Difference between revisions of "1997 AHSME Problems/Problem 18"
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<math> \textbf{(A)}\ 16\qquad\textbf{(B)}\ 17\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 19\qquad\textbf{(E)}\ 20 </math> | <math> \textbf{(A)}\ 16\qquad\textbf{(B)}\ 17\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 19\qquad\textbf{(E)}\ 20 </math> | ||
| + | |||
| + | ==Solution== | ||
| + | |||
| + | Let there be <math>n</math> integers on the list. The list of <math>n</math> integers has mean <math>22</math>, so the sum of the integers is <math>22n</math>. | ||
| + | |||
| + | Replacing <math>m</math> with <math>m+10</math> will increase the sum of the list from <math>22n</math> to <math>22n + 10</math>. | ||
| + | |||
| + | The new mean of the list is <math>24</math>, so the new sum of the list is also <math>24n</math>. | ||
| + | |||
| + | Thus, we get <math>22n + 10 = 24n</math>, leading to <math>n=5</math> numbers on the list. | ||
| + | |||
| + | If there are <math>5</math> numbers on the list with mode <math>32</math> and smallest number <math>10</math>, then the list is <math>\{10, x, m, 32, 32\}</math> | ||
| + | |||
| + | Since replacing <math>m</math> with <math>m-8</math> gives a new median of <math>m-4</math>, and <math>m-4</math> must be on the list of <math>5</math> integers since <math>5</math> is odd, <math>x = m-4</math>, and the list is now <math>\{10, m-4, m, 32, 32\}</math> | ||
| + | |||
| + | The sum of the numbers on this list is <math>22n = 22\cdot 5 = 110</math>, so we get: | ||
| + | |||
| + | <math>10 + m - 4 + m + 32 + 32 = 110</math> | ||
| + | |||
| + | <math>70 + 2m = 110</math> | ||
| + | |||
| + | <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 second list is <math>\{10, 16, 30, 32, 32\}</math>, with mean <math>\frac{10 + 16 + 30 + 32 + 32}{5} = 24</math> and median <math>m + 10 = 30</math>. | ||
| + | |||
| + | The third list is <math>\{10, 16, 12, 32, 32\} \rightarrow \{10, 12, 16, 32,32\}</math> with median <math>m-4 = 16</math>. | ||
== See also == | == See also == | ||
{{AHSME box|year=1997|num-b=17|num-a=19}} | {{AHSME box|year=1997|num-b=17|num-a=19}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 14:13, 5 July 2013
Problem
A list of integers has mode 
and mean 
. The smallest number in the list is 
. The median 
of the list is a member of the list. If the list member were replaced by 
, the mean and median of the new list would be 
and , respectively. If were instead replaced by 
, the median of the new list would be 
. What is ?
Solution
Let there be 
integers on the list. The list of integers has mean , so the sum of the integers is 
.
Replacing with will increase the sum of the list from to 
.
The new mean of the list is , so the new sum of the list is also 
.
Thus, we get 
, leading to 
numbers on the list.
If there are 
numbers on the list with mode and smallest number , then the list is 
Since replacing with gives a new median of , and must be on the list of integers since is odd, 
, and the list is now 
The sum of the numbers on this list is 
, so we get:
, giving answer 
.
The original list is 
, with mean 
and median 
and mode .
The second list is 
, with mean 
and median 
.
The third list is 
with median 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
