Difference between revisions of "2011 AMC 10A Problems/Problem 19"
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== Solution 3 == | == 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>400-1000</math>. | 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>. | ||
− | <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{\ | + | <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>. |
== See Also == | == See Also == |
Revision as of 21:11, 12 February 2019
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 . We see that and differ by , and can we can confirm that is the correct starting number by noting that . Thus, the answer is .
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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