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

Revision as of 00:10, 12 November 2022

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 positive values of $X$ ?

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

Solution

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 $X$

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

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

~MrThinker

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.

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 == See Also ==
 {{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}}

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

Revision as of 00:10, 12 November 2022

Problem

Solution

See Also