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

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{{duplicate|[[2006 AMC 12A Problems|2006 AMC 12A #3]] and [[2006 AMC 10A Problems/Problem 3|2006 AMC 10A #3]]}}
 
== Problem ==
 
== Problem ==
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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?
  
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<math> \mathrm{(A) \ } 15\qquad \mathrm{(B) \ } 18\qquad \mathrm{(C) \ } 20\qquad \mathrm{(D) \ } 24</math>
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<math>\textbf{(A)}\ 15\qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 24\qquad\textbf{(E)}\  50</math>
  
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<math>\mathrm{(E) \ } 50</math>
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== Solution 1 ==
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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=\boxed{\textbf{(B) }18}.</math>
  
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== Solution ==
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==Solution 2==
  
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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.
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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 ==
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* [[2006 AMC 12A Problems]]
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{{AMC12 box|year=2006|ab=A|num-b=2|num-a=4}}
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*[[2006 AMC 12A Problems/Problem 2|Previous Problem]]
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{{AMC10 box|year=2006|ab=A|num-b=2|num-a=4}}
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*[[2006 AMC 12A Problems/Problem 4|Next Problem]]
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{{MAA Notice}}
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]

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

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