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

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==Problem==
 
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?
  
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\end{tabular}
 
\end{tabular}
 
\endgroup</cmath>
 
\endgroup</cmath>
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==Solution 1==
 
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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>. ~icematrix Note: this is the result of realizing that the 1st one must be greater then the 2nd, <math>\frac{121}{5}=24.2 or {\textbf{(B)}}</math>. ~oceanxia
 
  
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<math>\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>
  
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==Solution 1==
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We will show that <math>22</math>, <math>11</math>, <math>22</math>, <math>44</math>, and <math>22</math> meters are the heights of the trees from left to right. The problem states that all tree heights are integers, so since Tree 2 has height <math>11</math> meters, we can deduce that Trees 1 and 3 both have a height of <math>22</math> 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 <math>11</math> and <math>22</math>, <math>44</math> and <math>88</math>, or <math>44</math> and <math>22</math>. Checking each of these, in the first case, the average is <math>17.6</math> meters, which doesn't end in <math>.2</math> as the problem requires. Therefore, we consider the other cases. With <math>44</math> and <math>88</math>, the average is <math>37.4</math> meters, which again does not end in <math>.2</math>, but with <math>44</math> and <math>22</math>, the average is <math>24.2</math> meters, which does. Consequently, the answer is <math>\boxed{\textbf{(B) }24.2}</math>.
  
 
==Solution 2==
 
==Solution 2==
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For the heights of the trees to be integers, we must have the height of Tree 1, Tree 2, Tree 3 as <math>22, 11, 22</math>. Now, there are two possible cases.
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Let <math>S</math> be the sum of the heights of the trees, so that the average height will be <math>\frac{S}{5}</math> meters. We note that <math>0.2 = \frac{1}{5}</math>, so in order for <math>\frac{S}{5}</math> to end in <math>.2</math>, <math>S</math> must be one more than a multiple of <math>5</math>. Moreover, as all the heights are integers, the heights of Tree 1 and Tree 3 are both <math>22</math> meters. At this point, our table looks as follows:
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<cmath>
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\begingroup
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\setlength{\tabcolsep}{10pt}
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Case 1: The length of tree 4 is <math>11</math>-
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\renewcommand{\arraystretch}{1.5}
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So, we have the first 4 tree lengths as <math>22, 11, 22, 11</math>. For the length of Tree 5 to be an integer, we must have the length of Tree <math>5</math> as <math>22</math>. But, when we take the average of these <math>5</math> integers, it results in <math>17.6</math>, which doesn't satisfy our conditions.
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\begin{tabular}{|c|c|}
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Case 2: The length of tree 4 is <math>44</math>-
 
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So, we have the first <math>4</math> tree lengths as <math>22, 11, 22, 44</math>. Now, using quick modular arithmetic, we see that when the length of Tree 5 is <math>88</math>, the average of the heights of the 5 trees is <math>\boxed{24.2}\rightarrow\boxed{\textbf{(B)}}</math>. This is where our condition is satisfied.
 
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~ATGY
 
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==Solution 3==
 
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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:
 
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<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 \\
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Tree 4 & \rule{0.2cm}{0.15mm} meters \\
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Tree 4 & \rule{0.5cm}{0.15mm} meters \\
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Tree 5 & \rule{0.2cm}{0.15mm} meters \\ \hline
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Tree 5 & \rule{0.5cm}{0.15mm} meters \\ \hline
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Average height & \rule{0.2cm}{0.15mm}.2 meters \\
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Average height & \rule{0.5cm}{0.15mm}\text{ .}2 meters \\
 
\hline
 
\hline
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\end{tabular}</cmath>
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\end{tabular}
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\endgroup</cmath>
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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 <math>S</math> would equal <math>88</math> which is not one more than a multiple of <math>5</math>. So we instead guess that Tree 4 has a height of <math>44</math>. Then we only need to try out heights of <math>22</math> and <math>88</math> for Tree 5. Using a height of <math>22</math> for Tree 5 gives us <math>S=121</math> which is <math>1</math> more than a multiple of <math>5</math>. Thus, the average height of the trees is <math>\frac{121}{5}=24.2</math> meters <math>\implies\boxed{\textbf{(B) }24.2}</math>.<br>
 
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~[http://artofproblemsolving.com/community/user/jmansuri junaidmansuri]
 
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==Solution 4==
 
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The problem states that all tree heights are integers. Therefore, we can deduce that the first and third trees have a height of <math>22</math> meters. Trees four and five must have heights of either <math>11,22</math> or <math>44,88</math> or <math>44,22</math>. Checking which ones match the answer choices, we find that the trees four and five have heights of <math>44</math> and <math>22</math> meters, respectively. Thus, the answer is <math>\frac{22+11+22+44+22}5=\textbf{(B) }24.2</math>.
 
  
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-franzliszt
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If Tree 4 now has a height of <math>11</math>, then Tree 5 would need to have height <math>22</math>, but in that case <math>S</math> would equal <math>88</math>, which is not <math>1</math> more than a multiple of <math>5</math>. So we instead take Tree 4 to have height <math>44</math>. Then the sum of the heights of the first 4 trees is <math>22+11+22+44 = 99</math>, so using a height of <math>22</math> for Tree 5 gives <math>S=121</math>, which is <math>1</math> more than a multiple of <math>5</math> (whereas <math>88</math> gives <math>S = 187</math>, which is not). Thus the average height of the trees is <math>\frac{121}{5} = \boxed{\textbf{(B) }24.2}</math> meters.
  
 
==See also==  
 
==See also==  
 
{{AMC8 box|year=2020|num-b=19|num-a=21}}
 
{{AMC8 box|year=2020|num-b=19|num-a=21}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 09:15, 20 November 2020

Problem

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. The problem states 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

Let $S$ be the sum of the heights of the trees, 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.

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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