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

Revision as of 16:51, 3 July 2013

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 B.

2006 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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

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

Revision as of 16:51, 3 July 2013

Problem

Solution

See also