Difference between revisions of "2000 AMC 8 Problems/Problem 1"

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</math>
 
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== Solution ==
 
== Solution ==
If Brianna is half as old as Aunt Anna, then Briana is <math>42/2</math> years old, or <math>21</math> years old.  
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If Brianna is half as old as Aunt Anna, then Briana is <math>\frac{42}{2}</math> years old, or <math>21</math> years old.  
  
 
If Caitlin is <math>5</math> years younger than Briana, she is <math>21-5</math> years old, or <math>16</math>.
 
If Caitlin is <math>5</math> years younger than Briana, she is <math>21-5</math> years old, or <math>16</math>.

Revision as of 17:15, 30 July 2011

Problem

Aunt Anna is $42$ years old. Caitlin is $5$ years younger than Brianna, and Brianna is half as old as Aunt Anna. How old is Caitlin?

$\mathrm{(A)}\ 15 \qquad \mathrm{(B)}\ 16 \qquad \mathrm{(C)}\ 17 \qquad \mathrm{(D)}\ 21 \qquad \mathrm{(E)}\ 37$

Solution

If Brianna is half as old as Aunt Anna, then Briana is $\frac{42}{2}$ years old, or $21$ years old.

If Caitlin is $5$ years younger than Briana, she is $21-5$ years old, or $16$.

So, the answer is $\boxed{B}$

See Also

2000 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
First
Question
Followed by
Problem 2
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