Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Video Solution)
(→‎Solution 1)
Line 21: Line 21:
 
==Solution 1==
 
==Solution 1==
 
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. We are given 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>.
 
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. We are given 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==
 +
Notice the average height of the trees ends with <math>0.2</math> therefore the sum of all five heights of the trees must end with <math>1</math>. (<math>0.2</math> * <math>5</math> = <math>1</math>) We already know one of the heights is <math>11</math> meters. Both trees right of the second tree must be <math>22</math> meters - since neither can be <math>5.5</math>. Once again, apply our observation for solving for the fourth tree's height. The fourth tree can't be <math>11</math> meters for the sum of the five tree heights to still end with <math>1</math>. Therefore, the fourth tree is <math>44</math> meters tall. Now try finding the average height for both cases of the fifth tree being either <math>22</math> or <math>88</math>. Doing this, we realize the fifth tree must be <math>22</math> for the average height to end with <math>0.2</math> and that the average height is <math>\boxed{\textbf{(B) }24.2}</math>.
  
 
==Solution 2==
 
==Solution 2==

Revision as of 15:22, 6 December 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. 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 $1$. ($0.2$ * $5$ = $1$) We already know one of the heights is $11$ meters. Both trees right of the second tree must be $22$ meters - since neither can be $5.5$. Once again, apply our observation for solving for the fourth tree's height. The fourth tree can't be $11$ meters for the sum of the five tree heights to still end with $1$. Therefore, the fourth tree is $44$ meters tall. Now try finding the average height for both cases of the fifth tree being either $22$ or $88$. Doing this, we realize the fifth tree must be $22$ for the average height to end with $0.2$ and that the average height is $\boxed{\textbf{(B) }24.2}$.

Solution 2

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.

Video Solution

https://youtu.be/VnOecUiP-SA

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_8_Problems/Problem_20&oldid=139146"