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 3== | ==Solution 3== | ||
Revision as of 15:42, 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 3
Since the first box had 5 pounds, and 50 percent off it had peanuts, we know there were 5 pounds of peanuts at the beginning. Adding the second mixture of nuts, we call this value x, as in x pounds. Of that 20% or x/5, are peanuts. Since the final percentage in 40 percent peanuts, (5+x/5)/(10+x)=2/5. Multiplying both sides by 5, we get, 25+x/10+x=2.
Multiplying both sides by 10+x, we get 25+x=20+2x. This gives us x=5. But the problem is asking us to solve for cashews. The first mixture was 1/5 cashews, there were 2 cashews in the first mix. In the second, there were 2x/5 cashews, or 2 pounds of cashews. Adding this together gives us a final total of 
cashews.
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. 
