A
list of integers has mode 32 and mean 22. The smallest number on the
list is 10. The median $m$ of the list is a member of the list. If the
list member $m$ were replaced by $(m+10)$, the mean and median of the
new list would be 24 and $(m+10)$, respectively. If $m$ were instead
replaced by $(m-8)$, the median of the new list would be $m-4$. What is
$m$?
Solution. Let the numbers on the list be denoted by
a_1=10\le a_2 \le \cdots \le m \le \cdots \le a_n.
Since the mean is 22, we have
22n=10+a_2+\cdots + m + \cdots + a_n.
The mean becomes 24 when $m$ is replaced by $m+10$ so
24n=10+a_2+\cdots + (m +10)+ \cdots + a_n.
Subtracting corresponding terms in the two equations leaves us with $2n=10$, so $n=5$. As a result, the original list of
terms becomes $10\le a_2\le m\le a_4\le a_5$, and
10+a_2 + m + a_4 + a_5 = 5\cdot 22=110.
We are told that the mode is 32, which must be the value of at
least two of the terms of the sequence. If the mode is both $a_2$ and
$m$, then the sequence would have the form 10, 32, 32, $a_4$, $a_5$,
with $32\le a_4\le a_5$. This would imply that
10+a_2+ m+a_4 + a_5\ge 10 + 4\cdot 32 = 138,
which is false since we've exceeded 110.
Suppose now that the mode is both $m$ and $a_4$. Then the sequence
would have the form 10, $a_2$, 32, 32, $a_5$ with $10\le a_2$ and
$32\le a_5$. This would imply that
10+a_2+m+a_4+a_5 \ge 20+3\cdot 32 = 116,
which is also false.
Hence the mode must be $32=a_4=a_5$. Thus the sequence is 10, $a_2$, $m$, 32, 32 and
110=10+a_2+m+a_4+a_5=10+a_2+m+32+32 \Longrightarrow a_2+m=36.
Replacing $m$ by $m-8$ changes the median to $m-4$, which means the ordered list is now
10, m-8,a_2, 32, 32,
and so $a_2=m-4$. So $36=a_2+m=(m-4)+m \Longrightarrow m=20$ (and $a_2=16$).