2025 AMC 10A 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 | ==Solution == | ||
Since the first box had 5 pounds, and 50 percent of it had peanuts, we know there were 5 pounds of peanuts at the beginning. | Since the first box had 5 pounds, and 50 percent of it had peanuts, we know there were 5 pounds of peanuts at the beginning. | ||
Revision as of 16:38, 6 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 
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
Since the first box had 5 pounds, and 50 percent of it had peanuts, we know there were 5 pounds of peanuts at the beginning.
Adding the second mixture of nuts, we call this value 
, as in pounds.
Of that, 20%, or 
, are peanuts.
Since the final percentage is 40 percent peanuts, we have
![\[\frac{5 + \frac{x}{5}}{10 + x} = \frac{2}{5}.\]](http://latex-new.aopstest.com/7/7/6/776314129416d02390b520c1ff00db9a1a9f9ea0.png)
Multiplying both sides by 
, we get
![\[25 + x = 20 + 2x.\]](http://latex-new.aopstest.com/d/a/8/da8b9c4f90f3fef47f5a9acb6326f5a173452ddb.png)
This gives us 
.
But the problem is asking us to solve for cashews.
The first mixture was 
cashews, so there were 
pounds of cashews in the first mix.
In the second, there were 
cashews, or 2 pounds of cashews.
Adding this together gives us a final total of
![\[2 + 2 = \boxed{4}\]](http://latex-new.aopstest.com/c/f/3/cf3921d27dc5c16215383695024c91b482d996f7.png)
pounds of cashews.
- ~Minor edits to LaTeX by WildSealVM/Vincent M
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. 
