2025 AMC 10A Problems/Problem 2: Difference between revisions
Cheltstudent (talk | contribs) No edit summary |
Stevens0209 (talk | contribs) No edit summary |
||
| Line 1: | Line 1: | ||
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 | 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> | <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 have <cmath>5</cmath> pounds of peanuts, <cmath>2</cmath> pounds of cashews, and <cmath>3</cmath> pounds of almonds in the first nut mix. | |||
Let there be <cmath>x</cmath> pounds of nuts in the second nut mix, thus we have <cmath>0.2x</cmath> worth of peanuts. That means: | |||
<cmath>\(\frac{5 + 0.2x}{10 + x} = 0.4\)</cmath> | |||
<cmath>5+0.2x=0.4(10+x) | |||
5+0.2x=4+0.4x | |||
1=0.2x | |||
x=5</cmath> | |||
That means we have 5 pounds of the second nut mix. We are trying to find the amount of cashews in pounds. | |||
<cmath>10*0.2+5*0.4=2+2=4</cmath> | |||
There are a total of <cmath>\fbox{\textbf{(B)} 4}</cmath> | |||
Revision as of 13:17, 6 November 2025
A box contains 
pounds of a nut mix that is 
percent peanuts, 
percent cashews, and 
percent almonds. A second nut mix containing percent peanuts, 
percent cashews, and percent almonds is added to the box resulting in a new nut mix that is percent peanuts. How many pounds of cashews are now in the box?
Solution 1
We have ![\[5\]](http://latex.artofproblemsolving.com/6/d/b/6dbe026a587f5a1dd1ab0ae93debc1d966959aa9.png)
pounds of peanuts, ![\[2\]](http://latex.artofproblemsolving.com/c/e/d/ced7949d3bc8ef8c7a09c4ec819f30277b50d2c2.png)
pounds of cashews, and ![\[3\]](http://latex.artofproblemsolving.com/c/d/2/cd2004d88d5be0144ba38aa331d32c36042e84ab.png)
pounds of almonds in the first nut mix.
Let there be ![\[x\]](http://latex.artofproblemsolving.com/8/e/7/8e7ba7578864781f5052ab8e190bd855a9a77c82.png)
pounds of nuts in the second nut mix, thus we have ![\[0.2x\]](http://latex.artofproblemsolving.com/9/3/d/93d9d7cfc9c56aef36e24619f00e19ba554e2ab1.png)
worth of peanuts. That means:
\[\(\frac{5 + 0.2x}{10 + x} = 0.4\)\] (Error compiling LaTeX. Unknown error_msg)
![\[5+0.2x=0.4(10+x) 5+0.2x=4+0.4x 1=0.2x x=5\]](http://latex-new.aopstest.com/6/8/9/689e5d01addfa27a151d3308bb8650689d8af03d.png)
That means we have 5 pounds of the second nut mix. We are trying to find the amount of cashews in pounds.
![\[10*0.2+5*0.4=2+2=4\]](http://latex-new.aopstest.com/2/7/8/278d53c9675c7030a7d35791981646e2174689a4.png)
There are a total of ![\[\fbox{\textbf{(B)} 4}\]](http://latex-new.aopstest.com/9/b/1/9b1e4c80a98ecb44a804c10023fb3efa2d201616.png)
