Art of Problem Solving

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

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==Solution 1==
==Solution 1==


We are given \(0.2 \cdot 10 = 2\) pounds of cashews in the first box.
We are given \(0.2(10) = 2\) pounds of cashews in the first box.  
Denote the pounds of nuts in the second nut mix as \(x\).
Denote the pounds of nuts in the second nut mix as \(x\).


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\[
\[
0.4(5) + 2 = 2 + 2 = \boxed{\text{(B) } 4}
0.4(5) + 2 = 2 + 2 = \boxed{\text{(B) }4}
\]
\]



Revision as of 19:59, 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

TOOO EZ

Solution 1

We are given \(0.2(10) = 2\) pounds of cashews in the first box. Denote the pounds of nuts in the second nut mix as \(x\).

\[ 5 + 0.2x = 0.4(10 + x) \]

\[ 0.2x = 1 \]

\[ x = 5 \]

Thus, we have 5 pounds of the second mix.

\[ 0.4(5) + 2 = 2 + 2 = \boxed{\text{(B) }4} \]


~pigwash

~yuvaG (Formatting)

Solution 2

Let the number of pounds of nuts in the second nut mix be $x$. Therefore, we get the equation $0.5 \cdot 10 + 0.2 \cdot x = 0.4(x+10)$. Solving it, we get $x=5$. Therefore the amount of cashews in the two bags is $0.2 \cdot 10 + 0.4  \cdot 5 = 4$, so out answer choice is $\boxed{\textbf{(B)} 4}$.

~iiiiiizh

~yuvaG - $\LaTeX$ Formatting ;)

Solution 3

The percent of peanuts in the first mix is 10% away from the total percentage of peanuts, and the percent of peanuts in the second mix is 20% away from the total percentage. This means the first mix has twice as many nuts as the second mix, so the second mix has 5 pounds. $0.20*10+0.40*5= 4$ pounds of cashews. So our answer is, $\boxed{\textbf{(B)}4}$

~LUCKYOKXIAO

Video Solution (Intuitive, Quick Explanation!)

https://youtu.be/Qb-9KDYDDX8

~ Education, the Study of Everything

Video Solution (Fast and Easy)

https://youtu.be/YpJ3QZTmDuw?si=ucvH15JKX2tw4SKZ ~ Pi Academy

Video Solution by Daily Dose of Math

https://youtu.be/LN5ofIcs1kY

~Thesmartgreekmathdude

Video Solution

https://youtu.be/gWSZeCKrOfU

~MK

See Also

2025 AMC 10A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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

These problems are copyrighted © by the Mathematical Association of America.