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

(Created page with "==Problem== The mean age of Amanda's <math>4</math> cousins is <math>8</math>, and their median age is <math>5</math>. What is the sum of the ages of Amanda's youngest and ol...")
 
Line 4: Line 4:
  
 
<math>\textbf{(A)}\ 13\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 19\qquad\textbf{(D)}\ 22\qquad\textbf{(E)}\ 25</math>
 
<math>\textbf{(A)}\ 13\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 19\qquad\textbf{(D)}\ 22\qquad\textbf{(E)}\ 25</math>
 +
 +
==Solution==
 +
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.
 +
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>.
 +
 +
==See Also==
 +
{{AMC10 box|year=2016|ab=B|num-b=4|num-a=6}}
 +
{{MAA Notice}}

Revision as of 11:13, 21 February 2016

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 $\textbf{(D)}\ 22$.

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS