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1990 AHSME Problems/Problem 1

Revision as of 03:03, 4 November 2013 by RavenclawPrefect (talk | contribs) (Changed the value under the square root symbol from x to x^2.)

Problem

If $\dfrac{x/4}{2}=\dfrac{4}{x/2}$, then $x=$

$\text{(A)}\ \pm\frac{1}{2}\qquad\text{(B)}\ \pm 1\qquad\text{(C)}\ \pm 2\qquad\text{(D)}\ \pm 4\qquad\text{(E)}\ \pm 8$

Solution

Cross-multiplying leaves

$<cmath> \begin{align*}\dfrac{x^2}{8} &= 8\\ x^2 &= 64\\ \sqrt{x^2} &= \sqrt{64}\\ x &= \pm 8\end{align*} </cmath>$ (Error compiling LaTeX. ! Package amsmath Error: \begin{align*} allowed only in paragraph mode.)

So the answer is $\boxed{\text{(E)} \, \pm 8}$. The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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