.2\cdot 10 = 2</imath> cashews originally. So we now have <imath>2+2 = \boxed{4}</imath> cashews.
-~lprado
+==See Also==
+{{AMC10 box|year=2025|ab=A|num-b=1|num-a=3}}
+{{AMC12 box|year=2025|ab=A|num-b=1|num-a=3}}
+{{MAA Notice}}