Difference between revisions of "2006 AMC 10A Problems/Problem 3"
Ragnarok23 (talk | contribs) |
Ragnarok23 (talk | contribs) (→Solution) |
||
| Line 4: | Line 4: | ||
<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 | + | We can create the equation |
<math>\frac{3}{5}=\frac{x}{30}</math> | <math>\frac{3}{5}=\frac{x}{30}</math> | ||
| Line 11: | Line 11: | ||
<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 == | == See Also == | ||
*[[2006 AMC 10A Problems]] | *[[2006 AMC 10A Problems]] | ||
Revision as of 14:31, 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?
Solution
We can create the equation
Where x is Mary's age. Simplfying yields
