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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 ==
  
−
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>.
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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>.
 +
 
 +
== Solution 2 ==
 +
Proceed through the difference of squares for <math>p</math> and <math>q</math>:
 +
<math>141=q^2-p^2=(q-p)(q+p)</math>
 +
 
 +
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.
 +
<math>r^2-p^2=300</math>
 +
<math>(r-p)(r+p)=300</math>
 +
Test through pairs of <math>r</math> and <math>p</math> that makes sure <math>p=22</math> or <math>p=70</math>.
 +
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 <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>.
 +
<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>.
 +
 
 +
==Video Solution==
 +
https://youtu.be/arsFJaUhsbs
 +
 
 +
~savannahsolver
 +
 
 +
== See Also ==
 +
 
 +
 
 +
{{AMC10 box|year=2011|ab=A|num-b=18|num-a=20}}
 +
{{MAA Notice}}

Revision as of 21:53, 8 January 2021

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 can we can confirm that $484$ is the correct starting number by noting that $484+150=634=25^2+9$. Thus, the answer is $784/484-1\approx \boxed{\textbf{(E) } 62\%}$.

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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