Difference between revisions of "1990 AHSME Problems/Problem 1"

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Cross-multiplying leaves  
 
Cross-multiplying leaves  
  
<math><cmath> \begin{align*}\dfrac{x^2}{8} &= 8\\ x^2 &= 64\\ \sqrt{x^2} &= \sqrt{64}\\ x &= \pm 8\end{align*} </cmath></math>
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<cmath> \begin{align*}\dfrac{x^2}{8} &= 8\\ x^2 &= 64\\ \sqrt{x^2} &= \sqrt{64}\\ x &= \pm 8\end{align*} </cmath>
  
 
So the answer is <math>\boxed{\text{(E)} \, \pm 8}</math>.
 
So the answer is <math>\boxed{\text{(E)} \, \pm 8}</math>.

Latest revision as of 03:35, 4 February 2016

Problem

If $\dfrac{\frac{x}{4}}{2}=\dfrac{4}{\frac{x}{2}}$, 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

\begin{align*}\dfrac{x^2}{8} &= 8\\ x^2 &= 64\\ \sqrt{x^2} &= \sqrt{64}\\ x &= \pm 8\end{align*}

So the answer is $\boxed{\text{(E)} \, \pm 8}$.

See also

1990 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 2
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All AHSME Problems and Solutions

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