2011 AMC 10A Problems/Problem 19
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 | |
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