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

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

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

Solving this quadratic yields $B=4.$ Moreover, $A=B^2=16.$

The answer is $16-4 = 12 \implies \boxed{D}.$

Solution by Baolan

2019 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

2019 AMC 10A Problems/Problem 3

Problem

Solution

See Also