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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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| We are given <imath>0.2(10) = 2</imath> lbs of cashews in the first box. Denote the lbs of nuts in the second nut mix as <imath>x</imath>. <imath>5+0.2(x) = 0.4(10+x), 0.2x = 1, x = 5</imath> so we have <imath>5</imath> lbs of the second mix. <imath>0.4(5)+2 = 2+2 = \boxed{\text{(B) }4}.</imath>
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| ~pigwash
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| Edit: please dont delete my solution
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| ==Solution 2==
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| Let the number of pounds of nuts in the second nut mix be <imath>x</imath>. Therefore, we get the equation <imath>0.5 * 10 + 0.2 * x = 0.4(x+10)</imath>. Solving it, we get <imath>x=5</imath>. Therefore the amount of cashews in the two bags is <imath>0.2 * 10 + 0.4 * 5 = 4</imath>, so out answer choice is <imath>\boxed{\textbf{(B)} 4}</imath>.
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| ~iiiiiizh
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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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