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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

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

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