Difference between revisions of "2014 AMC 8 Problems/Problem 10"
Hogwartsian (talk | contribs) (→Solution) |
m (→Solution) |
||
| Line 5: | Line 5: | ||
==Solution== | ==Solution== | ||
| − | The seventh AMC 8 would have been given in 1991. If Samantha was 12 then, that means she was born 12 years ago, so she was born in 1991-12= <math>\boxed{(\text{A})1979.}</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{A})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 20:05, 6 October 2020
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 
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. 
