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

Revision as of 02:35, 11 September 2007

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

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

We can create the equation

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

Where x is Mary's age. Simplfying yields

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

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

@@ Line 12: / Line 12: @@
 <math>(3)(30)=5x\Longrightarrow 90=5x\Longrightarrow x=18, (B)</math>
-== See Also ==
+== See also ==
-*[[2006 AMC 10A Problems]]
+{{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]]