2010 AMC 8 Problems/Problem 11

Revision as of 18:23, 8 November 2017 by Destroyer2000 (talk | contribs) (Solution)

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

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 $\frac{3}}$ (Error compiling LaTeX. Unknown error_msg), 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 $h=64 \Rightarrow \boxed{\textbf{(B)}\ 64}$


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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png