|
|
| 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]]}}
| | #redirect [[2025 AMC 12A Problems/Problem 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>
| |
| | |
| ==Video Solution==
| |
| https://youtu.be/l1RY_C20Q2M
| |
| | |
| | |
| 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 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>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
| |
| | |
| ==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}}
| |
| * [[AMC 10]]
| |
| * [[AMC 10 Problems and Solutions]]
| |
| * [[Mathematics competitions]]
| |
| * [[Mathematics competition resources]]
| |
| {{MAA Notice}}
| |
