Difference between revisions of "2020 AMC 8 Problems/Problem 20"
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==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 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== | ==Solution 2== | ||
− | Let <math>S</math> be the sum of the heights | + | 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> | ||
\begingroup | \begingroup |
Revision as of 10:07, 20 November 2020
Contents
Problem
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
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.
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.