2023 AMC 8 Problem 20

Attempt Problem 20 of the 2023 AMC 8 below, then check your answer against the video solution and professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2023 AMC 8 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

20.

Two integers are inserted into the list 3,3, 3,3, 8,8, 11,11, 2828 to double its range. The mode and median remain unchanged. What is the maximum possible sum of the two additional numbers?
5656
5757
5858
6060
6161
Answer: D
Concepts:rangemedian (data)modeoptimization
Small hint:
The original range is 2525, so the new range must be 5050
Big hint:
To keep the median 88, add one number below 88 and one above 88
Video solution:
Solution video thumbnail
Play video

Click to load, then click again to play

Written solution:
The original range is 28−3=25,28-3=25, so the new range must be 50.50. To keep the median 8,8, one added number xx must be less than 88 and the other, y,y, must be greater than 8.8.
If x≥3,x\ge3, the minimum remains 3,3, so the new maximum must be 53.53. The value x=3x=3 would change the mode, so x≤7.x\le7. Hence x+y≤7+53=60.x+y\le7+53=60.
If x<3x<3 and y>28,y>28, then y−x=50,y-x=50, so x+y=2x+50≤54.x+y=2x+50\le54. If y≤28,y\le28, the maximum remains 28,28, forcing x=−22x=-22 and giving an even smaller sum. Therefore no case exceeds 60.60.
The values x=7x=7 and y=53y=53 preserve the mode and median and give range 50,50, so the maximum sum is 60.60.
Thus, D is the correct answer.
LIVE by Po-Shen Loh
Programs

LIVE Classes

Self-Paced Classes

Stars

Scholarship

About Us

Who Is Po-Shen Loh?

Email Contact

Newsletter

Jobs

Become a Star

For a more Thought Full world.

© 2026 Expii, Inc.