Difference between revisions of "1968 AHSME Problems/Problem 16"
(Created page with "== Problem == If <math>x</math> is such that <math>\frac{1}{x}<2</math> and <math>\frac{1}{x}>-3</math>, then: <math>\text{(A) } -\frac{1}{3}<x<\frac{1}{2}\quad \text{(B) } -\f...") |
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== Solution == | == Solution == | ||
− | <math>\fbox{}</math> | + | |
+ | Because <math>\frac{1}{x}<2</math>, <math>x>\frac{1}{2}</math> when <math>x</math> is positive. Because <math>\frac{1}{x}>-3</math>, <math>x<\frac{-1}{3}</math> when <math>x</math> is negative. <math>x</math> can never be <math>0</math>, so these two inequalities cover all cases for the value of <math>x</math>. Thus, our answer is <math>\fbox{E}</math>. | ||
== See also == | == See also == | ||
− | {{AHSME box|year=1968|num-b=15|num-a=17}} | + | {{AHSME 35p box|year=1968|num-b=15|num-a=17}} |
[[Category: Introductory Algebra Problems]] | [[Category: Introductory Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 19:46, 17 July 2024
Problem
If is such that and , then:
Solution
Because , when is positive. Because , when is negative. can never be , so these two inequalities cover all cases for the value of . Thus, our answer is .
See also
1968 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
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All AHSME Problems and Solutions |
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