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
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 \( x \), as in \( x \) pounds. \\ Of that, 20\%, or \( \frac{x}{5} \), are peanuts. \\
Since the final percentage is 40 percent peanuts, we have: \[ \frac{5 + \frac{x}{5}}{10 + x} = \frac{2}{5}. \]
Multiplying both sides by \( 5(10 + x) \), 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 \( \frac{1}{5} \) cashews, so there were \( 2 \) pounds of cashews in the first mix. \\ In the second, there were \( \frac{2x}{5} \) cashews, or 2 pounds of cashews. \\
Adding this together gives us a final total of: \[ 2 + 2 = \boxed{4} \] pounds of cashews.
\smallskip \textit{~Minor edits to LaTeX by WildSealVM/Vincent M. (LaTeX compatible for AoPS)}
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 | |
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