Question

Ana and Bonita were born on the same date in different years, $n$ years apart. Last year Ana was $5$ times as old as Bonita. This year Ana's age is the square of Bonita's age. What is $n?$

$3$

$5$

$9$

$12$

$15$

Solution:

Let $a$ be Ana's age and $b$ be Bonita's age. The statement then gives us that $\begin{gather*} a - 1 = 5(b - 1), \\ a = b^2. \end{gather*}$

We can substitute the second equation into the first to get $\begin{gather*} b^2 - 1 = 5b - 5 \\ b^2 - 5b + 4 = 0 \\ (b - 4)(b - 1) = 0. \end{gather*}$

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.

2019 AMC 10A Problem 3

Solution:

Problem 3 in Other Years