2001 AMC 12 Problems/Problem 4
- The following problem is from both the 2001 AMC 12 #4 and 2001 AMC 10 #16, so both problems redirect to this page.
Problem
The mean of three numbers is more than the least of the numbers and less than the greatest. The median of the three numbers is . What is their sum?
Solution 1
Let be the mean of the three numbers. Then the least of the numbers is and the greatest is . The middle of the three numbers is the median, . So , which can be solved to get . Hence, the sum of the three numbers is .
Solution 2
Say the three numbers are , and . When we arrange them in ascending order then we can assume is in the middle therefore .
We can also assume that the smallest number is and the largest number of the three is . Therefore,
Taking up the first equation and simplifying we obtain doing so for the equation we obtain the equation
when solve the above obtained equation and we obtain the values of and
Therefore the sum of the three numbers is
Video Solution by Daily Dose of Math
https://youtu.be/t0UOhO9bJTo?si=MwvGzoT8Ya8uyx7n
~Thesmartgreekmathdude
See Also
2001 AMC 12 (Problems • Answer Key • Resources) | |
Preceded by Problem 3 |
Followed by Problem 5 |
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 | |
All AMC 12 Problems and Solutions |
2001 AMC 10 (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
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 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.