|
|
| (50 intermediate revisions by 18 users not shown) |
| Line 1: |
Line 1: |
| {{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]] |
| | |
| ==Problem==
| |
| | |
| 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?
| |
| | |
| <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==
| |
| | |
| 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>
| |
| | |
| ~iiiiiizh
| |
| | |
| ==See Also==
| |
| {{AMC10 box|year=2025|ab=A|num-b=1|num-a=3}}
| |
| {{AMC12 box|year=2025|ab=A|num-b=1|num-a=3}}
| |
| * [[AMC 10]]
| |
| * [[AMC 10 Problems and Solutions]]
| |
| * [[Mathematics competitions]]
| |
| * [[Mathematics competition resources]]
| |
| {{MAA Notice}}
| |
