2019 AMC 10A Problem 3

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

Let Bonita’s current age be

b

Use last year’s relation and this year’s square relation together

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.

Problem 3 in other years