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

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<math> \textbf{(A)}\ 42 \qquad\textbf{(B)}\ 47 \qquad\textbf{(C)}\ 52\qquad\textbf{(D)}\ 57\qquad\textbf{(E)}\ 62 </math>
 
<math> \textbf{(A)}\ 42 \qquad\textbf{(B)}\ 47 \qquad\textbf{(C)}\ 52\qquad\textbf{(D)}\ 57\qquad\textbf{(E)}\ 62 </math>
  
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=== Solution ===
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== 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>. 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>.
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== Solution 2 ==
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Proceed through the difference of squares for <math>p</math> and <math>q</math>:
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<math>141=q^2-p^2=(q-p)(q+p)</math>
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However, instead of testing both pairs of factors we take a more certain approach. Here <math>r^2</math> is the population of the town in 2011.
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<math>r^2-p^2=300</math>
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<math>(r-p)(r+p)=300</math>
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Test through pairs of <math>r</math> and <math>p</math> that makes sure <math>p=22</math> or <math>p=70</math>.
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Then go through the same routine as demonstrated above to finish this problem.
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 +
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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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 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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==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
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== See Also ==
 +
 
 +
 
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{{AMC10 box|year=2011|ab=A|num-b=18|num-a=20}}
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{{MAA Notice}}

Latest revision as of 19:54, 21 August 2023

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

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

~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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