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Inequality

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Source: — Data Sufficiency |
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by pemdas » Fri Jul 22, 2011 4:51 am
revised e

st(1) x² - y² > 1, translates into x² > y² + 1 which means only |x|>|y| Not Sufficient;
st(2) x/y + 1 > 0, x/y>-1 Not Sufficient;
combined st(1&2): |x|>|y| and x/y>-1 leaves x and y can be both either positive or negative, still not Sufficient
finance wrote:Is x + y > 0 ?
(I) x² - y² > 1
(II) x/y + 1 > 0
Last edited by pemdas on Fri Jul 22, 2011 5:04 am, edited 1 time in total.
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by knight247 » Fri Jul 22, 2011 5:03 am
Pemdas,
In statement 2 can't we resolve it further to get

(x+y)/y>0
x+y>0

?
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by pemdas » Fri Jul 22, 2011 5:09 am
x/y + 1>0 is the original inequality, and we need to know the sign of y to multiply safely without the sign revision

I left this untouched, as the sign of y is not known.
knight247 wrote:Pemdas,
In statement 2 can't we resolve it further to get

(x+y)/y>0
x+y>0

?
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by abhisays » Fri Jul 22, 2011 5:14 am
To quickly solve the problem, we can take x = -2 and y = -1. These values satisfy both 1 and 2. Also if you take x = 2 and y = 1, again these values satisfy both the inequalities so it not possible to tell x+y>0
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by finance » Fri Jul 22, 2011 6:21 am
I thought it would be B, but the OA is E. I missed the point that x and y are not given as positive.
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by GAMATO » Fri Jul 22, 2011 6:45 am
pemdas wrote:revised e

st(1) x² - y² > 1, translates into x² > y² + 1 which means only |x|>|y| Not Sufficient;
st(2) x/y + 1 > 0, x/y>-1 Not Sufficient;
combined st(1&2): |x|>|y| and x/y>-1 leaves x and y can be both either positive or negative, still not Sufficient
finance wrote:Is x + y > 0 ?
(I) x² - y² > 1
(II) x/y + 1 > 0
Hi pemdas,

Cd you please explain Statement 1 please. I am not able to understand how did you derive |x|>|y|

Thanks in advance.
GAMATO
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by pemdas » Fri Jul 22, 2011 9:03 am
since x² > y² + 1, I conclude that x² > y² must be true (x², y² are positive values)
That x² > y² means |x|>|y| only, as the signs are not given

GAMATO wrote:
pemdas wrote:revised e

st(1) x² - y² > 1, translates into x² > y² + 1 which means only |x|>|y| Not Sufficient;
st(2) x/y + 1 > 0, x/y>-1 Not Sufficient;
combined st(1&2): |x|>|y| and x/y>-1 leaves x and y can be both either positive or negative, still not Sufficient
finance wrote:Is x + y > 0 ?
(I) x² - y² > 1
(II) x/y + 1 > 0
Hi pemdas,

Cd you please explain Statement 1 please. I am not able to understand how did you derive |x|>|y|

Thanks in advance.
GAMATO
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