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>(B)</math>. |
== See also == | == See also == |
Revision as of 22:45, 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.