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AMC 10

Mean-median – Statistics – AMC 10B, 2019 Problem 13

The simplest example of coverting between septal (base 7)and decimal number system. Learn in this self-learning module for math olympiad

Mean-median of some numbers


Mean, median, and mode are three kinds of “averages”. … The “mean” is the “average” you’re used to, where you add up all the numbers and then divide by the number of numbers. The “median” is the “middle” value in the list of numbers, And the mode is the number repeated most number of times in the given list. Let’s see how to find the mean-median of some numbers.

Try the problem


What is the sum of all real numbers $x$ for which the median of the numbers $4,6,8,17,$ and $x$ is equal to the mean of those five numbers?

$\textbf{(A) } -5 \qquad\textbf{(B) } 0 \qquad\textbf{(C) } 5 \qquad\textbf{(D) } \frac{15}{4} \qquad\textbf{(E) } \frac{35}{4}$

AMC 10B, 2019 Problem 13

Statistics (Mean-median)

6 out of 10

challenges and thrills of pre college mathematics

Knowledge Graph


mean-median - knowledge graph

Use some hints


The mean is $\frac{4+6+8+17+x}{5}=\frac{35+x}{5}$.

Now there are only three possibilities for the median, It can be either 6,8 or x. It is because 4 is the smallest number and 17 cannot fit in the middle for any possible value of x.

Now if we consider 6 to be median then we must have to get 6 as the mean also. And we will verify this condition for each of the 6,8, and x.

See the final step for more hints.

So lets start with 6 and then 8 and x itelf.

$\frac{35+x}{5}=6$ has solution $x=-5$, and the sequence is $-5, 4, 6, 8, 17$, which does have median $6$, so this is a valid solution.

Now let the median be $8$.

$\frac{35+x}{5}=8$ gives $x=5$, so the sequence is $4, 5, 6, 8, 17$, which has median $6$, so this is not valid.

Finally we let the median be $x$.

Finally we let the median be $x$.

$\frac{35+x}{5}=x \implies 35+x=5x \implies x=\frac{35}{4}=8.75$

and the sequence is $4, 6, 8, 8.75, 17$, which has median $8$. This case is therefore again not valid.

Hence the only possible value of $x$ is \((A) -5\).

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