2006 AMC 12A Problems/Problem 3: Difference between revisions

Revision as of 22:32, 27 April 2008

The ratio of Mary's age to Alice's age is $3:5$ . Alice is $30$ years old. How old is Mary?

$\mathrm{(A)}\ 15\qquad\mathrm{(B)}\ 18\qquad\mathrm{(C)}\ 20\qquad\mathrm{(D)}\ 24\qquad\mathrm{(E)}\ 50$

Let $m$ be Mary's age. Then $\frac{m}{30}=\frac{3}{5}$ . Solving for $m$ , we obtain $m=18$ . The answer is $\mathrm{(B)}$ .

2006 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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 AMC 12 Problems and Solutions

@@ Line 1: / Line 1: @@
 == Problem ==
 The ratio of Mary's age to Alice's age is <math>3:5</math>. Alice is <math>30</math> years old. How old is Mary?
-<math> \mathrm{(A) \ } 15\qquad \mathrm{(B) \ } 18\qquad \mathrm{(C) \ } 20\qquad \mathrm{(D) \ } 24</math>
+<math>\mathrm{(A)}\ 15\qquad\mathrm{(B)}\ 18\qquad\mathrm{(C)}\ 20\qquad\mathrm{(D)}\ 24\qquad\mathrm{(E)}\  50</math>
-<math>\mathrm{(E) \ }  50</math>
 == Solution ==
+Let <math>m</math> be Mary's age. Then <math>\frac{m}{30}=\frac{3}{5}</math>. Solving for <math>m</math>, we obtain <math>m=18</math>. The answer is <math>\mathrm{(B)}</math>.
-Let <math>m</math> be Mary's age.  Then <math>\frac{m}{30}=\frac{3}{5}</math>.  Solving for <math>m</math>, we obtain <math>m=18</math>.  The answer is B.
 == See also ==
-* [[2006 AMC 12A Problems]]
 {{AMC12 box|year=2006|ab=A|num-b=2|num-a=4}}
 [[Category:Introductory Algebra Problems]]