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

Revision as of 15:20, 6 November 2025

The following problem is from both the 2025 AMC 10A #2 and 2025 AMC 12A #2, so both problems redirect to this page.

Problem

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 are given $0.5(10) = 5$ lbs of peanuts in the first box. Denote the number of nuts in the second box as $x$ . $5+0.2x = 0.4(10+x), 0.2x = 1, x = 5$ so we have $5$ lbs of the second mix. $0.4(5)+2 = 2+2 = \boxed{\text{(B) }4}.$

~pigwash

Solution 2

Let the number of pounds in the second nut mix be $x$ . Therefore $0.5 * 10 + 0.2 * x = 0.4 * (10+x)$ . Solving this, we get $x = 5$ . Therefore the number of pounds of cashews is $0.2 * 10 + 0.4 * 5 = 4$ pounds $=>$ $\boxed{\text{(B) }4}.$

~iiiiiizh

Solution 4

Since the first box had 5 pounds, and 50 percent off it had peanuts, we know there were 5 pounds of peanuts at the beginning. Adding the second mixture of nuts, we call this value x, as in x pounds. Of that 20% or x/5, are peanuts. Since the final percentage in 40 percent peanuts, (5+x/5)/(10+x)=2/5. Multiplying both sides by 5, we get, 25+x/10+x=2. Multiplying both sides by 10+x, we get 25+x=20+2x. This gives us x=5. But the problem is asking us to solve for cashews. The first mixture was 1/5 cashews, there were 2 cashews in the first mix. In the second, there were 2x/5 cashews, or 2 pounds of cashews. Adding this together gives us a final total of $2+2 = \boxed{4}$ cashews.

2025 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

2025 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

@@ Line 19: / Line 19: @@
 ~iiiiiizh
-==Solution 3==
-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.
+==Solution 4==
+Since the first box had 5 pounds, and 50 percent off it had peanuts, we know there were 5 pounds of peanuts at the beginning. Adding the second mixture of nuts, we call this value x, as in x pounds. Of that 20% or x/5, are peanuts. Since the final percentage in 40 percent peanuts, (5+x/5)/(10+x)=2/5. Multiplying both sides by 5, we get, 25+x/10+x=2.
-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
+Multiplying both sides by 10+x, we get 25+x=20+2x. This gives us x=5. But the problem is asking us to solve for cashews. The first mixture was 1/5 cashews, there were 2 cashews in the first mix. In the second, there were 2x/5 cashews, or 2 pounds of cashews. Adding this together gives us a final total of <imath>2+2 = \boxed{4}</imath> cashews.
-<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
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