2000 AMC 8 Problems/Problem 1

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 1

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 $\boxed{B}$

Solution 2

Since Brianna is half of Aunt Anna's age this means that Brianna is $21$ years old. Now we just find Caitlin's age by doing $21-5$ . This makes $16$ or $\boxed{B}$

2000 AMC 8 (Problems • Answer Key • Resources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

2000 AMC 8 Problems/Problem 1

Contents

Problem

Solution 1

Solution 2

See Also