Difference between revisions of "1995 AHSME Problems/Problem 25"
(solution) |
(No difference)
|
Revision as of 20:29, 8 January 2008
Problem
A list of five positive integers has mean 
and range 
. The mode and median are both 
. How many different values are possible for the second largest element of the list?
Solution
Let 
be the smallest element, so 
is the largest element. Since the mode is , at least two of the five numbers must be . The last number we denote as 
.
Then their average is 
. Clearly 
. Also we have 
. Thus there are a maximum of 
values of which corresponds to values of ; listing shows that all such values work.
See also
| 1995 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 24 |
Followed by Problem 26 | |
| 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 | ||
