2025 AMC 10A Problems/Problem 2: Difference between revisions
| Line 9: | Line 9: | ||
==Solution 1== | ==Solution 1== | ||
We are given <imath>0. | We are given <imath>0.2(10) = 2</imath> lbs of cashews in the first box. Denote the lbs of nuts in the second nut mix as <imath>x</imath>. <imath>5+0.2(x) = 0.4(10+x), 0.2x = 1, x = 5</imath> so we have <imath>5</imath> lbs of the second mix. <imath>0.4(5)+2 = 2+2 = \boxed{\text{(B) }4}.</imath> | ||
~pigwash | ~pigwash | ||
Revision as of 16:50, 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 
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 are given 
lbs of cashews in the first box. Denote the lbs of nuts in the second nut mix as 
. 
so we have 
lbs of the second mix. 
~pigwash
Edit: please dont delete my solution
Solution 2
Let the number of pounds of nuts in the second nut mix be . Therefore, we get the equation 
. Solving it, we get 
. Therefore the amount of cashews in the two bags is 
, so out answer choice is 
.
~iiiiiizh
See Also
| 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 | |
These problems are copyrighted © by the Mathematical Association of America. 
