Difference between revisions of "1981 AHSME Problems/Problem 12"
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<math>\textbf{(A)}\ p>q \qquad\textbf{(B)}\ p>\dfrac{q}{100-q}\qquad\textbf{(C)}\ p>\dfrac{q}{1-q}\qquad \textbf{(D)}\ p>\dfrac{100q}{100+q}\qquad\textbf{(E)}\ p>\dfrac{100q}{100-q}</math> | <math>\textbf{(A)}\ p>q \qquad\textbf{(B)}\ p>\dfrac{q}{100-q}\qquad\textbf{(C)}\ p>\dfrac{q}{1-q}\qquad \textbf{(D)}\ p>\dfrac{100q}{100+q}\qquad\textbf{(E)}\ p>\dfrac{100q}{100-q}</math> | ||
| − | ==Solution (Answer Choices)== | + | ==Solution 1== |
| + | |||
| + | The problem asks when <math>M\left(1+\frac{p}{100}\right)\left(1-\frac{q}{100}\left) > 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}\left) > 1</math> | ||
| + | |||
| + | <math>\left(\frac{100+p}{100}\right)\left({100-q}{100}\left) > 1</math> | ||
| + | |||
| + | <math>\frac{(100+p)(100-q)}{10000} > 1</math> | ||
| + | |||
| + | <math>100+p > \frac{10000}{100-q}</math> | ||
| + | |||
| + | <math>p > \frac{10000}{100-q} - 100 = \frac{100q}{100-q} \fbox{(E)}</math> | ||
| + | |||
| + | |||
| + | ==Solution 2 (Answer Choices)== | ||
Answer Choice <math>A</math>: It is obviously incorrect because if <math>M</math> is <math>50</math> and we increase by <math>50</math>% and then decrease <math>49</math>%, <math>M</math> will be around <math>37</math>. | Answer Choice <math>A</math>: It is obviously incorrect because if <math>M</math> is <math>50</math> and we increase by <math>50</math>% and then decrease <math>49</math>%, <math>M</math> will be around <math>37</math>. | ||
Revision as of 13:53, 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 $M\left(1+\frac{p}{100}\right)\left(1-\frac{q}{100}\left) > M$ (Error compiling LaTeX. Unknown error_msg). We can simplify this inequality and isolate as follows:
$\left(1+\frac{p}{100}\right)\left(1-\frac{q}{100}\left) > 1$ (Error compiling LaTeX. Unknown error_msg)
$\left(\frac{100+p}{100}\right)\left({100-q}{100}\left) > 1$ (Error compiling LaTeX. Unknown error_msg)
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. 
