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Difference between revisions of "2006 AMC 12A Problems/Problem 7"

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== Solution ==
 
== Solution ==
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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 B.
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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 ==
 
{{AMC12 box|year=2006|ab=A|num-b=6|num-a=8}}
 
{{AMC12 box|year=2006|ab=A|num-b=6|num-a=8}}
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{{MAA Notice}}
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]

Latest revision as of 21:46, 5 February 2016

Problem

Mary is $20\%$ older than Sally, and Sally is $40\%$ younger than Danielle. The sum of their ages is $23.2$ years. How old will Mary be on her next birthday?

$\mathrm{(A) \ } 7\qquad \mathrm{(B) \ } 8\qquad \mathrm{(C) \ } 9\qquad \mathrm{(D) \ } 10\qquad \mathrm{(E) \ } 11$

Solution

Let $m$ be Mary's age, let $s$ be Sally's age, and let $d$ be Danielle's age. We have $s=.6d$, and $m=1.2s=1.2(.6d)=.72d$. The sum of their ages is $m+s+d=.72d+.6d+d=2.32d$. Therefore, $2.32d=23.2$, and $d=10$. Then $m=.72(10)=7.2$. Mary will be $8$ on her next birthday. The answer is $\mathrm{(B)}$.

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

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