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

1. 2020 AMC 8 官方解析

The notebook shows that the second tree is 11 meters tall. The first and third trees must be twice as tall at 22 meters, otherwise if they were half as tall, their heights would not be integers. This means the first 3 trees have a total height of $22 + 11 + 22 = 55$ meters. Because the third tree is 22 meters tall, the only possibilities for the heights of the last 2 trees are

• 11 and 22 meters, which result in an average of $\frac{55 + 11 + 22}{5} = 17.6$ meters,
• 44 and 22 meters, which result in an average of $\frac{55 + 44 + 22}{5} = 24.2$ meters, or
• 44 and 88 meters, which result in an average of $\frac{55 + 44 + 88}{5} = 37.4$ meters.
  The average ends in ".2" so the second option must be the correct one. Thus it must be that the heights of the trees are 22, 11, 22, 44, and 22 meters, and the average height of all thetrees is 24.2 meters.

笔记本记录了第二棵树高11米. 第一棵和第三棵树的高度必须是11米的两倍，否则，如果它们的高度是第二棵树的一半，它们的高度将不是整数. 这意味着前3树的总高度为$22 + 11 + 22 = 55$米. 由于第三棵树的高度为22米，因此最后2棵树的高度只能是

• 11米和22米，平均高度为$\frac{55 + 11 + 22}{5} = 17.6$米，
• 44米和22米，平均高度为$\frac{55 + 44 + 22}{5} = 24.2$米，或
• 44米和88米，平均高度为$\frac{55 + 44 + 88}{5} = 37.4$米.
  平均值以“.2”结尾，故第二个选项是正确的. 因此，五颗树的高度必须为22，11，22，44，22和22米，所有树木的平均高度为$\frac{55 + 44 + 22}{5} = 24.2$米.

The average tree height equals the sum of the tree heights divided by 5. Because the average ends in ".2" which equals $\frac{2}{10} = \frac{1}{5}$, the sum of the tree heights must be one more than a multiple of 5. As shown above, the heights of the first 3 trees are 22, 11, and 22 meters, producing a total height of 55 meters, which is evenly divisible by 5. The sum of the heights of the last 2 trees must be one more than a multiple of 5. Because the third tree is 22 meters tall, the only possibilities for the heights of the last 2 trees are

• 11 and 22 meters, which sum to 33 meters,
• 44 and 22 meters, which sum to 66 meters, or
• 44 and 88 meters, which sum to 132 meters.
  Only the second option is one more than a multiple of 5. Thus it must be that the heights of the trees are 22, 11, 22, 44, and 22 meters, and the average height is $\frac{55 + 44 + 22}{5} = 24.2$ meters.

树木的平均高度等于树木的高度之和除以5. 因为平均数以“.2”结尾，等于$\frac{2}{10} = \frac{1}{5}$，所以树高的总和必须比5的倍数大1. 如上所示，前面3棵树的高度为22，11和22米，总高度为55米，可以被5整除. 后面2棵树的高度之和必须比5的倍数大1. 由于第三棵树的高度为22米，因此后面的2棵树为

• 11和22米，总计33米，
• 44和22米，总计66米，或
• 44和88米，总计132米.