# 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 (Problems • Answer Key • Resources) Preceded byProblem 6 Followed byProblem 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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