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

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==Solution 2 ==
 
==Solution 2 ==
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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 B(64)  \boxed{\textbf{(B)}\ 64}$
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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 B(64)   
 +
 
 +
[[File:http://latex.artofproblemsolving.com/8/9/5/8950693058536402bb0952f02564a8bf28016c16.png|thumb|Caption]]
  
 
==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}}

Revision as of 23:24, 3 November 2016

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

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 B(64)

File:Http://latex.artofproblemsolving.com/8/9/5/8950693058536402bb0952f02564a8bf28016c16.png
Caption

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. AMC logo.png

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