Difference between revisions of "2001 AMC 12 Problems/Problem 4"
(New page: the mean of 3 numbers is 10 more than the least of the numbers and 15 less than the greatest. the median of the three numbers is 5. What is their sum?) |
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− | + | == Problem == | |
+ | The mean of three numbers is <math>10</math> more than the least of the numbers and <math>15</math> | ||
+ | less than the greatest. The median of the three numbers is <math>5</math>. What is their | ||
+ | sum? | ||
+ | |||
+ | <math>\text{(A)}\ 5\qquad \text{(B)}\ 20\qquad \text{(C)}\ 25\qquad \text{(D)}\ 30\qquad \text{(E)}\ 36</math> | ||
+ | |||
+ | == Solution == | ||
+ | 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 | ||
+ | <math>\dfrac{1}{3}[(m-10) + 5 + (m + 15)] = m</math>, which implies that <math>m=10</math>. | ||
+ | Hence, the sum of the three numbers is <math>3(10) = 30</math>, and the answer is <math>\text{(D)}</math>. | ||
+ | |||
+ | == See Also == | ||
+ | {{AMC12 box|year=2001|num-b=3|num-a=5}} |
Revision as of 14:13, 16 February 2008
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 $m − 10$ (Error compiling LaTeX. ! Package inputenc Error: Unicode character − (U+2212)) 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 |