smc

This problem is very wordy. Nonetheless, let $a$ and $b$ be Ann and Barbara's current ages, respectively. We are given that $a+b=44$. Let $y$ equal the difference between their ages, so $y=a-b$. Know that $y$ is constant because the difference between their ages will always be the same. Now, let's tackle the equation: $b=$ Ann's age when Barbara was Ann's age when Barbara was $\frac{a}{2}$. When Barbara was $\frac{a}{2}$ years old, Ann was $\frac{a}{2}+y$ years old. So the equation becomes $b=$ Ann's age when Barbara was $\frac{a}{2}+y$. Adding on their age difference again, we get $b=\frac{a}{2}+y+y\Rightarrow b=\frac{a}{2}+2y$. Substitute $a-b$ back in for $y$ to get $b=\frac{a}{2}+2(a-b)$. Simplify: $2b=a+4(a-b)\Rightarrow 6b=5a$. Solving $b$ in terms of $a$, we have $b=\frac{5a}{6}$. Substitute that back into the first equation of $a+b=44$ to get $\frac{11a}{6}=44$. Solve for $a$, and the answer is $\boxed{24}$.