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

Revision as of 16:46, 8 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, $\boxed{\textbf{(B)}4}$ On the manifold M = Δ 2 M=Δ 2

, the line of mixtures parameterized by

x x is a 1D affine submanifold: γ ( x ) = 10 v 1 + x v 2 10 + x . γ(x)= 10+x 10v 1

+xv

2

The constraint surface S = { v ∈ Δ 2

p = 0.4 } S={v∈Δ 2

:p=0.4} is a codimension-1 affine submanifold (a plane slice).

The intersection S ∩ Im ( γ ) S∩Im(γ) is transversal because the derivative d π p ( γ ′ ( x ) ) ≠ 0 dπ p

(γ

′

(x))

 =0. Hence there exists a unique transverse intersection point x = 5 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.

~LUCKYOKXIAO

~LEONG2023-Latex

Solution 1(extremely long, overcomplicated, never use on the test)

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

Ω → [ 0 , 1 ] f i

:Ω→[0,1] representing the probability distribution (composition) of each mix

i i: f 1 ( peanuts ) = 0.5 , f 1 ( cashews ) = 0.2 , f 1 ( almonds ) = 0.3 , f 1

(peanuts)=0.5,f

1

(cashews)=0.2,f

1

(almonds)=0.3,

f 2 ( peanuts ) = 0.2 , f 2 ( cashews ) = 0.4 , f 2 ( almonds ) = 0.4. f 2

(peanuts)=0.2,f

2

(cashews)=0.4,f

2

(almonds)=0.4.

Each mix corresponds to a measure ν i = m i f i μ ν i

=m

i

μ,

where m i m i

 is the total mass (10 lb for

i = 1 i=1, unknown x x lb for i = 2 i=2). The combined measure is ν = ν 1 + ν 2 = ( m 1 f 1 + m 2 f 2 ) μ . ν=ν 1

+ν

2

=(m

1

+m

2

)μ.

The normalized mixture (probability measure for composition) is: f = m 1 f 1 + m 2 f 2 m 1 + m 2 . f= m 1

+m

2

m 1

1

+m

2

We are told that f ( peanuts ) = 0.4. f(peanuts)=0.4. 2️⃣ Functional Equation in Measure Form This is equivalent to: m 1 f 1 ( peanuts ) + m 2 f 2 ( peanuts ) m 1 + m 2 = 0.4. m 1

+m

2

m 1

1

(peanuts)+m

2

(peanuts)

=0.4.

Substitute m 1 = 10 m 1

=10:

10 ( 0.5 ) + x ( 0.2 ) 10 + x = 0.4. 10+x 10(0.5)+x(0.2)

=0.4.

Same as before — but this time we view x x as a scalar measure parameter in the space of signed measures. Solving yields: x = 5. x=5. 3️⃣ Abstract Affine Geometry View Let Δ 2 = { ( p , c , a ) ∈ R 3

p + c + a = 1 , p , c , a ≥ 0 } Δ 2

={(p,c,a)∈R

3

:p+c+a=1,p,c,a≥0}, the 2-simplex representing all possible nut compositions.

Each mix is a point in this simplex: v 1 = ( 0.5 , 0.2 , 0.3 ) , v 2 = ( 0.2 , 0.4 , 0.4 ) . v 1

=(0.5,0.2,0.3),v

2

=(0.2,0.4,0.4).

The combined mix lies on the affine line joining them: v = 10 v 1 + 5 v 2 15 . v= 15 10v 1

+5v

2

The map Φ

( R > 0 ) 2 → Δ 2 , ( m 1 , m 2 ) ↦ m 1 v 1 + m 2 v 2 m 1 + m 2 Φ:(R >0

2

→Δ

2

,(m

1

,m

2

)↦

m 1

+m

2

m 1

1

+m

2

is an affine morphism of positive cones that collapses scalar measures to compositions. The constraint π p ( v ) = 0.4 π p

(v)=0.4 defines a hyperplane section of the simplex, and the intersection with the line segment joining

v 1 , v 2 v 1

,v

2

 defines a unique barycentric coordinate

λ = 1 3 λ= 3 1

This corresponds to an affine convex combination: v = ( 1 − λ ) v 1 + λ v 2 , λ = 1 3 . v=(1−λ)v 1

+λv

2

,λ=

3 1

4️⃣ Categorical Abstract Algebra Interpretation We can view the mixing process as a functor: M i x

( F i n M e a s , + ) → ( Δ 2 , convex combinations ) , Mix:(FinMeas,+)→(Δ 2

,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

F i n M e a s → R , Φ ( ν ) = total mass of peanuts , Ψ ( ν ) = total mass . Φ,Ψ:FinMeas→R,Φ(ν)=total mass of peanuts,Ψ(ν)=total mass. We require Φ ( ν ) / Ψ ( ν ) = 0.4. Φ(ν)/Ψ(ν)=0.4. This induces a categorical equation that forces the unique morphism ratio ν 2

ν 1 = 1

2 ν 2

:ν

1

=1:2.

Hence x = 5. x=5. 5️⃣ Differential-Geometric / Tangent-Space Insight On the manifold M = Δ 2 M=Δ 2

, the line of mixtures parameterized by

x x is a 1D affine submanifold: γ ( x ) = 10 v 1 + x v 2 10 + x . γ(x)= 10+x 10v 1

+xv

2

The constraint surface S = { v ∈ Δ 2

p = 0.4 } S={v∈Δ 2

:p=0.4} is a codimension-1 affine submanifold (a plane slice).

The intersection S ∩ Im ( γ ) S∩Im(γ) is transversal because the derivative d π p ( γ ′ ( x ) ) ≠ 0 dπ p

(γ

′

(x))

 =0. Hence there exists a unique transverse intersection point x = 5 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. M cashew

=10(0.20)+5(0.40)=2+2=4.

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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Preceded by Problem 1	Followed by Problem 3
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2025 AMC 12A Problems/Problem 2: Difference between revisions

Revision as of 16:46, 8 November 2025

Contents

Problem

Solution 1

Solution 2

Solution 3

Solution 1(extremely long, overcomplicated, never use on the test)

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 6: / Line 6: @@
 <imath>\textbf{(A) } 3.5 \qquad\textbf{(B) } 4 \qquad\textbf{(C) } 4.5 \qquad\textbf{(D) } 5 \qquad\textbf{(E) } 6</imath>
-==Solution 2==
+==Solution 1==
 We are given <imath>0.2(10) = 2</imath> pounds of cashews in the first box.
@@ Line 27: / Line 27: @@
 ~LucasW (Minor LaTeX)
-==Solution 3==
+==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
@@ Line 38: / Line 38: @@
 ~Amon26(really minor edits)
-==Solution 4==
+==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.
@@ Line 132: / Line 132: @@
 ~LEONG2023-Latex
+==Solution 1(extremely long, overcomplicated, never use on the test)==
+Let
+(
+Ω
+,
+F
+,
+μ
+)
+(Ω,F,μ)
+be a finite measure space, where
+Ω
+=
+{
+peanuts
+,
+cashews
+,
+almonds
+}
+Ω={peanuts,cashews,almonds}.
+Define a density function
+f
+i
+:
+Ω
+→
+[
+,
+]
+f
+i
+ :Ω→[0,1] representing the probability distribution (composition) of each mix
+i
+i:
+f
+(
+peanuts
+)
+=
+.5
+,
+f
+(
+cashews
+)
+=
+.2
+,
+f
+(
+almonds
+)
+=
+.3
+,
+f
+ (peanuts)=0.5,f
+ (cashews)=0.2,f
+ (almonds)=0.3,
+f
+(
+peanuts
+)
+=
+.2
+,
+f
+(
+cashews
+)
+=
+.4
+,
+f
+(
+almonds
+)
+=
+.4.
+f
+ (peanuts)=0.2,f
+ (cashews)=0.4,f
+ (almonds)=0.4.
+Each mix corresponds to a measure
+ν
+i
+=
+m
+i
+f
+i
+μ
+ν
+i
+ =m
+i
+ f
+i
+ μ,
+where
+m
+i
+m
+i
+  is the total mass (10 lb for
+i
+=
+i=1, unknown
+x
+x lb for
+i
+=
+i=2).
+The combined measure is
+ν
+=
+ν
++
+ν
+=
+(
+m
+f
++
+m
+f
+)
+μ
+.
+ν=ν
+ +ν
+ =(m
+ f
+ +m
+ f
+ )μ.
+The normalized mixture (probability measure for composition) is:
+f
+=
+m
+f
++
+m
+f
+m
++
+m
+.
+f=
+m
+ +m
+m
+ f
+ +m
+ f
+ .
+We are told that
+f
+(
+peanuts
+)
+=
+.4.
+f(peanuts)=0.4.
+️⃣ Functional Equation in Measure Form
+This is equivalent to:
+m
+f
+(
+peanuts
+)
++
+m
+f
+(
+peanuts
+)
+m
++
+m
+=
+.4.
+m
+ +m
+m
+ f
+ (peanuts)+m
+ f
+ (peanuts)
+ =0.4.
+Substitute
+m
+=
+m
+ =10:
+(
+.5
+)
++
+x
+(
+.2
+)
++
+x
+=
+.4.
++x
+(0.5)+x(0.2)
+ =0.4.
+Same as before — but this time we view
+x
+x as a scalar measure parameter in the space of signed measures.
+Solving yields:
+x
+=
+.
+x=5.
+️⃣ Abstract Affine Geometry View
+Let
+Δ
+=
+{
+(
+p
+,
+c
+,
+a
+)
+∈
+R
+:
+p
++
+c
++
+a
+=
+,
+p
+,
+c
+,
+a
+≥
+}
+Δ
+ ={(p,c,a)∈R
+ :p+c+a=1,p,c,a≥0}, the 2-simplex representing all possible nut compositions.
+Each mix is a point in this simplex:
+v
+=
+(
+.5
+,
+.2
+,
+.3
+)
+,
+v
+=
+(
+.2
+,
+.4
+,
+.4
+)
+.
+v
+ =(0.5,0.2,0.3),v
+ =(0.2,0.4,0.4).
+The combined mix lies on the affine line joining them:
+v
+=
+v
++
+v
+.
+v=
+v
+ +5v
+ .
+The map
+Φ
+:
+(
+R
+>
+)
+→
+Δ
+,
+(
+m
+,
+m
+)
+↦
+m
+v
++
+m
+v
+m
++
+m
+Φ:(R
+>0
+ )
+ →Δ
+ ,(m
+ ,m
+ )↦
+m
+ +m
+m
+ v
+ +m
+ v
+is an affine morphism of positive cones that collapses scalar measures to compositions.
+The constraint
+π
+p
+(
+v
+)
+=
+.4
+π
+p
+ (v)=0.4 defines a hyperplane section of the simplex, and the intersection with the line segment joining
+v
+,
+v
+v
+ ,v
+  defines a unique barycentric coordinate
+λ
+=
+λ=
+ .
+This corresponds to an affine convex combination:
+v
+=
+(
+−
+λ
+)
+v
++
+λ
+v
+,
+λ
+=
+.
+v=(1−λ)v
+ +λv
+ ,λ=
+ .
+️⃣ Categorical Abstract Algebra Interpretation
+We can view the mixing process as a functor:
+M
+i
+x
+:
+(
+F
+i
+n
+M
+e
+a
+s
+,
++
+)
+→
+(
+Δ
+,
+convex combinations
+)
+,
+Mix:(FinMeas,+)→(Δ
+ ,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
+:
+F
+i
+n
+M
+e
+a
+s
+→
+R
+,
+Φ
+(
+ν
+)
+=
+total mass of peanuts
+,
+Ψ
+(
+ν
+)
+=
+total mass
+.
+Φ,Ψ:FinMeas→R,Φ(ν)=total mass of peanuts,Ψ(ν)=total mass.
+We require
+Φ
+(
+ν
+)
+/
+Ψ
+(
+ν
+)
+=
+.4.
+Φ(ν)/Ψ(ν)=0.4.
+This induces a categorical equation that forces the unique morphism ratio
+ν
+:
+ν
+=
+:
+ν
+ :ν
+ =1:2.
+Hence
+x
+=
+.
+x=5.
+️⃣ Differential-Geometric / Tangent-Space Insight
+On the manifold
+M
+=
+Δ
+M=Δ
+ , the line of mixtures parameterized by
+x
+x is a 1D affine submanifold:
+γ
+(
+x
+)
+=
+v
++
+x
+v
++
+x
+.
+γ(x)=
++x
+v
+ +xv
+ .
+The constraint surface
+S
+=
+{
+v
+∈
+Δ
+:
+p
+=
+.4
+}
+S={v∈Δ
+ :p=0.4} is a codimension-1 affine submanifold (a plane slice).
+The intersection
+S
+∩
+Im
+(
+γ
+)
+S∩Im(γ) is transversal because the derivative
+d
+π
+p
+(
+γ
+′
+(
+x
+)
+)
+≠
+dπ
+p
+ (γ
+′
+ (x))
+
+=0.
+Hence there exists a unique transverse intersection point
+x
+=
+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
+=
+(
+.20
+)
++
+(
+.40
+)
+=
++
+=
+.
+M
+cashew
+ =10(0.20)+5(0.40)=2+2=4.
 ==Chinese Video Solution==