2010 AMC 8 Problems/Problem 11

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

Solution 3 (another algebra solution)

$s + 16 = t$

$3t = 4s$

Solving by substitution, $\frac{3}{4}t + 16 = t$

$16 = \frac{1}{4}t$

$t=\boxed{\textbf{(B)}\ 64}$

Video by MathTalks

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

2010 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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All AJHSME/AMC 8 Problems and Solutions

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

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