2015 AMC 8 Problem 13

Attempt Problem 13 of the 2015 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 2015 AMC 8 solutions, or check the answer key.

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

13.

How many subsets of two elements can be removed from the set {1,2,3,4,5,6,7,8,9,10,11}\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11\} so that the mean (average) of the nine remaining numbers is 6?6?
11
22
33
55
66
Answer: D
Concepts:meansystematic listing
Small hint:
Find the sum the remaining nine numbers must have.
Big hint:
So the two removed numbers must have a fixed sum.
Video solution:
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Written solution:
The original set has sum 1+2+⋯+11=66.1+2+\cdots+11=66. After removing two numbers, 99 numbers remain and must have mean 6,6, so their sum must be 9⋅6=54.9\cdot6=54.
Therefore the two removed numbers must have sum 66−54=12.66-54=12. The possible two-element subsets are {1,11},\{1,11\}, {2,10},\{2,10\}, {3,9},\{3,9\}, {4,8},\{4,8\}, {5,7},\{5,7\}, so there are 55 choices.
Thus, D is the correct answer.
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