Question 587·Medium·Math

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When the numbers $4,\,9,\,11,\,15,\,17$ , and $x$ are arranged in ascending order, the median of the six numbers is $13$ . What is the least possible value of $x$ ?

Your answer

A given median with one unknown value is a question about positions in an ordered list. With an even number of values, average the two middle ones. Use the known values to show how small the unknown can be, then check that boundary with a Desmos list. The trap is treating the median as a value the unknown must equal.

Hints

Hint 1
For six ordered numbers, the median is the average of positions $3$ and $4$ , not one number you pick from the list. What must those two middle numbers add up to?
Hint 2
To find the least possible value, consider what happens if $x$ is below $15$ . The fixed values $4$ , $9$ , and $11$ would all come before $15$ . Where would that leave the fourth position?

Step-by-step

Bound the middle positions
Step 1
Find the required sum of the middle values
With six numbers, the median is the average of the third and fourth numbers after sorting. Call them $m_3$ and $m_4$ . Write their average as the given median:
$\frac{m_3+m_4}{2}=13$
Multiply by $2$ :
$m_3+m_4=26.$
Step 2
Rule out every smaller value
If $x<15$ , then $4$ , $9$ , $11$ , and $x$ all come before $15$ . So the third number is at most $11$ , and the fourth is below $15$ . Their sum is less than $11+15=26$ , but the middle numbers must add up to $26$ . No value of $x$ below $15$ works.
Step 3
Check that the boundary works
Try the smallest value not ruled out, $x=15$ . Type $a=[4,9,11,15,17,15]$ , then $\operatorname{median}(a)$ . Desmos shows $13$ , so $15$ works. Since every smaller value fails, the least possible value of $x$ is $15$ . Grid in 15.

Question 587·Medium·Math

Strategy

Hints

Step-by-step

Lessons that teach this

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Strategy

Hints

Step-by-step

Lessons that teach this