Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2006 AMC 12A Problems/Problem 3"

(→‎Solution 1)
(→‎Solution 2)
 
(One intermediate revision by the same user not shown)
Line 10: Line 10:
 
==Solution 2==
 
==Solution 2==
  
−
We can see this is a combined ratio of <math>8</math>, <math>(5+3)</math>. We can equalize by doing <math>30\div5=6</math>, and <math>6\cdot3=18</math>. With the common ratio of <math>8</math> and difference ratio of <math>6</math>, we see <math>6\cdot8=30+18</math>. Therefore, we can see our answer is correct.
+
We can see this is a combined ratio of <math>8</math>, <math>(5+3)</math>. We can equalize by doing <math>30\div5=6</math>, and <math>6\cdot3=\boxed{\textbf{(B) }18}</math>. With the common ratio of <math>8</math> and difference ratio of <math>6</math>, we see <math>6\cdot8=30+18</math>. Therefore, we can see our answer is correct.
  
 
== See also ==
 
== See also ==

Latest revision as of 16:30, 16 December 2021

The following problem is from both the 2006 AMC 12A #3 and 2006 AMC 10A #3, so both problems redirect to this page.

Problem

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

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

Solution 1

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

Solution 2

We can see this is a combined ratio of $8$, $(5+3)$. We can equalize by doing $30\div5=6$, and $6\cdot3=\boxed{\textbf{(B) }18}$. With the common ratio of $8$ and difference ratio of $6$, we see $6\cdot8=30+18$. Therefore, we can see our answer is correct.

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
2006 AMC 10A (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 10 Problems and Solutions

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_12A_Problems/Problem_3&oldid=167838"