2025 AMC 12A Problems/Problem 2: Difference between revisions
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==Problem== | |||
== Solution 1== | |||
The initial box has 10 pounds. With <imath>50</imath> percent of it being peanuts, there are <imath>0.5\cdot10 = 5</imath> pounds of peanuts. | |||
We then add <imath>x</imath> pounds of a second mix, which is <imath>20</imath> percent peanuts, causing the peanuts to now be <imath>40</imath> percent of the total. We write the equation | |||
<cmath>\frac{5+0.2x}{10+x} = 0.4</cmath> | |||
<cmath>5+0.2x = 4+x</cmath> | |||
<cmath>x=5.</cmath> | |||
This means the second mix was a total of <imath>5</imath> pounds. Because <imath>40</imath> percent of that is cashews, there are <imath>0.4\cdot 5 = 2</imath> cashews in the second mix. The original mixture was <imath>20</imath> percent cashews, so there were <imath>0.2\cdot 10 = 2</imath> cashews originally. So we now have <imath>2+2 = \boxed{4}</imath> cashews. | |||
~lprado | |||
Revision as of 14:47, 6 November 2025
Problem
Solution 1
The initial box has 10 pounds. With 
percent of it being peanuts, there are 
pounds of peanuts.
We then add 
pounds of a second mix, which is 
percent peanuts, causing the peanuts to now be 
percent of the total. We write the equation
![\[\frac{5+0.2x}{10+x} = 0.4\]](http://latex-new.aopstest.com/4/b/d/4bd9e05b5fd27ba8c2bd95de12ad1fdeafabc84d.png)
![\[5+0.2x = 4+x\]](http://latex-new.aopstest.com/a/3/3/a33dcdecdd4c55b2433fa08b02ca78c895842f56.png)
![\[x=5.\]](http://latex-new.aopstest.com/e/e/3/ee33b1da647959b5f97c125074585e18f55aac73.png)
This means the second mix was a total of 
pounds. Because percent of that is cashews, there are 
cashews in the second mix. The original mixture was percent cashews, so there were 
cashews originally. So we now have 
cashews.
~lprado
