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
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
![$\dfrac{1}{3}[(m-10) + 5 + (m + 15)] = m$](http://latex.artofproblemsolving.com/c/a/c/cac4efd76d95dce2367e45a25811bc7142c1a3b8.png)
, 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. 
