Difference between revisions of "2001 AMC 12 Problems/Problem 4"
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== Solution == | == Solution == | ||
− | Let <math>m</math> be the mean of the three numbers. Then the least of the numbers is <math>m | + | 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 implies that <math>m=10</math>. | <math>\dfrac{1}{3}[(m-10) + 5 + (m + 15)] = m</math>, which implies that <math>m=10</math>. |
Revision as of 11:58, 13 October 2010
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 implies that . Hence, the sum of the three numbers is , and the answer 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 |