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
.
Alternate Solution
Proceed through the difference of squares for and
:
However, instead of testing both pairs of factors we take a more certain approach.
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.
See Also
2011 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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All AMC 10 Problems and Solutions |
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