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Revision as of 11:01, 4 July 2013

Problem 19

In 1991 the population of a town was a perfect square. Ten years later, after an increase of 150 people, the population was 9 more than a perfect square. Now, in 2011, with an increase of another 150 people, the population is once again a perfect square. Which of the following is closest to the percent growth of the town's population during this twenty-year period?

$\textbf{(A)}\ 42 \qquad\textbf{(B)}\ 47 \qquad\textbf{(C)}\ 52\qquad\textbf{(D)}\ 57\qquad\textbf{(E)}\ 62$

Solution

Let the population of the town in $1991$ be $p^2$. Let the population in $2001$ be $q^2+9$. Let the population in 2011 be $r^2$. It follows that $p^2+150=q^2+9$. Rearrange this equation to get $141=q^2-p^2=(q-p)(q+p)$. Since $q$ and $p$ are both positive integers with $q>p$, $(q-p)$ and $(q+p)$ also must be, and thus, they are both factors of $141$. We have two choices for pairs of factors of $141$: $1$ and $141$, and $3$ and $47$. Assuming the former pair, since $(q-p)$ must be less than $(q+p)$, $q-p=1$ and $q+p=141$. Solve to get $p=70, q=71$. Since $p^2+300$ is not a perfect square, this is not the correct pair. Solve for the other pair to get $p=22, q=25$. This time, $p^2+300=22^2+300=784=28^2$. This is the correct pair. Now, we find the percent increase from $22^2=484$ to $28^2=784$. Since the increase is $300$, the percent increase is $\frac{300}{484}\times100\%\approx\boxed{\textbf{(E)}\ 62\%}$.


See Also

2011 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
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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