Difference between revisions of "2013 AMC 10A Problems/Problem 6"
m |
m (→Solution 2) |
||
Line 12: | Line 12: | ||
==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 | + | 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> |
==See Also== | ==See Also== |
Revision as of 22:25, 11 November 2019
Contents
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 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 . 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
Solution 2
There are only kids who are under but since the 5-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 11-year old is left, which is answer choice
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.