Art of Problem Solving

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

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==Solution 2==
==Solution 2==

Revision as of 19:58, 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.