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

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== Solution ==
 
== Solution ==
 
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, 5. 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(10) = \boxed{(\text{D})30}</math>.
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Hence, the sum of the three numbers is <math>3\cdot 10 = \boxed{\textbf{(D) }30}</math>.
  
 
== See Also ==
 
== See Also ==

Revision as of 11:05, 8 November 2021

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?

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

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 can be solved to get $m=10$. Hence, the sum of the three numbers is $3\cdot 10 = \boxed{\textbf{(D) }30}$.

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

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