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

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Let the sum of the heights of the trees be denoted by <math>S</math>. The average height will then be <math>\frac{S}{5}</math>. Since the average height has decimal part <math>.2</math>, it follows that <math>S</math> must have a remainder of <math>1</math> when divided by <math>5</math>. Thus, <math>S</math> must be <math>1</math> more than a multiple of 5. Since <math>S</math> is an integer, it follows that the heights of Tree 1 and Tree 3 are likely both 22. At this point, our table looks as follows:
 
Let the sum of the heights of the trees be denoted by <math>S</math>. The average height will then be <math>\frac{S}{5}</math>. Since the average height has decimal part <math>.2</math>, it follows that <math>S</math> must have a remainder of <math>1</math> when divided by <math>5</math>. Thus, <math>S</math> must be <math>1</math> more than a multiple of 5. Since <math>S</math> is an integer, it follows that the heights of Tree 1 and Tree 3 are likely both 22. At this point, our table looks as follows:
 
<cmath>\begin{tabular}{|c|c|}
 
<cmath>\begin{tabular}{|c|c|}
−
\hline Tree 1 & \22 meters \\
+
\hline Tree 1 & 22 meters \\
 
Tree 2 & 11 meters \\
 
Tree 2 & 11 meters \\
 
Tree 3 & 22 meters \\
 
Tree 3 & 22 meters \\

Revision as of 07:16, 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 $\frac{121}{5}=24.2\rightarrow\boxed{\textbf{(B)}}$.


Solution 2

For the heights of the trees to be integers, we must have the height of Tree 1, Tree 2, Tree 3 as $22, 11, 22$. Now, there are two possible cases.


Case 1: The length of tree 4 is $11$- So, we have the first 4 tree lengths as $22, 11, 22, 11$. For the length of Tree 5 to be an integer, we must have the length of Tree $5$ as $22$. But, when we take the average of these $5$ integers, it results in $17.6$, which doesn't satisfy our conditions.


Case 2: The length of tree 4 is $44$- So, we have the first $4$ tree lengths as $22, 11, 22, 44$. Now, using quick modular arithmetic, we see that when the length of Tree 5 is $88$, the average of the heights of the 5 trees is $\boxed{24.2}\rightarrow\boxed{\textbf{(B)}}$. This is where our condition is satisfied.

~ATGY

Solution 3

Let the sum of the heights of the trees be denoted by $S$. The average height will then be $\frac{S}{5}$. Since the average height has decimal part $.2$, it follows that $S$ must have a remainder of $1$ when divided by $5$. Thus, $S$ must be $1$ more than a multiple of 5. Since $S$ is an integer, it follows that the heights of Tree 1 and Tree 3 are likely both 22. At this point, our table looks as follows: \[\begin{tabular}{|c|c|} \hline Tree 1 & 22 meters \\ Tree 2 & 11 meters \\ Tree 3 & 22 meters \\ Tree 4 & \rule{0.2cm}{0.15mm} meters \\ Tree 5 & \rule{0.2cm}{0.15mm} meters \\ \hline Average height & \rule{0.2cm}{0.15mm}.2 meters \\ \hline \end{tabular}\]

If we guess that Tree 4 has a height of 11, we would need to make Tree 5 have a height of 22. In this case $S$ would equal $88$ which is not one more than a multiple of $5$. So we instead guess that Tree 4 has a height of $44$. Then we only need to try out heights of $22$ and $88$ for Tree 5. Using a height of $22$ for Tree 5 gives us $S=121$ which is $1$ more than a multiple of $5$. Thus, $S=121$ and the average height of the trees is $\frac{121}{5}=24.2$ meters $\implies\boxed{\textbf{(B) }24.2}$.
~ junaidmansuri

See also

2020 AMC 8 (Problems • Answer Key • Resources)
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

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