2019 AMC 10A Problem 3

Let Bonita be $b$ years old this year, so Ana is $b^2$. Last year their ages were $b - 1$ and $b^2 - 1$, and the first condition says
$$b^2 - 1 = 5(b - 1).$$
Factor the left side as $(b-1)(b+1)$. Since $b \ne 1$ (that would make Bonita $0$ last year and Ana the same age, not "years apart"), divide by $b - 1$:
$$b + 1 = 5 \quad\Longrightarrow\quad b = 4.$$
So Bonita is $4$, Ana is $16$, and the gap is $n = 16 - 4 = 12$. Check last year: $15 = 5 \cdot 3$.

The answer is $\boxed{\textbf{(D)}\ 12}$.

Age problems reduce to one variable because everyone ages at the same rate. Here the "square" condition makes the equation quadratic, but $b^2 - 1$ and $5(b-1)$ share the factor $b - 1$, so the difference of squares collapses it to a linear equation instantly.

Test small squares. Bonita $b$, Ana $b^2$, last year ratio $\frac{b^2-1}{b-1} = b + 1$ must equal $5$, so $b = 4$ immediately. Or try the choices: $n = 12$ means ages $(b, b+12)$ with $b + 12 = b^2$, giving $b = 4$, and $15 = 5 \cdot 3$ confirms.

• Reporting Bonita's age 4 or Ana's age 16 instead of the difference n = 12, or accepting the degenerate root b = 1.
• Writing last year's condition with this year's ages, $b^2 = 5b$, which gives $b = 5$ and the wrong answer $20$ (not even a choice, a sign to recheck).