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

Solution 3 (Answer Choices)

The second sentence of the problem says that Ana's age was once $5$ times Bonita's age. Therefore, the difference of the ages $n$ must be divisible by $4.$ The only answer choice which is divisible by $4$ is $12 \rightarrow \boxed{\textbf(D)}.$

- awesome_weisur

Video Solution

https://youtu.be/rbKRwUubUWo

~savannahsolver

See Also

2019 AMC 10A (ProblemsAnswer KeyResources)
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

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