Difference between revisions of "2022 AMC 10A Problems/Problem 8"

Revision as of 13:20, 29 October 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 $6$ (not distinct) positive integers: $1$ , $7$ , $5$ , $2$ , $5$ , and $X$ . The average (arithmetic mean) of the $6$ numbers equals a value in the data set. What is the sum of all possible values of $X$ ?

$\textbf{(A) } 10 \qquad \textbf{(B) } 26 \qquad \textbf{(C) } 32 \qquad \textbf{(D) } 36 \qquad \textbf{(E) } 40$

Solution (Casework)

First, note that $1+7+5+2+5=20$ . There are $3$ possible cases:

Case 1: the mean is $5$ .

$X = 5 \cdot 6 - 20 = 10$ .

Case 2: the mean is $7$ .

$X = 7 \cdot 6 - 20 = 22$ .

Case 3: the mean is $X$ .

$X= \frac{20+X}{6} \Rightarrow X=4$ .

Therefore, the answer is $10+22+4=\boxed{\textbf{(D) }36}$ .

~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

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.

@@ Line 38: / Line 38: @@
 ~Math-X
+==Video Solution 3==
+https://www.youtube.com/watch?v=GX-jmRUadik
+-paixiao
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