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is x+y > 0?

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Intern
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is x+y > 0? [#permalink] New post 22 Jan 2006, 22:14
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
Hi guyz,

came across this qn. in one of the diagnostic tests:

is x+y > 0?

(I) x² - y² > 1
(II) x/y + 1 > 0


My reasoning:

(I) x² - y² > 1
---> (x+y)(x-y) > 1
---> not sufficient.

(II) x/y + 1 > 0
---> x/y > -1
---> x > -y (<<< here, i'm assuming that y!= 0, is this ok?)
---> x+y > 0
---> SUFFICIENT!

So, the ans should be (B).

However, the test claims that the ans is (E)!!!
Here's their reasoning:

From statement I x² - y² > 1
This can be written as (x + y)(x - y) > 1
There is no other available information to determine if (x + y) is greater than or less than 0.
Thus statement 1 alone is not sufficient to answer the question.

From statement 2, x/y + 1 > 0 or (x + y)/y > 0.
This will be true when either both x + y > 0 and y > 0 or x + y < 0 and y < 0.
Since there is no other available information, this statement alone is also not sufficient to answer the question.

Combining the two statements also we do not get any definite information about, x, y or x + y.
Thus the two statements together are also not sufficient to answer the question.
Therefore, we cannot say for sure whether x + y> 0.

Hence (E) is the correct answer.


i'm stumped!!

please help me out here....i'm still not convinced i'm wrong! :P
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Re: DS: what the..?! [#permalink] New post 22 Jan 2006, 23:00
sandy007 wrote:
Hi guyz,

came across this qn. in one of the diagnostic tests:

is x+y > 0?

(I) x² - y² > 1
(II) x/y + 1 > 0


My reasoning:

(I) x² - y² > 1
---> (x+y)(x-y) > 1
---> not sufficient.

(II) x/y + 1 > 0
---> x/y > -1
---> x > -y (<<< here, i'm assuming that y!= 0, is this ok?)
---> x+y > 0
---> SUFFICIENT!


II) x/y + 1 > 0
---> x/y > -1
---> x > -y (<<< here, i'm assuming that y!= 0, is this ok?)
No you can't assume that Y=0, Y can be -1 or -2
---> x+y > 0 Let's take X=0 hence -1>0 wrong hence
---> INSUFFICIENT!
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 [#permalink] New post 22 Jan 2006, 23:10
The OA and OE are correct. x/y+1>0, does not automatically mean that x+y>0. for instance,
assume x=2, y=-3, then x/y+1=1/3 > 0 but x+y=-1<0.

So basically if y>0, then x=y>0 and if y<0 then x+y<0.

Hope this helps.
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 [#permalink] New post 22 Jan 2006, 23:49
Quote:
The OA and OE are correct. x/y+1>0, does not automatically mean that x+y>0. for instance,
assume x=2, y=-3, then x/y+1=1/3 > 0 but x+y=-1<0.

So basically if y>0, then x=y>0 and if y<0 then x+y<0.



sure does help...tks a lot, whateva! so i guess the ans is indeed E.

however, if i do not substitute values, and simplify the algebric eqn like i did earlier, i do get x+y > 0. So does that mean that my simplification is not correct? If yes, where did i go wrong?
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 [#permalink] New post 23 Jan 2006, 01:56
sandy007 wrote:
Quote:
The OA and OE are correct. x/y+1>0, does not automatically mean that x+y>0. for instance,
assume x=2, y=-3, then x/y+1=1/3 > 0 but x+y=-1<0.

So basically if y>0, then x=y>0 and if y<0 then x+y<0.



sure does help...tks a lot, whateva! so i guess the ans is indeed E.

however, if i do not substitute values, and simplify the algebric eqn like i did earlier, i do get x+y > 0. So does that mean that my simplification is not correct? If yes, where did i go wrong?


Never do a multiplication by a variable in inequalities if that variable can be either +ve or -ve. You can do multiplications if that variable can be +ve only.

Like x/y > -1 doesn't necessarily mean x > -y.

You can do additions in inequalities without any fear. :-D
_________________

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 [#permalink] New post 23 Jan 2006, 13:17
just a nitpick, you can do a multipication as long as you know the sign of the multiplier ( in this case y). If you are sure that y is negative, then multiply and "flip" the sign.

ps_dahiya wrote:
sandy007 wrote:
Quote:
The OA and OE are correct. x/y+1>0, does not automatically mean that x+y>0. for instance,
assume x=2, y=-3, then x/y+1=1/3 > 0 but x+y=-1<0.

So basically if y>0, then x=y>0 and if y<0 then x+y<0.



sure does help...tks a lot, whateva! so i guess the ans is indeed E.

however, if i do not substitute values, and simplify the algebric eqn like i did earlier, i do get x+y > 0. So does that mean that my simplification is not correct? If yes, where did i go wrong?


Never do a multiplication by a variable in inequalities if that variable can be either +ve or -ve. You can do multiplications if that variable can be +ve only.

Like x/y > -1 doesn't necessarily mean x > -y.

You can do additions in inequalities without any fear. :-D
Senior Manager
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 [#permalink] New post 23 Jan 2006, 22:11
ps_dahiya wrote:
sandy007 wrote:
Quote:
The OA and OE are correct. x/y+1>0, does not automatically mean that x+y>0. for instance,
assume x=2, y=-3, then x/y+1=1/3 > 0 but x+y=-1<0.

So basically if y>0, then x=y>0 and if y<0 then x+y<0.



sure does help...tks a lot, whateva! so i guess the ans is indeed E.

however, if i do not substitute values, and simplify the algebric eqn like i did earlier, i do get x+y > 0. So does that mean that my simplification is not correct? If yes, where did i go wrong?


Never do a multiplication by a variable in inequalities if that variable can be either +ve or -ve. You can do multiplications if that variable can be +ve only.

Like x/y > -1 doesn't necessarily mean x > -y.

You can do additions in inequalities without any fear. :-D


Thnks Sandy and Dhiya ! :) Now i know a few tips on inequalities! ( I hate these :( )
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  [#permalink] 23 Jan 2006, 22:11
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