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