Difference between revisions of "2014 AMC 8 Problems/Problem 10"
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Since she was 12 when she took the seventh AMC 8, she should be <math>12-(7-1)=12-6=6</math> years old when the first AMC 8 occurred. Therefore, she was born or was 'age 0' in <math>1985-6=\boxed{\left(\text{A}\right)1979}</math>. | Since she was 12 when she took the seventh AMC 8, she should be <math>12-(7-1)=12-6=6</math> years old when the first AMC 8 occurred. Therefore, she was born or was 'age 0' in <math>1985-6=\boxed{\left(\text{A}\right)1979}</math>. | ||
~SweetMango77 | ~SweetMango77 | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/oFibh3B60FU ~savannahsolver | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2014|num-b=9|num-a=11}} | {{AMC8 box|year=2014|num-b=9|num-a=11}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 08:15, 27 April 2022
Problem
The first AMC 
was given in 
and it has been given annually since that time. Samantha turned 
years old the year that she took the seventh AMC . In what year was Samantha born?
Solution
The seventh AMC 8 would have been given in 
. If Samantha was 12 then, that means she was born 12 years ago, so she was born in 
.
Our answer is 
Solution 2
Since she was 12 when she took the seventh AMC 8, she should be 
years old when the first AMC 8 occurred. Therefore, she was born or was 'age 0' in 
.
~SweetMango77
Video Solution
https://youtu.be/oFibh3B60FU ~savannahsolver
See Also
| 2014 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 9 |
Followed by Problem 11 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
