2019 AMC 10A Problem 3

Let $a$ be Ana’s current age and $b$ be Bonita’s current age. Then $$\begin{aligned} a-1&=5(b-1),\\ a&=b^2. \end{aligned}$$ Substitution gives $$\begin{aligned} b^2-1&=5b-5,\\ b^2-5b+4&=0,\\ (b-4)(b-1)&=0. \end{aligned}$$ We can see that $b \neq 1$ since that would make Ana and Bonita the same age, so we know that $b = 4.$ This gives us that $a = 4^2 = 16$ and $n = 16 - 4 = 12.$ Thus, D is the correct answer.