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 1== | ||
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. | \documentclass{article} | ||
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 <imath>2+2 = \boxed{4}</imath> cashews. | \begin{document} | ||
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 <imath>x</imath>, as in <imath>x</imath> pounds. \\ | |||
Of that 20\% or <imath>x/5</imath>, are peanuts. \\ | |||
Since the final percentage in 40 percent peanuts, <imath>(5 + x/5)/(10 + x) = 2/5</imath>. \\ | |||
Multiplying both sides by 5, we get, <imath>25 + x/10 + x = 2</imath>. \\ | |||
Multiplying both sides by <imath>10 + x</imath>, we get <imath>25 + x = 20 + 2x</imath>. \\ | |||
This gives us <imath>x = 5</imath>. \\ | |||
But the problem is asking us to solve for cashews. \\ | |||
The first mixture was <imath>1/5</imath> cashews, there were 2 cashews in the first mix. \\ | |||
In the second, there were <imath>2x/5</imath> cashews, or 2 pounds of cashews. \\ | |||
Adding this together gives us a final total of <imath>2 + 2 = \boxed{4}</imath> cashews. | |||
\end{document} | |||
~Minor edits to latex by WildSealVM/Vincent M. | |||
==See Also== | ==See Also== | ||
Revision as of 16:33, 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 1
\documentclass{article} \begin{document}
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 
, as in pounds. \\
Of that 20\% or 
, are peanuts. \\
Since the final percentage in 40 percent peanuts, 
. \\
Multiplying both sides by 5, we get, 
. \\
Multiplying both sides by 
, we get 
. \\
This gives us 
. \\
But the problem is asking us to solve for cashews. \\
The first mixture was 
cashews, there were 2 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 
cashews.
\end{document} ~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. 
