Difference between revisions of "2001 AMC 12 Problems/Problem 4"

Revision as of 20:51, 16 November 2013

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 $10$ 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?

$\text{(A)}\ 5\qquad \text{(B)}\ 20\qquad \text{(C)}\ 25\qquad \text{(D)}\ 30\qquad \text{(E)}\ 36$

Solution

Let $m$ be the mean of the three numbers. Then the least of the numbers is $m-10$ and the greatest is $m + 15$ . The middle of the three numbers is the median, 5. So $\dfrac{1}{3}[(m-10) + 5 + (m + 15)] = m$ , which implies that $m=10$ . Hence, the sum of the three numbers is $3(10) = \boxed{(\text{D})30}$ .

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.

@@ Line 12: / Line 12: @@
 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>.
+Hence, the sum of the three numbers is <math>3(10) = \boxed{(\text{D})30}</math>.
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Difference between revisions of "2001 AMC 12 Problems/Problem 4"

Revision as of 20:51, 16 November 2013

Problem

Solution

See Also