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

Revision as of 20:26, 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

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 $1992-12=1980$ .

Our answer is $\boxed{(\text{B})1980}$

Solution 2

Since she was 12 when she took the seventh AMC 8, she should be $12-7=5$ years old when the first AMC 8 occurred. Therefore, she was born or was 'age 0' in $1985-5=\boxed{\left(\text{B}\right)1980}$ ~SweetMango77

Video Solution (CREATIVE THINKING)

https://youtu.be/Z_TWPk0NBrU

~Education, the Study of Everything

Video Solution

https://youtu.be/oFibh3B60FU ~savannahsolver

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.

@@ Line 10: / Line 10: @@
 ==Solution 2==
-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>.
+Since she was 12 when she took the seventh AMC 8, she should be <math>12-7=5</math> years old when the first AMC 8 occurred. Therefore, she was born or was 'age 0' in <math>1985-5=\boxed{\left(\text{B}\right)1980}</math>
 ~SweetMango77

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