Difference between revisions of "1981 AHSME Problems/Problem 12"
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==Solution 1== | ==Solution 1== | ||
− | The problem asks when <math>M\left(1+\frac{p}{100}\right)\left(1-\frac{q}{100}\ | + | The problem asks when <math>M\left(1+\frac{p}{100}\right)\left(1-\frac{q}{100}\right) > M</math>. We can simplify this inequality and isolate <math>p</math> as follows: |
− | <math>\left(1+\frac{p}{100}\right)\left(1-\frac{q}{100}\ | + | <math>\left(1+\frac{p}{100}\right)\left(1-\frac{q}{100}\right) > 1</math> |
− | <math>\left(\frac{100+p}{100}\right)\left({100-q}{100}\ | + | <math>\left(\frac{100+p}{100}\right)\left({100-q}{100}\right) > 1</math> |
<math>\frac{(100+p)(100-q)}{10000} > 1</math> | <math>\frac{(100+p)(100-q)}{10000} > 1</math> | ||
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<math>p > \frac{10000}{100-q} - 100 = \frac{100q}{100-q} \fbox{(E)}</math> | <math>p > \frac{10000}{100-q} - 100 = \frac{100q}{100-q} \fbox{(E)}</math> | ||
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==Solution 2 (Answer Choices)== | ==Solution 2 (Answer Choices)== |
Revision as of 13:54, 28 June 2025
Problem 12
If ,
, and
are positive numbers and
, then the number obtained by increasing
by
and decreasing the result by
exceeds
if and only if
Solution 1
The problem asks when . We can simplify this inequality and isolate
as follows:
Solution 2 (Answer Choices)
Answer Choice : It is obviously incorrect because if
is
and we increase by
% and then decrease
%,
will be around
.
Answer Choice : If
is
and
is
, it should be equal but instead we get
is more than
. This is therefore also incorrect.
Answer Choice : Obviously incorrect since if
is larger than
, this is always valid since
is less than
which is obviously false.
Answer Choice : If
is
and
is
, it should be equal but instead we get
is less than
. Therefore,
is also incorrect.
Answer Choice : If
is
and
is
, it should be equal, and if we check our equation, we get
=
. Therefore, our answer is
~Arcticturn
See also
1981 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
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 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.