2020 AMC 10B Problems/Problem 3
- The following problem is from both the 2020 AMC 10B #3 and 2020 AMC 12B #3, so both problems redirect to this page.
Contents
Problem
The ratio of to is , the ratio of to is , and the ratio of to is . What is the ratio of to
Solution 1 (One Sentence)
We have from which
~MRENTHUSIASM
Solution 2
We need to somehow link all three of the ratios together. We can start by connecting the last two ratios together by multiplying the last ratio by two.
, and since , we can link them together to get .
Finally, since , we can link this again to get: , so .
~quacker88
Solution 3
We have the equations , , and . Clearing denominators, we have , , and . Since we want , we look to find in terms of since we know the relationship between and . We begin by multiplying both sides of by two, obtaining . We then substitute that into to get . Now, to be able to substitute this into out first equation, we need to have on the RHS. Multiplying both sides by , we have . Substituting this into our first equation, we have , or , so our answer is .
~Binderclips1
Solution 4
WLOG, let and .
Since the ratio of to is , we can substitute in the value of to get .
The ratio of to is , so .
The ratio of to is then so our answer is .
~quacker88
Video Solution (HOW TO THINK CREATIVELY!!!)
https://www.youtube.com/watch?v=QqkNnsNEgiA (for AMC 10)
https://youtu.be/Lz6XmHr8tJE (for AMC 12)
~Education, the Study of Everything
Video Solution
https://youtu.be/Gkm5rU5MlOU (for AMC 10)
https://youtu.be/WfTty8Fe5Fo (for AMC 12)
~savannahsolver
~AlexExplains
See Also
2020 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
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 |
2020 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 2 |
Followed by Problem 4 |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.