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 $n?$

$3$

$5$

$9$

$12$

$15$

Answer: D

Concepts:ages system of equations quadratic

Difficulty rating: 960

Solution:

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.

2019 AMC 10A Problem 3

Answer: D

Difficulty rating: 960

Solution:

Problem 3 in Other Years