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

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==Solution 2==
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==Solution 2 -SweetMango77==
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
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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}}

Latest revision as of 16:34, 28 October 2020

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 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 $1991-12=1979$.

Our answer is $\boxed{(\text{A})1979.}$


Solution 2 -SweetMango77

Since she was 12 when she took the seventh AMC 8, she should be $12-(7-1)=12-6=6$ years old when the first AMC 8 occurred. Therefore, she was born or was 'age 0' in $1985-6=\boxed{\left(\text{A}\right)1979}$.

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

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