Difference between revisions of "2000 AMC 12 Problems/Problem 14"

Revision as of 17:18, 22 April 2023

The following problem is from both the 2000 AMC 12 #14 and 2000 AMC 10 #23, so both problems redirect to this page.

When the mean, median, and mode of the list

$10,2,5,2,4,2,x$

are arranged in increasing order, they form a non-constant arithmetic progression. What is the sum of all possible real values of $x$ ?

$\text {(A)}\ 3 \qquad \text {(B)}\ 6 \qquad \text {(C)}\ 9 \qquad \text {(D)}\ 17 \qquad \text {(E)}\ 20$

The mean is $\frac{10+2+5+2+4+2+x}{7} = \frac{25+x}{7}$ .
Arranged in increasing order, the list is $2,2,2,4,5,10$ , so the median is either $2,4$ or $x$ depending upon the value of $x$ .
The mode is $2$ , since it appears three times.

We apply casework upon the median:

If the median is $2$ ( $x \le 2$ ), then the arithmetic progression must be constant.
If the median is $4$ ( $x \ge 4$ ), because the mode is $2$ , the mean can either be $0,3,6$ to form an arithmetic progression. Solving for $x$ yields $-25,-4,17$ respectively, of which only $17$ works because it is larger than $4$ .
If the median is $x$ ( $2 \le x \le 4$ ), we must have the arithmetic progression $2, x, \frac{25+x}{7}$ . Thus, we find that $2x=2+\frac{25+x}{7}$ so $x=3$ .

The answer is $3 + 17 = 20\ \mathrm{(E)}$ .

~ pi_is_3.14

2000 AMC 12 (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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

2000 AMC 10 (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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

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 == Video Solution ==
 https://youtu.be/OpgSamhXPTg
+https://www.youtube.com/watch?v=fjOc_c7rpy0      ~David
 == See also ==