2019 AMC 10A Problems/Problem 3

Problem

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 $n?$

$\textbf{(A) } 3 \qquad\textbf{(B) } 5 \qquad\textbf{(C) } 9 \qquad\textbf{(D) } 12 \qquad\textbf{(E) } 15$

Solution

Solution 1

Let $A$ be the age of Ana and $B$ be the age of Bonita. Then,

$A-1 = 5(B-1)$ and $A = B^2.$

Substituting the second equation into the first gives us

$B^2-1 = 5(B-1).$

By using difference of squares and dividing, $B=4.$ Moreover, $A=B^2=16.$

The answer is $16-4 = 12 \implies \boxed{\textbf{(D)}}$

Solution 2 (Guess and Check)

Simple guess and check works. Start with all the square numbers - $1$ , $4$ , $9$ , $16$ , $25$ , $36$ , etc. (probably stop at around $100$ since at that point it wouldn't make sense). If Ana is $9$ , then Bonita is $3$ , so in the previous year, Ana's age was $4$ times greater than Bonita's. If Ana is $16$ , then Bonita is $4$ , and Ana's age was $5$ times greater than Bonita's in the previous year, as required. The difference in the ages is $16 - 4 = 12. \boxed{D}$

2019 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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2019 AMC 10A Problems/Problem 3

Contents

Problem

Solution

Solution 1

Solution 2 (Guess and Check)

See Also