Difference between revisions of "2022 AMC 10A Problems/Problem 8"
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<math>X= \frac{20+X}{6} \Rightarrow X=4</math>. | <math>X= \frac{20+X}{6} \Rightarrow X=4</math>. | ||
− | + | Therefore, the answer is <math>10+22+4=\boxed{\textbf{(D) }36}</math>. | |
~MrThinker | ~MrThinker |
Revision as of 16:39, 23 June 2023
- The following problem is from both the 2022 AMC 10A #8 and 2022 AMC 12A #6, so both problems redirect to this page.
Problem
A data set consists of (not distinct) positive integers:
,
,
,
,
, and
. The average (arithmetic mean) of the
numbers equals a value in the data set. What is the sum of all positive values of
?
Solution (Casework)
First, note that . There are
possible cases:
Case 1: the mean is .
.
Case 2: the mean is .
.
Case 3: the mean is .
.
Therefore, the answer is .
~MrThinker
Video Solution 1 (Quick and Simple)
~Education, the Study of Everything
See Also
2022 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
2022 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 5 |
Followed by Problem 7 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.