Difference between revisions of "1966 AHSME Problems/Problem 15"

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== Solution ==
 
== Solution ==
 
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From <math>x-y>x</math>, we get that <math>-y>0\implies y<0</math>.
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From <math>x+y<y</math>, we get that <math>x<0</math>.
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So, our final answer is <math>\fbox{D}</math>.
  
 
== See also ==
 
== See also ==
 
{{AHSME box|year=1966|num-b=14|num-a=16}}   
 
{{AHSME box|year=1966|num-b=14|num-a=16}}   
  
[[Category:]]
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[[Category:Introductory Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 21:11, 14 January 2018

Problem

If $x-y>x$ and $x+y<y$, then

$\text{(A) } y<x \quad \text{(B) } x<y \quad \text{(C) } x<y<0 \quad \text{(D) } x<0,y<0 \quad \text{(E) } x<0,y>0$

Solution

From $x-y>x$, we get that $-y>0\implies y<0$. From $x+y<y$, we get that $x<0$. So, our final answer is $\fbox{D}$.

See also

1966 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
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