Art of Problem Solving

2025 AMC 10A Problems/Problem 2: Difference between revisions

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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?
#redirect [[2025 AMC 12A Problems/Problem 2]]
 
<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==
 
We are given <imath>0.5(10) = 5</imath> lbs of peanuts in the first box. Denote the number of nuts in the second box as x. <imath>5+0.2x = 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>
 
~pigwash
 
==Solution 2==
 
Let the number of pounds in the second nut mix be <imath>x</imath>. Therefore <imath>0.5 * 10 + 0.2 * x = 0.4 * (10+x)</imath>. Solving this, we get <imath>x = 5</imath>. Therefore the number of pounds of cashews is <imath>0.2 * 10 + 0.4 * 5 = 4</imath> pounds <imath>=></imath> <imath>\boxed{\text{(B) }4}.</imath>
 
~iiiiiizh

Latest revision as of 02:15, 8 November 2025