Difference between revisions of "2013 AMC 10A Problems/Problem 6"
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==Solution== | ==Solution== | ||
| − | 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> | + | 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> |
==See Also== | ==See Also== | ||
{{AMC10 box|year=2013|ab=A|num-b=5|num-a=7}} | {{AMC10 box|year=2013|ab=A|num-b=5|num-a=7}} | ||
Revision as of 14:22, 8 February 2013
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?
Solution
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 
. 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 
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 | ||
