Difference between revisions of "2013 AMC 10A Problems/Problem 6"
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| − | + | ==Problem== | |
| + | 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? | ||
| − | 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 11 | + | <math> \textbf{(A)}\ 3 \qquad\textbf{(B)}\ 7 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 11 \qquad\textbf{(E)}\ 13 </math> |
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| + | ==Solution 1== | ||
| + | |||
| + | 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> | ||
| + | |||
| + | ==Video Solution (CREATIVE THINKING)== | ||
| + | https://youtu.be/ekM_0ec2Hdo | ||
| + | |||
| + | ~Education, the Study of Everything | ||
| + | |||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/OTRnrByN_Ng | ||
| + | |||
| + | ~savannahsolver | ||
| + | |||
| + | ==See Also== | ||
| + | |||
| + | {{AMC10 box|year=2013|ab=A|num-b=5|num-a=7}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 12:08, 1 July 2023
Contents
Problem
Joey and his five brothers are ages 
, 
, 
, 
, 
, and 
. One afternoon two of his brothers whose ages sum to 
went to the movies, two brothers younger than 
went to play baseball, and Joey and the -year-old stayed home. How old is Joey?
Solution 1
Because the -year-old stayed home, we know that the -year-old did not go to the movies, as the -year-old did not and 
. Also, the -year-old could not have gone to play baseball, as he is older than . Thus, the -year-old must have stayed home, so Joey is 
Solution 2
There are only 
kids who are under but since the -year old stayed home, the only possible ages who went to play baseball are the brothers who are 
, either 
or 
is but since we need 
kids to go to baseball who are under , 
must have been the pair to go to the movies and 
must have went to baseball, so only the -year old is left, which is answer choice
Video Solution (CREATIVE THINKING)
~Education, the Study of Everything
Video Solution
~savannahsolver
See Also
| 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. 
