2020 AMC 8

Complete problem set with solutions and individual problem pages

Problem 20 Hard

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?

  • A.

    22.2

  • B.

    24.2

  • C.

    33.2

  • D.

    35.2

  • E.

    37.2

Answer:B

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 or 6. (0.2 \cdot 5 = 1) We already know Tree 2 is 11 meters tall. Both Tree 1 and Tree 3 must 22 meters tall - since neither can be 5.5. Once again, apply our observation for solving for the Tree 4's height. Tree 4 can't be 11 meters for the sum of the five tree heights to still end with 1. Therefore, the Tree 4 is 44 meters tall. Now, Tree 5 can either be 22 or 88. Find the average height for both cases of Tree 5. Doing this, we realize the Tree 5 must be 22 for the average height to end with 0.2 and that the average height is \boxed{\textbf{(B)}\ 24.2}.