Difference between revisions of "2020 AMC 8 Problems/Problem 20"
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− | ==Problem== | + | ==Problem 20== |
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> | ||
− | + | <math>\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</math> | |
− | <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> | ||
==Solution 1== | ==Solution 1== | ||
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==Solution 2== | ==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 Tree <math>2</math> is <math>11</math> meters tall. Both Tree <math>1</math> and Tree <math>3</math> must <math>22</math> meters tall - since neither can be <math>5.5</math>. | ||
+ | Once again, apply our observation for solving for the Tree <math>4</math>'s height. Tree <math>4</math> can't be <math>11</math> meters for the sum of the five tree heights to still end with <math>1</math>. Therefore, the Tree <math>4</math> is <math>44</math> meters tall. | ||
+ | Now the Tree <math>5</math> can either be <math>22</math> or <math>88</math>. Find the average height for both cases of Tree <math>5</math>. Doing this, we realize the Tree <math>5</math> 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>. | ||
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+ | ==Solution 3== | ||
As in Solution 1, we shall show that the heights of the trees are <math>22</math>, <math>11</math>, <math>22</math>, <math>44</math>, and <math>22</math> meters. Let <math>S</math> be the sum of the heights, 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: | As in Solution 1, we shall show that the heights of the trees are <math>22</math>, <math>11</math>, <math>22</math>, <math>44</math>, and <math>22</math> meters. Let <math>S</math> be the sum of the heights, 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: | ||
<cmath> | <cmath> | ||
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==Video Solution== | ==Video Solution== | ||
− | https://youtu.be/ | + | https://youtu.be/VnOecUiP-SA |
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==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}} |
Latest revision as of 18:52, 6 January 2021
Problem 20
A scientist walking through a forest recorded as integers the heights of 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?
Solution 1
We will show that , , , , and 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 meters, we can deduce that Trees 1 and 3 both have a height of 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 and , and , or and . Checking each of these, in the first case, the average is meters, which doesn't end in as the problem requires. Therefore, we consider the other cases. With and , the average is meters, which again does not end in , but with and , the average is meters, which does. Consequently, the answer is .
Solution 2
Notice the average height of the trees ends with therefore the sum of all five heights of the trees must end with . ( * = ) We already know Tree is meters tall. Both Tree and Tree must meters tall - since neither can be . Once again, apply our observation for solving for the Tree 's height. Tree can't be meters for the sum of the five tree heights to still end with . Therefore, the Tree is meters tall. Now the Tree can either be or . Find the average height for both cases of Tree . Doing this, we realize the Tree must be for the average height to end with and that the average height is .
Solution 3
As in Solution 1, we shall show that the heights of the trees are , , , , and meters. Let be the sum of the heights, so that the average height will be meters. We note that , so in order for to end in , must be one more than a multiple of . Moreover, as all the heights are integers, the heights of Tree 1 and Tree 3 are both meters. At this point, our table looks as follows:
If Tree 4 now has a height of , then Tree 5 would need to have height , but in that case would equal , which is not more than a multiple of . So we instead take Tree 4 to have height . Then the sum of the heights of the first 4 trees is , so using a height of for Tree 5 gives , which is more than a multiple of (whereas gives , which is not). Thus the average height of the trees is meters.
Video Solution
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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