2013 AMC 10A Problems/Problem 6

Problem

Joey and his five brothers are ages $3$ , $5$ , $7$ , $9$ , $11$ , and $13$ . One afternoon two of his brothers whose ages sum to $16$ went to the movies, two brothers younger than $10$ went to play baseball, and Joey and the $5$ -year-old stayed home. How old is Joey?

$\textbf{(A)}\ 3 \qquad\textbf{(B)}\ 7 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 11 \qquad\textbf{(E)}\ 13$

Solution 1

Because the $5$ -year-old stayed home, we know that the $11$ -year-old did not go to the movies, as the $5$ -year-old did not and $11+5=16$ . Also, the $11$ -year-old could not have gone to play baseball, as he is older than $10$ . Thus, the $11$ -year-old must have stayed home, so Joey is $\boxed{\textbf{(D) }11}$

Solution 2

There are only $4$ kids who are under $10$ but since the $5$ -year old stayed home, the only possible ages who went to play baseball are the brothers who are $3,7,9$ , either $13+3$ or $7+9$ is $16$ but since we need $2$ kids to go to baseball who are under $10$ , $13,3$ must have been the pair to go to the movies and $9,7$ must have went to baseball, so only the $11$ -year old is left, which is answer choice $\boxed{\textbf{(D) }11}$

Video Solution (CREATIVE THINKING)

https://youtu.be/ekM_0ec2Hdo

~Education, the Study of Everything

Video Solution

https://youtu.be/OTRnrByN_Ng

~savannahsolver

2013 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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.

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