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Difference between revisions of "1967 AHSME Problems/Problem 7"

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== Solution ==
 
== Solution ==
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<math>\fbox{E}</math>
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Notice that <math>b</math> and <math>d</math> are irrelevant to the problem.  Also notice that the negative sign is irrelevant since <math>c</math> can be any real number.  So the question is equivalent to asking what values <math>a</math> can take on given the inequality <math>a<c</math>.  The answer, is of course, that <math>a</math> could be anything, or <math>\boxed{\textbf{(E) } \ a \; \text{can be positive, negative, or zero}}</math>
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~ proloto
  
 
== See also ==
 
== See also ==

Latest revision as of 18:41, 28 September 2023

Problem

If $\frac{a}{b}<-\frac{c}{d}$ where $a$, $b$, $c$, and $d$ are real numbers and $bd \not= 0$, then:

$\text{(A)}\ a \; \text{must be negative} \qquad \text{(B)}\ a \; \text{must be positive} \qquad$ $\text{(C)}\ a \; \text{must not be zero} \qquad \text{(D)}\ a \; \text{can be negative or zero, but not positive } \\ \text{(E)}\ a \; \text{can be positive, negative, or zero}$

Solution

Notice that $b$ and $d$ are irrelevant to the problem. Also notice that the negative sign is irrelevant since $c$ can be any real number. So the question is equivalent to asking what values $a$ can take on given the inequality $a<c$. The answer, is of course, that $a$ could be anything, or $\boxed{\textbf{(E) } \ a \; \text{can be positive, negative, or zero}}$

~ proloto

See also

1967 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
All AHSME Problems and Solutions

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