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

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

See Also