Difference between revisions of "2006 AMC 12A Problems/Problem 7"
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== Solution == | == Solution == | ||
| − | Let <math>m</math> be Mary's age, let <math>s</math> be Sally's age, and let <math>d</math> be Danielle's age. We have <math>s=.6d</math>, and <math>m=1.2s=1.2(.6d)=.72d</math>. The sum of their ages is <math>m+s+d=.72d+.6d+d=2.32d</math>. Therefore, <math>2.32d=23.2</math>, and <math>d=10</math>. Then <math>m=.72(10)=7.2</math>. Mary will be <math>8</math> on her next birthday. The answer is <math>B</math>. | + | Let <math>m</math> be Mary's age, let <math>s</math> be Sally's age, and let <math>d</math> be Danielle's age. We have <math>s=.6d</math>, and <math>m=1.2s=1.2(.6d)=.72d</math>. The sum of their ages is <math>m+s+d=.72d+.6d+d=2.32d</math>. Therefore, <math>2.32d=23.2</math>, and <math>d=10</math>. Then <math>m=.72(10)=7.2</math>. Mary will be <math>8</math> on her next birthday. The answer is <math>\mathrm{(B)}</math>. |
== See also == | == See also == | ||
Latest revision as of 21:46, 5 February 2016
Problem
Mary is 
older than Sally, and Sally is 
younger than Danielle. The sum of their ages is 
years. How old will Mary be on her next birthday?
Solution
Let 
be Mary's age, let 
be Sally's age, and let 
be Danielle's age. We have 
, and 
. The sum of their ages is 
. Therefore, 
, and 
. Then 
. Mary will be 
on her next birthday. The answer is 
.
See also
| 2006 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 6 |
Followed by Problem 8 |
| 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
