Difference between revisions of "2013 AMC 10A Problems/Problem 6"

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Revision as of 14:07, 7 January 2018

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 basketball 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 basketball who are under $10$, $13,3$ must have been the pair to go to the movies and $9,7$ must have went to basketball, 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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