1968 AHSME Problems/Problem 16

Problem

If $x$ is such that $\frac{1}{x}<2$ and $\frac{1}{x}>-3$ , then:

$\text{(A) } -\frac{1}{3}<x<\frac{1}{2}\quad \text{(B) } -\frac{1}{2}<x<3\quad \text{(C) } x>\frac{1}{2}\quad\\ \text{(D) } x>\frac{1}{2} \text{ or} -\frac{1}{3}<x<0\quad \text{(E) } x>\frac{1}{2} \text{ or } x<-\frac{1}{3}$

Solution

Because $\frac{1}{x}<2$ , $x>\frac{1}{2}$ when $x$ is positive. Because $\frac{1}{x}>-3$ , $x<\frac{-1}{3}$ when $x$ is negative. $x$ can never be $0$ , so these two inequalities cover all cases for the value of $x$ . Thus, our answer is $\fbox{E}$ .

1968 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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1968 AHSME Problems/Problem 16

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