Difference between revisions of "2006 AMC 10A Problems/Problem 3"

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<math>(3)(30)=5x\Longrightarrow 90=5x\Longrightarrow x=18, (B)</math>
 
<math>(3)(30)=5x\Longrightarrow 90=5x\Longrightarrow x=18, (B)</math>
  
== See Also ==
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== See also ==
*[[2006 AMC 10A Problems]]
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{{AMC10 box|year=2006|ab=A|num-b=2|num-a=4}}
 
 
*[[2006 AMC 10A Problems/Problem 2|Previous Problem]]
 
 
 
*[[2006 AMC 10A Problems/Problem 4|Next Problem]]
 
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]

Revision as of 01:35, 11 September 2007

Problem

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

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

Solution

We can create the equation

$\frac{3}{5}=\frac{x}{30}$

Where x is Mary's age. Simplfying yields

$(3)(30)=5x\Longrightarrow 90=5x\Longrightarrow x=18, (B)$

See also

2006 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
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All AMC 10 Problems and Solutions
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