Difference between revisions of "2020 AMC 8 Problems/Problem 20"

Line 7: Line 7:
  
 
==See also==  
 
==See also==  
{{AMC8 box|year=2020|before=First problem|num-a=2}}
+
{{AMC8 box|year=2020|num-b=19|num-a=21}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 00:19, 18 November 2020

A scientist walking through a forest recorded as integers the heights of $5$ 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?

\[\begin{tabular}{|c|c|} \hline Tree 1 & \rule{0.2cm}{0.15mm} meters \\ Tree 2 & 11 meters \\ Tree 3 & \rule{0.2cm}{0.15mm} meters \\ Tree 4 & \rule{0.2cm}{0.15mm} meters \\ Tree 5 & \rule{0.2cm}{0.15mm} meters \\ \hline Average height & \rule{0.2cm}{0.15mm}.2 meters \\ \hline \end{tabular}\]$\newline \textbf{(A) }22.2 \qquad \textbf{(B) }24.2 \qquad \textbf{(C) }33.2 \qquad \textbf{(D) }35.2 \qquad \textbf{(E) }37.2$

Solution 1

It is not too hard to construct $22, 11, 22, 44, 22$ as the heights of the trees from left to right. The average is $\frac{121}{5}=24.2\rightarrow\boxed{\textbf{(B)}}$.

See also

2020 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
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