Question

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?

$42$

$47$

$52$

$57$

$62$

Solution:

Let the population in $1991$ be $p^2.$ Then let the population in $2001$ be $q^2 + 9.$

Using these values, we have $p^2 + 150 = q^2 + 9$ $q^2 - p^2 = 141.$

Factoring, we get $(q - p)(q + p) = 141.$

As $p$ and $q$ are integers, we have that the only possible values for $q - p$ and $q + p$ are $(1, 141)$ and $(3, 47).$

Trying the first pair, we have $q - p = 1 \text{ and } q + p = 141,$ which adding together and dividing gives us $q = 71$ and $p = 70.$

We have that $p^2 + 300$ is not a square number, which means that this pair is the wrong one.

Trying the other pair and using the same strategy gives us $q = 25$ and $p = 22.$

Now, $p^2 + 300 = 784,$ which is a perfect square. The percent increase in population is then $\dfrac{300}{22^2} \cdot 100 \% \approx 62 \%.$

Thus, E is the correct answer.

2011 AMC 10A Problem 19

Solution:

Problem 19 in Other Years