2025 AMC 10A 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 1
We are given
pounds of cashews in the first box.
Denote the pounds of nuts in the second nut mix as
Thus, we have 5 pounds of the second mix.
~pigwash
~yuvaG (Formatting)
Solution 2
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 out answer choice is
.
~iiiiiizh
~yuvaG -
Formatting ;)
Solution 3
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.
0.20 \cdot 10 + 0.40 \cdot 5 = 4 pounds of cashews. So our answer is,
~LUCKYOKXIAO ~LEONG2023-Latex
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 Daily Dose of Math
~Thesmartgreekmathdude
Video Solution
~MK
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.