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==
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==Solution 1==
The seventh AMC 8 would have been given in 1992. If Samantha was 12 then, that means she was born 12 years ago, so she was born in 1992-12= <math>\boxed{(\text{B})1980.}</math>
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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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corrections made by DrDominic
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==Solution 2==
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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>
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~SweetMango77
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corrections made by DrDominic
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==Video Solution (CREATIVE THINKING)==
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https://youtu.be/Z_TWPk0NBrU
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~Education, the Study of Everything
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==Video Solution==
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https://youtu.be/oFibh3B60FU ~savannahsolver
  
 
==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 21:43, 6 January 2024

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 1

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 $\boxed{(\text{A})1979}$ corrections made by DrDominic

Solution 2

Since she was 12 when she took the seventh AMC 8, she should be $12-6=6$ years old when the first AMC 8 occurred. Therefore, she was born or was 'age 0' in $1985-6=\boxed{\left(\text{A}\right)1979}$ ~SweetMango77 corrections made by DrDominic

Video Solution (CREATIVE THINKING)

https://youtu.be/Z_TWPk0NBrU

~Education, the Study of Everything


Video Solution

https://youtu.be/oFibh3B60FU ~savannahsolver

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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