2010 AMC 8 Problems/Problem 11: Difference between revisions
| (16 intermediate revisions by 8 users not shown) | |||
| Line 3: | Line 3: | ||
<math> \textbf{(A)}\ 48 \qquad\textbf{(B)}\ 64 \qquad\textbf{(C)}\ 80 \qquad\textbf{(D)}\ 96\qquad\textbf{(E)}\ 112 </math> | <math> \textbf{(A)}\ 48 \qquad\textbf{(B)}\ 64 \qquad\textbf{(C)}\ 80 \qquad\textbf{(D)}\ 96\qquad\textbf{(E)}\ 112 </math> | ||
== Solution == | == Solution 1(algebra solution)== | ||
Let the height of the taller tree be <math>h</math> and let the height of the smaller tree be <math>h-16</math>. Since the ratio of the smaller tree to the larger tree is <math>\frac{3}{4}</math>, we have <math>\frac{h-16}{h}=\frac{3}{4}</math>. Solving for <math>h</math> gives us <math>h=64 \Rightarrow \boxed{\textbf{(B)}\ 64}</math> | Let the height of the taller tree be <math>h</math> and let the height of the smaller tree be <math>h-16</math>. Since the ratio of the smaller tree to the larger tree is <math>\frac{3}{4}</math>, we have <math>\frac{h-16}{h}=\frac{3}{4}</math>. Solving for <math>h</math> gives us <math>h=64 \Rightarrow \boxed{\textbf{(B)}\ 64}</math> | ||
==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 | 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> | ||
== Solution 3 (another algebra solution)== | |||
<math>s + 16 = t</math> | |||
<math>3t = 4s</math> | |||
Solving by substitution, | |||
<math>\frac{3}{4}t + 16 = t</math> | |||
<math>16 = \frac{1}{4}t</math> | |||
<math>t=\boxed{\textbf{(B)}\ 64}</math> | |||
==Video by MathTalks== | |||
https://www.youtube.com/watch?v=6hRHZxSieKc | |||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2010|num-b=10|num-a=12}} | {{AMC8 box|year=2010|num-b=10|num-a=12}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 19:47, 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
Solution 3 (another algebra solution)
Solving by substitution,
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing
