Difference between revisions of "1997 AHSME Problems/Problem 18"
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<math>70 + 2m = 110</math> | <math>70 + 2m = 110</math> | ||
| − | <math>m = 20</math>, giving answer <math>\boxed{ | + | <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>. | ||
Revision as of 16:52, 9 December 2011
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 | ||
