1986 AHSME Problems/Problem 20

Revision as of 18:50, 9 October 2017 by GeronimoStilton (talk | contribs) (Solution)

Problem

Suppose $x$ and $y$ are inversely proportional and positive. If $x$ increases by $p\%$, then $y$ decreases by

$\textbf{(A)}\ p\%\qquad \textbf{(B)}\ \frac{p}{1+p}\%\qquad \textbf{(C)}\ \frac{100}{p}\%\qquad \textbf{(D)}\ \frac{p}{100+p}\%\qquad \textbf{(E)}\ \frac{100p}{100+p}\%$

Solution

We see that $x$ is multiplied by $\frac{100+p}{100}$ when it is increased by $p%$ (Error compiling LaTeX. Unknown error_msg). Therefore, $y$ is multiplied by $\frac{100}{100+p}$, and so it is decreased by $\frac{p}{100+p}$ times itself. Therefore, it is decreased by $\frac{100p}{100+p}%$ (Error compiling LaTeX. Unknown error_msg), and so the answer is $\boxed{E}$.

See also

1986 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png