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

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<imath>\textbf{(A) } 3.5 \qquad\textbf{(B) } 4 \qquad\textbf{(C) } 4.5 \qquad\textbf{(D) } 5 \qquad\textbf{(E) } 6</imath>
<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(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
​
f
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
​
f
1
​
+m
2
​
f
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
​
f
1
​
+m
2
​
f
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
​
f
1
​
(peanuts)+m
2
​
f
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
​
v
1
​
+m
2
​
v
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.


==Solution 2==
==Solution 2==

Revision as of 16:45, 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 2

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 3

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 4

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

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

See Also

2025 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
2025 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.

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