1981 AHSME Problems/Problem 12

Problem 12

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 1

The goal is to find when $M\left(1+\frac{p}{100}\right)\left(1-\frac{q}{100}\right) > M$ . Isolate $p$ as follows:

$\left(1+\frac{p}{100}\right)\left(1-\frac{q}{100}\right) > 1$

$\left(\frac{100+p}{100}\right)\left(\frac{100-q}{100}\right) > 1$

$\frac{(100+p)(100-q)}{10000} > 1$

$100+p > \frac{10000}{100-q}$

$p > \frac{10000}{100-q} - 100 = \boxed{(\textbf{E})\ \frac{100q}{100-q}}$

-j314andrews

Solution 2 (Answer Choices)

Choice A is incorrect. If $(M, p, q) = (100, 50, 40)$ , then $p > q$ , but $M$ will not increase, as $100 \cdot 1.5 \cdot 0.6 = 90$ .

Choice B is incorrect. If $(M, p, q) = (100, 50, 50)$ , then $p > \frac{q}{100-q}$ , but $M$ will not increase, as $100 \cdot 1.5 \cdot 0.5 = 75$ .

Choice C is incorrect. If $(M, p, q) = (100, 50, 50)$ , then $p > \frac{q}{1-q}$ , but $M$ will not increase, as $100 \cdot 1.5 \cdot 0.5 = 75$ .

Choice D is incorrect. If $(M, p, q) = (100, 100, 60)$ , then $p > \frac{100q}{100+q}$ , but $M$ will not increase, as $100 \cdot 2 \cdot 0.4 = 80$ .

By process of elimination, $\boxed{(\textbf{E})\ \dfrac{100q}{100-q}}$ must be correct.

-Arcticturn, edited by j314andrews

1981 AHSME (Problems • Answer Key • Resources)
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1981 AHSME Problems/Problem 12

Contents

Problem 12

Solution 1

Solution 2 (Answer Choices)

See also