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

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{{duplicate|[[2001 AMC 12 Problems|2001 AMC 12 #4]] and [[2001 AMC 10 Problems|2002 AMC 10A #16]]}}
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{{duplicate|[[2001 AMC 12 Problems|2001 AMC 12 #4]] and [[2001 AMC 10 Problems|2001 AMC 10 #16]]}}
  
 
== Problem ==
 
== Problem ==

Revision as of 17:38, 16 March 2011

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 $3(10) = 30$, and the answer is $\text{(D)}$.

See Also

2001 AMC 12 (ProblemsAnswer KeyResources)
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
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