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Difference between revisions of "1981 AHSME Problems/Problem 12"

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If <math>p</math>, <math>q</math>, and <math>M</math> are positive numbers and <math>q<100</math>, then the number obtained by increasing <math>M</math> by <math>p\%</math> and decreasing the result by <math>q\%</math> exceeds <math>M</math> if and only if
 
If <math>p</math>, <math>q</math>, and <math>M</math> are positive numbers and <math>q<100</math>, then the number obtained by increasing <math>M</math> by <math>p\%</math> and decreasing the result by <math>q\%</math> exceeds <math>M</math> if and only if
  
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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>
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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>
  
 
==Solution (Answer Choices)==
 
==Solution (Answer Choices)==

Latest revision as of 05:41, 11 September 2023

Problem

If $p$, $q$, and $M$ are positive numbers and $q<100$, then the number obtained by increasing $M$ by $p\%$ and decreasing the result by $q\%$ exceeds $M$ if and only if

$\textbf{(A)}\ p>q \qquad\textbf{(B)}\ p>\dfrac{q}{100-q}\qquad\textbf{(C)}\ p>\dfrac{q}{1-q}\qquad \textbf{(D)}\ p>\dfrac{100q}{100+q}\qquad\textbf{(E)}\ p>\dfrac{100q}{100-q}$

Solution (Answer Choices)

Answer Choice $A$: It is obviously incorrect because if $M$ is $50$ and we increase by $50$% and then decrease $49$%, $M$ will be around $37$.

Answer Choice $B$: If $p$ is $100$ and $q$ is $50$, it should be equal but instead we get $100$ is more than $1$. This is therefore also incorrect.

Answer Choice $C$: Obviously incorrect since if $q$ is larger than $1$, this is always valid since $\frac {1}{1-q}$ is less than $0$ which is obviously false.

Answer Choice $D$: If $p$ is $100$ and $q$ is $50$, it should be equal but instead we get $100$ is less than $\frac {5000}{150}$. Therefore, $D$ is also incorrect.

Answer Choice $E$: If $p$ is $100$ and $q$ is $50$, it should be equal, and if we check our equation, we get $\frac {5000}{50}$ = $100$. Therefore, our answer is $\boxed {(E)\dfrac{100q}{100-q}}$

~Arcticturn

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