Difference between revisions of "2014 AMC 8 Problems/Problem 10"

(Solution)
Line 5: Line 5:
  
 
==Solution==
 
==Solution==
 +
The seventh AMC8 would have been given in 1991, when Samantha was 12. Therefore, she was born in <math>1991-12=1979</math>, or <math>\fbox{A}</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 06:40, 27 November 2014

Problem

The first AMC $8$ was given in $1985$ and it has been given annually since that time. Samantha turned $12$ years old the year that she took the seventh AMC $8$. In what year was Samantha born?

$\textbf{(A) }1979\qquad\textbf{(B) }1980\qquad\textbf{(C) }1981\qquad\textbf{(D) }1982\qquad \textbf{(E) }1983$

Solution

The seventh AMC8 would have been given in 1991, when Samantha was 12. Therefore, she was born in $1991-12=1979$, or $\fbox{A}$.

See Also

2014 AMC 8 (ProblemsAnswer KeyResources)
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. AMC logo.png

Invalid username
Login to AoPS