Difference between revisions of "2016 AMC 10B Problems/Problem 5"
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<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> | ||
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| + | ==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 
cousins is 
, and their median age is 
. What is the sum of the ages of Amanda's youngest and oldest cousins?
Solution
The sum of the ages of the cousins is times the mean, or 
.
There are an even number of cousins, so there is no single median, so must be the mean of the two in the middle.
Therefore the sum of the ages of the two in the middle is 
. Subtracting from produces 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
