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

(Created page with "==Problem== 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...")
 
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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>
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==Solution==

Revision as of 17:55, 23 October 2021

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

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