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 $\boxed{\textbf{(D) }11}$

See Also

2013 AMC 10A (ProblemsAnswer KeyResources)
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

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