Difference between revisions of "2006 AMC 12A 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 1 == |
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 <math>\mathrm{(B)}</math>. | 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 <math>\mathrm{(B)}</math>. | ||
| + | |||
| + | ==Solution 2== | ||
| + | |||
| + | 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=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 == | ||
{{AMC12 box|year=2006|ab=A|num-b=2|num-a=4}} | {{AMC12 box|year=2006|ab=A|num-b=2|num-a=4}} | ||
{{AMC10 box|year=2006|ab=A|num-b=2|num-a=4}} | {{AMC10 box|year=2006|ab=A|num-b=2|num-a=4}} | ||
| + | {{MAA Notice}} | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
Revision as of 17:01, 1 June 2021
- The following problem is from both the 2006 AMC 12A #3 and 2006 AMC 10A #3, so both problems redirect to this page.
Contents
Problem
The ratio of Mary's age to Alice's age is 
. Alice is 
years old. How old is Mary?
Solution 1
Let 
be Mary's age. Then 
. Solving for , we obtain 
. The answer is 
.
Solution 2
We can see this is a combined ratio of 
, 
. We can equalize by doing 
, and 
. With the common ratio of and difference ratio of 
, we see 
. 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. 
