Difference between revisions of "2011 AMC 10A Problems/Problem 19"
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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>. | 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>: | 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> | <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. | + | 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^2-p^2=300</math> | ||
<math>(r-p)(r+p)=300</math> | <math>(r-p)(r+p)=300</math> | ||
Line 19: | Line 19: | ||
Then go through the same routine as demonstrated above to finish this problem. | 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. | + | 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 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>. | ||
+ | |||
+ | ==Solution 4== | ||
+ | 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>. | ||
+ | |||
+ | ~MrThinker | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/arsFJaUhsbs | ||
+ | |||
+ | ~savannahsolver | ||
+ | |||
== See Also == | == See Also == | ||
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?
Solution
Let the population of the town in be . Let the population in be . It follows that . Rearrange this equation to get . Since and are both positive integers with , and also must be, and thus, they are both factors of . We have two choices for pairs of factors of : and , and and . Assuming the former pair, since must be less than , and . Solve to get . Since is not a perfect square, this is not the correct pair. Solve for the other pair to get . This time, . This is the correct pair. Now, we find the percent increase from to . Since the increase is , the percent increase is .
Solution 2
Proceed through the difference of squares for and :
However, instead of testing both pairs of factors we take a more certain approach. Here is the population of the town in 2011. Test through pairs of and that makes sure or . 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 , we know the town's starting population must be around . We list perfect squares from to . We see that and differ by , and we can confirm that is the correct starting number by noting that . Thus, the answer is .
Solution 4
Let the population of the town in 1991 be and the population in 2011 be . We know that . Note that must be even. Testing, we see that and works, as , so .
~MrThinker
Video Solution
~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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