Difference between revisions of "2014 AMC 8 Problems/Problem 10"
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<math>\textbf{(A) }1979\qquad\textbf{(B) }1980\qquad\textbf{(C) }1981\qquad\textbf{(D) }1982\qquad \textbf{(E) }1983</math> | <math>\textbf{(A) }1979\qquad\textbf{(B) }1980\qquad\textbf{(C) }1981\qquad\textbf{(D) }1982\qquad \textbf{(E) }1983</math> | ||
− | ==Solution== | + | ==Solution 1== |
− | The seventh AMC 8 would have been given in <math> | + | 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>. |
− | Our answer is <math>\boxed{(\text{ | + | Our answer is <math>\boxed{(\text{A})1979}</math> |
+ | corrections made by DrDominic | ||
==Solution 2== | ==Solution 2== | ||
− | Since she was 12 when she took the seventh AMC 8, she should be <math> | + | Since she was 12 when she took the seventh AMC 8, she should be <math>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 | ||
+ | corrections made by DrDominic | ||
==Video Solution (CREATIVE THINKING)== | ==Video Solution (CREATIVE THINKING)== |
Latest revision as of 20:43, 6 January 2024
Contents
[hide]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 1
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 corrections made by DrDominic
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 corrections made by DrDominic
Video Solution (CREATIVE THINKING)
~Education, the Study of Everything
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.