Difference between revisions of "2022 AMC 10A Problems/Problem 8"
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| + | {{duplicate|[[2022 AMC 10A Problems/Problem 8|2022 AMC 10A #8]] and [[2022 AMC 12A Problems/Problem 6|2022 AMC 12A #6]]}} | ||
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==Problem== | ==Problem== | ||
| − | A data set consists of <math>6</math> not distinct) positive integers: <math>1</math>, <math>7</math>, <math>5</math>, <math>2</math>, <math>5</math>, and <math>X</math>. The | + | A data set consists of <math>6</math> (not distinct) positive integers: <math>1</math>, <math>7</math>, <math>5</math>, <math>2</math>, <math>5</math>, and <math>X</math>. The average (arithmetic mean) of the <math>6</math> numbers equals a value in the data set. What is the sum of all positive values of <math>X</math>? |
| − | average (arithmetic mean) of the <math>6</math> numbers equals a value in the data set. What is | ||
| − | the sum of all positive values of <math>X</math>? | ||
<math>\textbf{(A) } 10 \qquad \textbf{(B) } 26 \qquad \textbf{(C) } 32 \qquad \textbf{(D) } 36 \qquad \textbf{(E) } 40</math> | <math>\textbf{(A) } 10 \qquad \textbf{(B) } 26 \qquad \textbf{(C) } 32 \qquad \textbf{(D) } 36 \qquad \textbf{(E) } 40</math> | ||
| − | ==Solution== | + | ==Solution (Casework)== |
| − | '''Case 1: the mean is <math>5</math>''' | + | First, note that <math>1+7+5+2+5=20</math>. There are <math>3</math> possible cases: |
| + | |||
| + | '''Case 1: the mean is <math>5</math>.''' | ||
<math>X = 5 \cdot 6 - 20 = 10</math>. | <math>X = 5 \cdot 6 - 20 = 10</math>. | ||
| − | '''Case 2: the mean is <math>7</math>''' | + | '''Case 2: the mean is <math>7</math>.''' |
<math>X = 7 \cdot 6 - 20 = 22</math>. | <math>X = 7 \cdot 6 - 20 = 22</math>. | ||
| − | '''Case 3: the mean is <math>X</math>''' | + | '''Case 3: the mean is <math>X</math>.''' |
| − | <math>\frac{20+X}{6} | + | <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 | ||
| + | |||
| + | ==Video Solution 1 (Quick and Simple)== | ||
| + | |||
| + | https://youtu.be/8s6SngtEBY4 | ||
| + | |||
| + | ~Education, the Study of Everything | ||
| + | |||
| + | ==Video Solution 2 (Don't fall into the trap)== | ||
| + | |||
| + | https://youtu.be/7yAh4MtJ8a8?si=r_qxJ_xhfngz4Xu6&t=1143 | ||
| + | |||
| + | ~Math-X | ||
| + | ==Video Solution 3== | ||
| + | https://www.youtube.com/watch?v=GX-jmRUadik | ||
| + | |||
| + | -paixiao | ||
| + | |||
| + | ==Video Solution 4== | ||
| + | https://youtu.be/35cuytqS9iw | ||
== See Also == | == See Also == | ||
| − | {{AMC10 box|year=2022|ab=A|num-b= | + | {{AMC10 box|year=2022|ab=A|num-b=7|num-a=9}} |
| + | {{AMC12 box|year=2022|ab=A|num-b=5|num-a=7}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 18:55, 27 September 2024
- The following problem is from both the 2022 AMC 10A #8 and 2022 AMC 12A #6, so both problems redirect to this page.
Contents
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
Video Solution 2 (Don't fall into the trap)
https://youtu.be/7yAh4MtJ8a8?si=r_qxJ_xhfngz4Xu6&t=1143
~Math-X
Video Solution 3
https://www.youtube.com/watch?v=GX-jmRUadik
-paixiao
Video Solution 4
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. 
