Difference between revisions of "2001 AMC 12 Problems/Problem 4"
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Let <math>m</math> be the mean of the three numbers. Then the least of the numbers is <math>m-10</math> | Let <math>m</math> be the mean of the three numbers. Then the least of the numbers is <math>m-10</math> | ||
and the greatest is <math>m + 15</math>. The middle of the three numbers is the median, 5. So | and the greatest is <math>m + 15</math>. The middle of the three numbers is the median, 5. So | ||
− | <math>\dfrac{1}{3}[(m-10) + 5 + (m + 15)] = m</math>, which | + | <math>\dfrac{1}{3}[(m-10) + 5 + (m + 15)] = m</math>, which can be solved to get <math>m=10</math>. |
Hence, the sum of the three numbers is <math>3(10) = \boxed{(\text{D})30}</math>. | Hence, the sum of the three numbers is <math>3(10) = \boxed{(\text{D})30}</math>. | ||
Latest revision as of 20:12, 25 January 2021
- 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
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, 5. So , which can be solved to get . Hence, the sum of the three numbers is .
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.