Difference between revisions of "2020 AMC 8 Problems/Problem 20"
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A scientist walking through a forest recorded as integers the heights of <math>5</math> trees standing in a row. She observed that each tree was either twice as tall or half as tall as the one to its right. Unfortunately some of her data was lost when rain fell on her notebook. Her notes are shown below, with blanks indicating the missing numbers. Based on her observations, the scientist was able to reconstruct the lost data. What was the average height of the trees, in meters? | A scientist walking through a forest recorded as integers the heights of <math>5</math> trees standing in a row. She observed that each tree was either twice as tall or half as tall as the one to its right. Unfortunately some of her data was lost when rain fell on her notebook. Her notes are shown below, with blanks indicating the missing numbers. Based on her observations, the scientist was able to reconstruct the lost data. What was the average height of the trees, in meters? | ||
<cmath>\begin{tabular}{|c|c|} \hline Tree 1 & \rule{0.2cm}{0.15mm} meters \\ Tree 2 & 11 meters \\ Tree 3 & \rule{0.2cm}{0.15mm} meters \\ Tree 4 & \rule{0.2cm}{0.15mm} meters \\ Tree 5 & \rule{0.2cm}{0.15mm} meters \\ \hline Average height & \rule{0.2cm}{0.15mm}.2 meters \\ \hline \end{tabular}</cmath><math>\newline \textbf{(A) }22.2 \qquad \textbf{(B) }24.2 \qquad \textbf{(C) }33.2 \qquad \textbf{(D) }35.2 \qquad \textbf{(E) }37.2</math> | <cmath>\begin{tabular}{|c|c|} \hline Tree 1 & \rule{0.2cm}{0.15mm} meters \\ Tree 2 & 11 meters \\ Tree 3 & \rule{0.2cm}{0.15mm} meters \\ Tree 4 & \rule{0.2cm}{0.15mm} meters \\ Tree 5 & \rule{0.2cm}{0.15mm} meters \\ \hline Average height & \rule{0.2cm}{0.15mm}.2 meters \\ \hline \end{tabular}</cmath><math>\newline \textbf{(A) }22.2 \qquad \textbf{(B) }24.2 \qquad \textbf{(C) }33.2 \qquad \textbf{(D) }35.2 \qquad \textbf{(E) }37.2</math> | ||
| − | ==See also== {{AMC8 box|year=2020|before= | + | ==Solution 1== |
| + | It is not too hard to construct <math>22, 11, 22, 44, 22</math> as the heights of the trees from left to right. The average is <math>\frac{121}{5}=24.2\rightarrow\boxed{\textbf{(B)}}</math>. | ||
| + | |||
| + | ==See also== | ||
| + | {{AMC8 box|year=2020|before=First problem|num-a=2}} | ||
| + | {{MAA Notice}} | ||
Revision as of 00:18, 18 November 2020
A scientist walking through a forest recorded as integers the heights of 
trees standing in a row. She observed that each tree was either twice as tall or half as tall as the one to its right. Unfortunately some of her data was lost when rain fell on her notebook. Her notes are shown below, with blanks indicating the missing numbers. Based on her observations, the scientist was able to reconstruct the lost data. What was the average height of the trees, in meters?
![\[\begin{tabular}{|c|c|} \hline Tree 1 & \rule{0.2cm}{0.15mm} meters \\ Tree 2 & 11 meters \\ Tree 3 & \rule{0.2cm}{0.15mm} meters \\ Tree 4 & \rule{0.2cm}{0.15mm} meters \\ Tree 5 & \rule{0.2cm}{0.15mm} meters \\ \hline Average height & \rule{0.2cm}{0.15mm}.2 meters \\ \hline \end{tabular}\]](http://latex.artofproblemsolving.com/2/b/4/2b4aafba4ccdbff56d5c5253d9219b4261fe5c92.png)
Solution 1
It is not too hard to construct 
as the heights of the trees from left to right. The average is 
.
See also
| 2020 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by First problem |
Followed by Problem 2 | |
| 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. 
