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Difference between revisions of "1997 AHSME Problems/Problem 18"
 
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<math>70 + 2m = 110</math>
 
<math>70 + 2m = 110</math>
  
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<math>m = 20</math>, giving answer <math>\boxed{C}</math>.
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<math>m = 20</math>, giving answer <math>\boxed{E}</math>.
  
 
The original list is <math>\{10, 16, 20, 32, 32\}</math>, with mean <math>\frac{10 + 16 + 20 + 32 + 32}{5} = 22</math> and median <math>20</math> and mode <math>32</math>.
 
The original list is <math>\{10, 16, 20, 32, 32\}</math>, with mean <math>\frac{10 + 16 + 20 + 32 + 32}{5} = 22</math> and median <math>20</math> and mode <math>32</math>.
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== See also ==
 
== See also ==
 
{{AHSME box|year=1997|num-b=17|num-a=19}}
 
{{AHSME box|year=1997|num-b=17|num-a=19}}
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{{MAA Notice}}

Latest revision as of 13:13, 5 July 2013

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 (Problems • Answer Key • Resources)
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

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