Difference between revisions of "2006 AMC 12A Problems/Problem 3"
m (→See also: box) |
I like pie (talk | contribs) m (Style) |
||
| Line 1: | Line 1: | ||
== Problem == | == 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? | 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) \ | + | <math>\mathrm{(A)}\ 15\qquad\mathrm{(B)}\ 18\qquad\mathrm{(C)}\ 20\qquad\mathrm{(D)}\ 24\qquad\mathrm{(E)}\ 50</math> |
| − | |||
| − | |||
== Solution == | == 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. | ||
== See also == | == See also == | ||
| − | |||
| − | |||
{{AMC12 box|year=2006|ab=A|num-b=2|num-a=4}} | {{AMC12 box|year=2006|ab=A|num-b=2|num-a=4}} | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
Revision as of 23:32, 27 April 2008
Problem
The ratio of Mary's age to Alice's age is 
. Alice is 
years old. How old is Mary?
Solution
Let 
be Mary's age. Then 
. Solving for , we obtain 
. The answer is 
.
See also
| 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 | |
