2006 AMC 12A Problems/Problem 7

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 (ProblemsAnswer KeyResources)
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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