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

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==Solution 1==
 
==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>\boxed{\textbf{(D) }11}</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>
  
 
==Solution 2==
 
==Solution 2==
  
There are only <math>4</math> kids who are under <math>10</math> but since the 5-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 11-year old is left, which is answer choice <math>\boxed{\textbf{(D) }11}</math>
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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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==Video Solution (CREATIVE THINKING)==
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https://youtu.be/ekM_0ec2Hdo
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~Education, the Study of Everything
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==Video Solution==
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https://youtu.be/OTRnrByN_Ng
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~savannahsolver
  
 
==See Also==
 
==See Also==

Latest revision as of 12:08, 1 July 2023

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

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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