|
|
| (57 intermediate revisions by 20 users not shown) |
| Line 1: |
Line 1: |
| {{duplicate|[[2025 AMC 10A Problems/Problem 2|2025 AMC 10A #2]] and [[2025 AMC 12A Problems/Problem 2|2025 AMC 12A #2]]}}
| | #redirect [[2025 AMC 12A Problems/Problem 2]] |
| | |
| ==Problem==
| |
| | |
| A box contains <imath>10</imath> pounds of a nut mix that is <imath>50</imath> percent peanuts, <imath>20</imath> percent cashews, and <imath>30</imath> percent almonds. A second nut mix containing <imath>20</imath> percent peanuts, <imath>40</imath> percent cashews, and <imath>40</imath> percent almonds is added to the box resulting in a new nut mix that is <imath>40</imath> percent peanuts. How many pounds of cashews are now in the box?
| |
| | |
| <imath>\textbf{(A)}~3.5\qquad\textbf{(B)}~4\qquad\textbf{(C)}~4.5\qquad\textbf{(D)}~5\qquad\textbf{(E)}~6</imath>
| |
| | |
| ==Solution 1==
| |
| \documentclass{article}
| |
| \begin{document}
| |
| | |
| 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 <imath>x</imath>, as in <imath>x</imath> pounds. \\
| |
| Of that, 20\%, or <imath>x/5</imath>, are peanuts. \\
| |
| | |
| Since the final percentage is 40 percent peanuts, we have:
| |
| \[
| |
| \frac{5 + x/5}{10 + x} = \frac{2}{5}.
| |
| \]
| |
| | |
| Multiplying both sides by <imath>5(10 + x)</imath>, we get:
| |
| \[
| |
| 25 + x = 20 + 2x.
| |
| \]
| |
| | |
| This gives us <imath>x = 5</imath>. \\
| |
| | |
| But the problem is asking us to solve for cashews. \\
| |
| | |
| The first mixture was <imath>\tfrac{1}{5}</imath> cashews, so there were <imath>2</imath> pounds of cashews in the first mix. \\
| |
| In the second, there were <imath>\tfrac{2x}{5}</imath> cashews, or <imath>2</imath> pounds of cashews. \\
| |
| | |
| Adding this together gives us a final total of:
| |
| \[
| |
| 2 + 2 = \boxed{4}
| |
| \]
| |
| pounds of cashews.
| |
| | |
| \end{document}
| |
| | |
| ==See Also==
| |
| {{AMC10 box|year=2025|ab=A|num-b=1|num-a=3}}
| |
| {{AMC12 box|year=2025|ab=A|num-b=1|num-a=3}}
| |
| * [[AMC 10]]
| |
| * [[AMC 10 Problems and Solutions]]
| |
| * [[Mathematics competitions]]
| |
| * [[Mathematics competition resources]]
| |
| {{MAA Notice}}
| |
