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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 mean 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 mean of the two in the middle.
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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>.
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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>.
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==Video Solution (CREATIVE THINKING)==
 +
https://youtu.be/PHsQWnCkg0Q
 +
 
 +
~Education, the Study of Everything
 +
 
 +
 
 +
 
 +
==Video Solution==
 +
https://youtu.be/8TwlpiWwnJY
 +
 
 +
~savannahsolver
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2016|ab=B|num-b=4|num-a=6}}
 
{{AMC10 box|year=2016|ab=B|num-b=4|num-a=6}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 11:48, 2 July 2023

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 mean 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 (CREATIVE THINKING)

https://youtu.be/PHsQWnCkg0Q

~Education, the Study of Everything


Video Solution

https://youtu.be/8TwlpiWwnJY

~savannahsolver

See Also

2016 AMC 10B (Problems • Answer Key • Resources)
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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