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

Latest revision as of 02:35, 4 February 2016

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

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

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 $\boxed{\text{(E)} \, \pm 8}$ .

1990 AHSME (Problems • Answer Key • Resources)
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

@@ Line 8: / Line 8: @@
 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>
+<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>.
+== See also ==
+{{AHSME box|year=1990|num-b=1|num-a=2}}
+[[Category: Introductory Algebra Problems]]
 {{MAA Notice}}