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Difference between revisions of "2011 AMC 10A Problems/Problem 19"

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

Revision as of 20:11, 10 February 2011

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 $\frac{300}{484}\times100\%\approx\boxed{\textbf{(E)}\ 62\%}$.

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