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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 problem asks when $M\left(1+\frac{p}{100}\right)\left(1-\frac{q}{100}\right) > M$. We can simplify this inequality and isolate $p$ as follows:

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

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

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

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

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

-j314andrews

Solution 2 (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

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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