2000 AMC 8 Problems/Problem 1: Difference between revisions

Latest revision as of 01:58, 28 June 2024

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?

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

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

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

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

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

@@ Line 2: / Line 2: @@
 Aunt Anna is <math>42</math> years old. Caitlin is <math>5</math> years younger than Brianna, and Brianna is half as old as Aunt Anna. How old is Caitlin?
-<math>
+<math>\mathrm{(A)}\ 15\qquad\mathrm{(B)}\ 16\qquad\mathrm{(C)}\ 17\qquad\mathrm{(D)}\ 21\qquad\mathrm{(E)}\ 37</math>
-\mathrm{(A)}\ 15
-\qquad
-\mathrm{(B)}\ 16
-\qquad
-\mathrm{(C)}\ 17
-\qquad
-\mathrm{(D)}\ 21
-\qquad
-\mathrm{(E)}\ 37
-</math>
-== 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.
-If Caitlin is <math>5</math> years younger than Briana, she is <math>21-5</math> years old, or <math>16</math>.
+==Solution 1==
+If Brianna is half as old as Aunt Anna, then Brianna is <math>\frac{42}{2}</math> years old, or <math>21</math> years old.
+If Caitlin is <math>5</math> years younger than Brianna, she is <math>21-5</math> years old, or <math>16</math>.
 So, the answer is <math>\boxed{B}</math>
 ==See Also==
+{{AMC8 box|year=2000|before=First<br />Question|num-a=2}}
-[[2000 AMC 8 Problems/Problem 2|Next Problem]]
+{{MAA Notice}}
-[[2000 AMC 8]]