2011 AMC 10A Problems/Problem 19

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?

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


Let the population of the town in $1991$ be $p^2$. Let the population in $2001$ be $q^2+9$. 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 $\frac{300}{484}\times100\%\approx\boxed{\textbf{(E)}\ 62\%}$.

Solution 2

Proceed through the difference of squares for $p$ and $q$: $141=q^2-p^2=(q-p)(q+p)$

However, instead of testing both pairs of factors we take a more certain approach. Here $r^2$ is the population of the town in 2011. $r^2-p^2=300$ $(r-p)(r+p)=300$ Test through pairs of $r$ and $p$ that makes sure $p=22$ or $p=70$. Then go through the same routine as demonstrated above to finish this problem.

Note that this approach might take more testing if one is not familiar with finding factors.

Solution 3

Since all the answer choices are around $50\%$, we know the town's starting population must be around $600$. We list perfect squares from $400$ to $1000$. \[441, 484, 529,576,625,676,729,784,841,900,961\]We see that $484$ and $784$ differ by $300$, and we can confirm that $484$ is the correct starting number by noting that $484+150=634=25^2+9$. Thus, the answer is $784/484-1\approx \boxed{\textbf{(E) } 62\%}$.

Solution 4

Let the population of the town in 1991 be $a^2$ and the population in 2011 be $b^2$. We know that $a^2+150+150=b^2 \implies a^2-b^2=-300 \implies b^2-a^2=300 \implies (b-a)(b+a)=300$. Note that $b-a$ must be even. Testing, we see that $a=22$ and $b=28$ works, as $484+150-9=625=25^2$, so $\frac{784-484}{484} \approx \boxed{\textbf{(E) } 62\%}$.


Video Solution



See Also

2011 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 18
Followed by
Problem 20
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All AMC 10 Problems and Solutions

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