1981 AHSME Problems/Problem 12

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 $0$ .

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 $\boxed {(E)\dfrac{100q}{100-q}}$

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1981 AHSME Problems/Problem 12

Problem

Solution (Answer Choices)