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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 3==
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| 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.
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| 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.
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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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