2020 AMC 8 Problems/Problem 20

Revision as of 01:40, 14 June 2022 by Spook (talk | contribs)

Problem 20

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?

\[\begingroup \setlength{\tabcolsep}{10pt} \renewcommand{\arraystretch}{1.5} \begin{tabular}{|c|c|} \hline Tree 1 & \rule{0.4cm}{0.15mm} meters \\ Tree 2 & 11 meters \\ Tree 3 & \rule{0.5cm}{0.15mm} meters \\ Tree 4 & \rule{0.5cm}{0.15mm} meters \\ Tree 5 & \rule{0.5cm}{0.15mm} meters \\ \hline Average height & \rule{0.5cm}{0.15mm}\text{ .}2 meters \\ \hline \end{tabular} \endgroup\] $\textbf{(A) }22.2 \qquad \textbf{(B) }24.2 \qquad \textbf{(C) }33.2 \qquad \textbf{(D) }35.2 \qquad \textbf{(E) }37.2$

Solution 1

We will show that $22$, $11$, $22$, $44$, and $22$ meters are the heights of the trees from left to right. We are given that all tree heights are integers, so since Tree 2 has height $11$ meters, we can deduce that Trees 1 and 3 both have a height of $22$ meters. There are now three possible cases for the heights of Trees 4 and 5 (in order for them to be integers), namely heights of $11$ and $22$, $44$ and $88$, or $44$ and $22$. Checking each of these, in the first case, the average is $17.6$ meters, which doesn't end in $.2$ as the problem requires. Therefore, we consider the other cases. With $44$ and $88$, the average is $37.4$ meters, which again does not end in $.2$, but with $44$ and $22$, the average is $24.2$ meters, which does. Consequently, the answer is $\boxed{\textbf{(B) }24.2}$.

Solution 2

Notice the average height of the trees ends with $0.2$ therefore the sum of all five heights of the trees must end with $1$ or $6$. ($0.2$ $\cdot$ $5$ = $1$) We already know Tree $2$ is $11$ meters tall. Both Tree $1$ and Tree $3$ must $22$ meters tall - since neither can be $5.5$. Once again, apply our observation for solving for the Tree $4$'s height. Tree $4$ can't be $11$ meters for the sum of the five tree heights to still end with $1$. Therefore, the Tree $4$ is $44$ meters tall. Now the Tree $5$ can either be $22$ or $88$. Find the average height for both cases of Tree $5$. Doing this, we realize the Tree $5$ must be $22$ for the average height to end with $0.2$ and that the average height is $\boxed{\textbf{(B) }24.2}$.

Solution 3

As in Solution 1, we shall show that the heights of the trees are $22$, $11$, $22$, $44$, and $22$ meters. Let $S$ be the sum of the heights, so that the average height will be $\frac{S}{5}$ meters. We note that $0.2 = \frac{1}{5}$, so in order for $\frac{S}{5}$ to end in $.2$, $S$ must be one more than a multiple of $5$. Moreover, as all the heights are integers, the heights of Tree 1 and Tree 3 are both $22$ meters. At this point, our table looks as follows: \[\begingroup \setlength{\tabcolsep}{10pt} \renewcommand{\arraystretch}{1.5} \begin{tabular}{|c|c|} \hline Tree 1 & 22 meters \\ Tree 2 & 11 meters \\ Tree 3 & 22 meters \\ Tree 4 & \rule{0.5cm}{0.15mm} meters \\ Tree 5 & \rule{0.5cm}{0.15mm} meters \\ \hline Average height & \rule{0.5cm}{0.15mm}\text{ .}2 meters \\ \hline \end{tabular} \endgroup\]

If Tree 4 now has a height of $11$, then Tree 5 would need to have height $22$, but in that case $S$ would equal $88$, which is not $1$ more than a multiple of $5$. So we instead take Tree 4 to have height $44$. Then the sum of the heights of the first 4 trees is $22+11+22+44 = 99$, so using a height of $22$ for Tree 5 gives $S=121$, which is $1$ more than a multiple of $5$ (whereas $88$ gives $S = 187$, which is not). Thus the average height of the trees is $\frac{121}{5} = \boxed{\textbf{(B) }24.2}$ meters.

Solution 4

Since we know that the tree heights have to be integers, then it is immediate that Tree 1 and 3 have a height of $22$. Now using the information given by the last column (that the average of the heights of the trees ends in $.2$), we can tell that the sum of all the heights of the trees ends in either $1$ or $6$, because those are the only numbers from $0$ to $9$ that are congruent to $1$ after taking modulo $5$. The two multiples of eleven (eleven because all of the tree heights have to be a multiple of eleven if they are integers) that come to mind are $66$ and $121$. Since the sum of the heights of Tree 1, 2, and 3 is already $55$, we know that $66$ is impossible to obtain. Then, we can decide with relative confidence that the answer should be $\frac{121}{5} = \boxed{\textbf{(B) }24.2}$.


P.S. : We can check our solution by trying to obtain $121$. Surely enough, we have Tree 1 = $22$, Tree 2 = $11$, Tree 3 = $22$, Tree 4 = $44$, and Tree 5 = $22$. These all add up to get $121$.

Video Solution by WhyMath

https://youtu.be/uCV9-WpaIjw

~savannahsolver

Video Solution

https://youtu.be/VnOecUiP-SA

Video Solution by Interstigation

https://youtu.be/YnwkBZTv5Fw?t=1045

~Interstigation

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

Invalid username
Login to AoPS