2025 AMC 12A Problems/Problem 2
- 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 
pounds of a nut mix that is 
percent peanuts, 
percent cashews, and 
percent almonds. A second nut mix containing percent peanuts, 
percent cashews, and percent almonds is added to the box resulting in a new nut mix that is percent peanuts. How many pounds of cashews are now in the box?
Solution 2
We are given 
pounds of cashews in the first box.
Denote the pounds of nuts in the second nut mix as 
![\[5 + 0.2x = 0.4(10 + x)\]](http://latex-new.aopstest.com/4/6/5/465abe12e05bb15bc19b04d9c82116ed04c90e91.png)
![\[0.2x = 1\]](http://latex-new.aopstest.com/c/4/a/c4afc4b51c5f6fa0f9bf78a34e438ededa7fd19e.png)
![\[x = 5\]](http://latex-new.aopstest.com/6/7/7/6773f565392dfa07209879afbc209b09615f5048.png)
Thus, we have 
pounds of the second mix.
![\[0.4(5) + 2 = 2 + 2 = \boxed{\text{(B) }4}\]](http://latex-new.aopstest.com/b/4/3/b43ffd87b3276968f1dad2a69b98033075c839d7.png)
~pigwash
~yuvaG (Formatting)
~LucasW (Minor LaTeX)
Solution 3
Let the number of pounds of nuts in the second nut mix be 
. Therefore, we get the equation 
. Solving it, we get 
. Therefore the amount of cashews in the two bags is 
, so our answer choice is 
.
~iiiiiizh
~yuvaG - 
Formatting ;)
~Amon26(really minor edits)
Solution 4
The percent of peanuts in the first mix is 
away from the total percentage of peanuts, and the percent of peanuts in the second mix is 
away from the total percentage. This means the first mix has twice as many nuts as the second mix, so the second mix has pounds.
pounds of cashews. So our answer is, 
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!)
~ 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
~Thesmartgreekmathdude
Video Solution
~MK
Video Solution
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. 
