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

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Note that this approach might take more testing if one is not familiar with finding factors.
 
Note that this approach might take more testing if one is not familiar with finding factors.
  
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== Solution 3 ==
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== Solution 3 (Answer choices)==
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Since all the answer choices are around <math>50\%</math>, we know the town's starting population must be around <math>600</math>. We list perfect squares from <math>400-1000</math>.
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Since all the answer choices are around <math>50\%</math>, we know the town's starting population must be around <math>600</math>. We list perfect squares from <math>400</math> to <math>1000</math>.
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<cmath>441, 484, 529,576,625,676,729,784,841,900,961</cmath>We see that <math>484</math> and <math>784</math> differ by <math>300</math>, and can we can confirm that <math>484</math> is the correct starting number by noting that <math>484+150=634=25^2+9</math>. Thus, the answer is <math>784/484-1\approx \boxed{\textbf{(E) } 62\%}</math>.
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<cmath>441, 484, 529,576,625,676,729,784,841,900,961</cmath>We see that <math>484</math> and <math>784</math> differ by <math>300</math>, and we can confirm that <math>484</math> is the correct starting number by noting that <math>484+150=634=25^2+9</math>. Thus, the answer is <math>484/784\approx \boxed{\textbf{(E) } 62\%}</math>.
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==Solution 4==
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Let the population of the town in 1991 be <math>a^2</math> and the population in 2011 be <math>b^2</math>. We know that <math>a^2+150+150=b^2 \implies a^2-b^2=-300 \implies b^2-a^2=300 \implies (b-a)(b+a)=300</math>. Note that <math>b-a</math> must be even. Testing, we see that <math>a=22</math> and <math>b=28</math> works, as <math>484+150-9=625=25^2</math>, so <math>\frac{784-484}{484} \approx \boxed{\textbf{(E) } 62\%}</math>.
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~MrThinker
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==Video Solution==
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https://youtu.be/arsFJaUhsbs
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~savannahsolver
  
 
== See Also ==
 
== See Also ==

Latest revision as of 20:25, 7 October 2025

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 $q^2+9$. 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 (Answer choices)

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 $484/784\approx \boxed{\textbf{(E) } 62\%}$.

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 $\frac{784-484}{484} \approx \boxed{\textbf{(E) } 62\%}$.

~MrThinker

Video Solution

https://youtu.be/arsFJaUhsbs

~savannahsolver

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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