Difference between revisions of "2010 AMC 8 Problems/Problem 11"

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<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 ==
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== 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 http://latex.artofproblemsolving.com/8/9/5/8950693058536402bb0952f02564a8bf28016c16.png
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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>
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==Video by MathTalks==
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https://www.youtube.com/watch?v=6hRHZxSieKc
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==See Also==
 
==See Also==
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{{AMC8 box|year=2010|num-b=10|num-a=12}}
 
{{AMC8 box|year=2010|num-b=10|num-a=12}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 19:52, 22 January 2023

Problem

The top of one tree is $16$ feet higher than the top of another tree. The heights of the two trees are in the ratio $3:4$. In feet, how tall is the taller tree?

$\textbf{(A)}\ 48 \qquad\textbf{(B)}\ 64 \qquad\textbf{(C)}\ 80 \qquad\textbf{(D)}\ 96\qquad\textbf{(E)}\ 112$

Solution 1(algebra solution)

Let the height of the taller tree be $h$ and let the height of the smaller tree be $h-16$. Since the ratio of the smaller tree to the larger tree is $\frac{3}{4}$, we have $\frac{h-16}{h}=\frac{3}{4}$. Solving for $h$ gives us $h=64 \Rightarrow \boxed{\textbf{(B)}\ 64}$

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 $h=64 \Rightarrow \boxed{\textbf{(B)}\ 64}$


Video by MathTalks

https://www.youtube.com/watch?v=6hRHZxSieKc




See Also

2010 AMC 8 (ProblemsAnswer KeyResources)
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

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