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

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So the answer is <math>\boxed{\text{(E)} \, \pm 8}</math>.
 
So the answer is <math>\boxed{\text{(E)} \, \pm 8}</math>.
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== See also ==
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{{AHSME box|year=1990|num-b=1|num-a=2}} 
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[[Category: Introductory Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 15:44, 28 September 2014

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

$<cmath> \begin{align*}\dfrac{x^2}{8} &= 8\\ x^2 &= 64\\ \sqrt{x^2} &= \sqrt{64}\\ x &= \pm 8\end{align*} </cmath>$ (Error compiling LaTeX. Unknown error_msg)

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

See also

1990 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 2
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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