1990 AHSME Problems/Problem 1

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. Unknown error_msg)

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.

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1990 AHSME Problems/Problem 1

Problem

Solution