Difference between revisions of "2020 AMC 8 Problems/Problem 25"
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Since each pair of boxes has a sum of <math>3322</math> or <math>2020</math> and a difference of <math>S_2</math>, we see that the answer is <math>\dfrac{3322 - 2020}{2} = \dfrac{1302}{2} = \boxed{(\text{A}) 651}.</math> | Since each pair of boxes has a sum of <math>3322</math> or <math>2020</math> and a difference of <math>S_2</math>, we see that the answer is <math>\dfrac{3322 - 2020}{2} = \dfrac{1302}{2} = \boxed{(\text{A}) 651}.</math> | ||
− | -A_MatheMagician. Note this is just a more quicker way to do it to get \boxed{(\text{A}) 651}. | + | -A_MatheMagician. Note this is just a more quicker way to do it to get <math>\boxed{(\text{A}) 651}.</math> |
https://artofproblemsolving.com/community/my-aops | https://artofproblemsolving.com/community/my-aops | ||
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==See also== | ==See also== | ||
{{AMC8 box|year=2020|num-b=24|after=Last Problem}} | {{AMC8 box|year=2020|num-b=24|after=Last Problem}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 10:11, 20 November 2020
Rectangles and and squares and shown below, combine to form a rectangle that is 3322 units wide and 2020 units high. What is the side length of in units?
Contents
Solution 1
For each square , let the sidelength of this square be denoted by .
As the diagram shows, We subtract the second equation from the first, getting , and thus , so the answer is ~icematrix, edits by starrynight7210
Solution 2
WLOG, assume that and . Let the sum of the lengths of and be and let the length of be . We have the system
which we solve to find that .
-franzliszt
Solution 3
Since each pair of boxes has a sum of or and a difference of , we see that the answer is
-A_MatheMagician. Note this is just a more quicker way to do it to get https://artofproblemsolving.com/community/my-aops
See also
2020 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last Problem | |
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 AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.