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| {{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]] |
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| ==Problem==
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| 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?
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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>
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| ==Solution 1==
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| 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. \\
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| Adding the second mixture of nuts, we call this value \( x \), as in \( x \) pounds. \\
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| Of that, 20\%, or \( \frac{x}{5} \), are peanuts. \\
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| Since the final percentage is 40 percent peanuts, we have:
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| \[
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| \frac{5 + \frac{x}{5}}{10 + x} = \frac{2}{5}.
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| \]
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| Multiplying both sides by \( 5(10 + x) \), we get:
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| \[
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| 25 + x = 20 + 2x.
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| \]
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| This gives us \( x = 5. \) \\
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| But the problem is asking us to solve for cashews. \\
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| The first mixture was \( \frac{1}{5} \) cashews, so there were \( 2 \) pounds of cashews in the first mix. \\
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| In the second, there were \( \frac{2x}{5} \) cashews, or 2 pounds of cashews. \\
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| Adding this together gives us a final total of:
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| \[
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| 2 + 2 = \boxed{4}
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| \]
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| pounds of cashews.
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| \smallskip
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| \textit{~Minor edits to LaTeX by WildSealVM/Vincent M. (LaTeX compatible for AoPS)}
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| ==See Also==
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| {{AMC10 box|year=2025|ab=A|num-b=1|num-a=3}}
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| {{AMC12 box|year=2025|ab=A|num-b=1|num-a=3}}
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| * [[AMC 10]]
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| * [[AMC 10 Problems and Solutions]]
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| * [[Mathematics competitions]]
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| * [[Mathematics competition resources]]
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| {{MAA Notice}}
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