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 | + | 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 | + | <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>. |
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+ | ==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 20:54, 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.