Difference between revisions of "2001 AMC 12 Problems/Problem 4"
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<math>\textbf{(A)}\ 5\qquad \textbf{(B)}\ 20\qquad \textbf{(C)}\ 25\qquad \textbf{(D)}\ 30\qquad \textbf{(E)}\ 36</math> | <math>\textbf{(A)}\ 5\qquad \textbf{(B)}\ 20\qquad \textbf{(C)}\ 25\qquad \textbf{(D)}\ 30\qquad \textbf{(E)}\ 36</math> | ||
− | == Solution == | + | == Solution 1== |
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, <math>5</math>. So | and the greatest is <math>m + 15</math>. The middle of the three numbers is the median, <math>5</math>. So | ||
<math>\dfrac{1}{3}[(m-10) + 5 + (m + 15)] = m</math>, which can be solved to get <math>m=10</math>. | <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\cdot 10 = \boxed{\textbf{(D) }30}</math>. | Hence, the sum of the three numbers is <math>3\cdot 10 = \boxed{\textbf{(D) }30}</math>. | ||
+ | |||
+ | ==Solution 2== | ||
== See Also == | == See Also == |
Revision as of 20:46, 10 February 2024
- The following problem is from both the 2001 AMC 12 #4 and 2001 AMC 10 #16, so both problems redirect to this page.
Contents
[hide]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
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.