Difference between revisions of "1990 AHSME Problems/Problem 1"
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== Problem== | == Problem== | ||
| − | If <math>\dfrac{x | + | If <math>\dfrac{\frac{x}{4}}{2}=\dfrac{4}{\frac{x}{2}}</math>, then <math>x=</math> |
<math>\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</math> | <math>\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</math> | ||
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Cross-multiplying leaves | 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> | |
So the answer is <math>\boxed{\text{(E)} \, \pm 8}</math>. | 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}} | ||
Latest revision as of 02:35, 4 February 2016
Problem
If 
, then 
Solution
Cross-multiplying leaves
So the answer is 
.
See also
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
