Difference between revisions of "2010 AMC 8 Problems/Problem 11"
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==Solution 2 == | ==Solution 2 == | ||
| − | To answer this problem, you have to make it so that we have the same proportion as 3:4, but the difference between them is 16. Since the two numbers are consecutive, if we multiply both of them by 16, we would get a difference of 16 between them. So, it would be 48:64 and since we need to find the height of the taller tree, we get <math>h=64 \Rightarrow \boxed{\textbf{(B)}\ 64}</math> | + | To answer this problem, you have to make it so that we have the same proportion as 3:4, but the difference between them is 16. Since the two numbers in the ratio are consecutive (difference of 1), if we multiply both of them by 16, we would get a difference of 16 between them. So, it would be 48:64 and since we need to find the height of the taller tree, we get <math>h=64 \Rightarrow \boxed{\textbf{(B)}\ 64}</math> |
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==Video by MathTalks== | ==Video by MathTalks== | ||
Revision as of 19:30, 23 October 2024
Problem
The top of one tree is 
feet higher than the top of another tree. The heights of the two trees are in the ratio 
. In feet, how tall is the taller tree?
Solution 1(algebra solution)
Let the height of the taller tree be 
and let the height of the smaller tree be 
. Since the ratio of the smaller tree to the larger tree is 
, we have 
. Solving for gives us 
Solution 2
To answer this problem, you have to make it so that we have the same proportion as 3:4, but the difference between them is 16. Since the two numbers in the ratio are consecutive (difference of 1), if we multiply both of them by 16, we would get a difference of 16 between them. So, it would be 48:64 and since we need to find the height of the taller tree, we get
Video by MathTalks
https://www.youtube.com/watch?v=6hRHZxSieKc
See Also
| 2010 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 10 |
Followed by Problem 12 | |
| 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. 
