Difference between revisions of "2016 AMC 10B Problems/Problem 5"

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The sum of the ages of the cousins is <math>4</math> times the mean, or <math>32</math>.
 
The sum of the ages of the cousins is <math>4</math> times the mean, or <math>32</math>.
 
There are an even number of cousins, so there is no single median, so <math>5</math> must be the median of the two in the middle.
 
There are an even number of cousins, so there is no single median, so <math>5</math> must be the median of the two in the middle.
Therefore the sum of the ages of the two in the middle is <math>10</math>. Subtracting <math>10</math> from <math>32</math> produces <math>\textbf{(D)}\ 22</math>.
+
Therefore the sum of the ages of the two in the middle is <math>10</math>. Subtracting <math>10</math> from <math>32</math> produces <math>\textbf{(D)}\ \boxed{22}</math>.
  
 
==Video Solution==
 
==Video Solution==

Revision as of 21:36, 25 January 2021

Problem

The mean age of Amanda's $4$ cousins is $8$, and their median age is $5$. What is the sum of the ages of Amanda's youngest and oldest cousins?

$\textbf{(A)}\ 13\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 19\qquad\textbf{(D)}\ 22\qquad\textbf{(E)}\ 25$

Solution

The sum of the ages of the cousins is $4$ times the mean, or $32$. There are an even number of cousins, so there is no single median, so $5$ must be the median of the two in the middle. Therefore the sum of the ages of the two in the middle is $10$. Subtracting $10$ from $32$ produces $\textbf{(D)}\ \boxed{22}$.

Video Solution

https://youtu.be/8TwlpiWwnJY

~savannahsolver

See Also

2016 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
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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