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

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<math> \mathrm{(A) \ } 15\qquad \mathrm{(B) \ } 18\qquad \mathrm{(C) \ } 20\qquad \mathrm{(D) \ } 24\qquad \mathrm{(E) \ } 50 </math>
 
<math> \mathrm{(A) \ } 15\qquad \mathrm{(B) \ } 18\qquad \mathrm{(C) \ } 20\qquad \mathrm{(D) \ } 24\qquad \mathrm{(E) \ } 50 </math>
 
== Solution ==
 
== Solution ==
 +
We can create the ratio
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 +
<math>\frac{3}{5}=\frac{x}{30}</math>
 +
 +
Where ''x'' is Mary's age. Simplfying yields
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<math>(3)(30)=5x\Longrightarrow 90=5x\Longrightarrow x=18, (B)</math>
 
== See Also ==
 
== See Also ==
 
*[[2006 AMC 10A Problems]]
 
*[[2006 AMC 10A Problems]]

Revision as of 13:31, 3 July 2006

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 ratio

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

Where x is Mary's age. Simplfying yields

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

See Also

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