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

Revision as of 13:17, 6 November 2025

A box contains $10$ pounds of a nut mix that is $50$ percent peanuts, $20$ percent cashews, and $30$ percent almonds. A second nut mix containing $20$ percent peanuts, $40$ percent cashews, and $40$ percent almonds is added to the box resulting in a new nut mix that is $40$ percent peanuts. How many pounds of cashews are now in the box?

$\textbf{(A)}~3.5\qquad\textbf{(B)}~4\qquad\textbf{(C)}~4.5\qquad\textbf{(D)}~5\qquad\textbf{(E)}~6$

Solution 1

We have $5$ pounds of peanuts, $2$ pounds of cashews, and $3$ pounds of almonds in the first nut mix.

Let there be $x$ pounds of nuts in the second nut mix, thus we have $0.2x$ worth of peanuts. That means:

$$\frac{5 + 0.2x}{10 + x} = 0.4$$ (Error compiling LaTeX. Unknown error_msg)

$5+0.2x=0.4(10+x) 5+0.2x=4+0.4x 1=0.2x x=5$

That means we have 5 pounds of the second nut mix. We are trying to find the amount of cashews in pounds.

$10*0.2+5*0.4=2+2=4$

There are a total of $\fbox{\textbf{(B)} 4}$

@@ Line 5: / Line 5: @@
 ==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.
+We have <imath>5</imath> pounds of peanuts, <imath>2</imath> pounds of cashews, and <imath>3</imath> 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:
+Let there be <cmath>x</cmath> pounds of nuts in the second nut mix, thus we have <imath>0.2x</imath> worth of peanuts. That means:
-<cmath>\(\frac{5 + 0.2x}{10 + x} = 0.4\)</cmath>
+<imath>\(\frac{5 + 0.2x}{10 + x} = 0.4\)</imath>
 <cmath>5+0.2x=0.4(10+x)

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

Revision as of 13:17, 6 November 2025

Solution 1