Difference between revisions of "2014 AMC 8 Problems/Problem 10"
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Our answer is <math>\boxed{(\text{A})1979.}</math> | Our answer is <math>\boxed{(\text{A})1979.}</math> | ||
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| + | ==Solution 2== | ||
| + | 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>. | ||
==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 15:02, 21 October 2020
Contents
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 
.
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. 
