1997 AHSME Problems/Problem 18

Revision as of 13:13, 5 July 2013 by Nathan wailes (talk | contribs)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Problem

A list of integers has mode $32$ and mean $22$. The smallest number in 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 were $m$ instead replaced by $m-8$, the median of the new list would be $m-4$. What is $m$?

$\textbf{(A)}\ 16\qquad\textbf{(B)}\ 17\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 19\qquad\textbf{(E)}\ 20$

Solution

Let there be $n$ integers on the list. The list of $n$ integers has mean $22$, so the sum of the integers is $22n$.

Replacing $m$ with $m+10$ will increase the sum of the list from $22n$ to $22n + 10$.

The new mean of the list is $24$, so the new sum of the list is also $24n$.

Thus, we get $22n + 10 = 24n$, leading to $n=5$ numbers on the list.

If there are $5$ numbers on the list with mode $32$ and smallest number $10$, then the list is $\{10, x, m, 32, 32\}$

Since replacing $m$ with $m-8$ gives a new median of $m-4$, and $m-4$ must be on the list of $5$ integers since $5$ is odd, $x = m-4$, and the list is now $\{10, m-4, m, 32, 32\}$

The sum of the numbers on this list is $22n = 22\cdot 5 = 110$, so we get:

$10 + m - 4 + m + 32 + 32 = 110$

$70 + 2m = 110$

$m = 20$, giving answer $\boxed{E}$.

The original list is $\{10, 16, 20, 32, 32\}$, with mean $\frac{10 + 16 + 20 + 32 + 32}{5} = 22$ and median $20$ and mode $32$.

The second list is $\{10, 16, 30, 32, 32\}$, with mean $\frac{10 + 16 + 30 + 32 + 32}{5} = 24$ and median $m + 10 = 30$.

The third list is $\{10, 16, 12, 32, 32\} \rightarrow \{10, 12, 16, 32,32\}$ with median $m-4 = 16$.

See also

1997 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 17
Followed by
Problem 19
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 26 27 28 29 30
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS