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

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==Problem==
 
==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?
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Joey and his five brothers are ages <math>3</math>, <math>5</math>, <math>7</math>, <math>9</math>, <math>11</math>, and <math>13</math>.  One afternoon two of his brothers whose ages sum to <math>16</math> went to the movies, two brothers younger than <math>10</math> went to play baseball, and Joey and the <math>5</math>-year-old stayed home.  How old is Joey?
  
  
 
<math> \textbf{(A)}\ 3 \qquad\textbf{(B)}\ 7  \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 11 \qquad\textbf{(E)}\ 13 </math>
 
<math> \textbf{(A)}\ 3 \qquad\textbf{(B)}\ 7  \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 11 \qquad\textbf{(E)}\ 13 </math>
  
==Solution==
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==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 <math>11+5=16</math>.  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 <math>11</math> years old, <math>\textbf{(D)}</math>
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Because the <math>5</math>-year-old stayed home, we know that the <math>11</math>-year-old did not go to the movies, as the <math>5</math>-year-old did not and <math>11+5=16</math>.  Also, the <math>11</math>-year-old could not have gone to play baseball, as he is older than <math>10</math>.  Thus, the <math>11</math>-year-old must have stayed home, so Joey is <math>\boxed{\textbf{(D) }11}</math>
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==Solution 2==
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There are only <math>4</math> kids who are under <math>10</math> but since the <math>5</math>-year old stayed home, the only possible ages who went to play baseball are the brothers who are <math>3,7,9</math>, either <math>13+3</math> or <math>7+9</math> is <math>16</math> but since we need <math>2</math> kids to go to baseball who are under <math>10</math>, <math>13,3</math> must have been the pair to go to the movies and <math>9,7</math> must have went to baseball, so only the <math>11</math>-year old is left, which is answer choice <math>\boxed{\textbf{(D) }11}</math>
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==See Also==
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{{AMC10 box|year=2013|ab=A|num-b=5|num-a=7}}
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{{MAA Notice}}

Revision as of 19:18, 28 December 2019

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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