2020 AMC 8 Problem 20

Below is the video solution and professionally curated solution for Problem 20 of the 2020 AMC 8, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2020 AMC 8 solutions, or check the answer key.

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Concepts:caseworkmean

Difficulty rating: 1370

20.

A scientist walking through a forest recorded as integers the heights of 55 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?

Tree 100 metersTree 211 metersTree 300 meters Tree 4 00 metersTree 500 metersAverage00.2 meters \begin{array}{|c|c|} \hline \text{Tree 1} & \underline{\phantom{00}} \text{ meters} \\ \text{Tree 2} & 11 \text{ meters} \\ \text{Tree 3} & \underline{\phantom{00}} \text{ meters} \\ \text{ Tree 4 }& \underline{\phantom{00}} \text{ meters} \\ \text{Tree 5} & \underline{\phantom{00}} \text{ meters} \\ \hline \text{Average} & \underline{\phantom{00}} \text{.2 meters} \\ \hline \end{array}

22.222.2

24.224.2

33.233.2

35.235.2

37.237.2

Video solution:
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Written solution:

Tree 22 is 1111 meters tall. Since all heights are integers and neighboring trees differ by a factor of 2,2, trees 11 and 33 must both be 2222 meters tall.

The first three trees total 22+11+22=5522+11+22=55 meters. Tree 44 is either 1111 or 4444 meters. If tree 44 were 11,11, then tree 55 would be 2222 or nonintegral 5.5,5.5, giving averages ending in .6.6 or invalid.

If tree 4=44,4=44, then tree 55 can be 2222 or 88.88. These give averages 24.224.2 and 37.4,37.4, respectively. The only listed average ending in .2.2 is 24.2.24.2.

Thus, the correct answer is B.

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