Difference between revisions of "2014 AMC 8 Problems/Problem 10"
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==Solution== | ==Solution== | ||
− | The seventh | + | The seventh AMC 8 would have been given in <math>1991</math>. If Samantha was 12 then, that means she was born 12 years ago, so she was born in <math>1991-12=1979</math>. |
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+ | 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>. | ||
+ | ~SweetMango77 | ||
==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}} |
Latest revision as of 12:01, 11 June 2021
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 . ~SweetMango77
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.