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If x + y > 0, is x > y? [#permalink]
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07 May 2010, 22:05
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If x + y > 0, is x > y? (1) x > y (2) y < 0
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Re: If x + y > 0, is x > y? [#permalink]
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Re: If x + y > 0, is x > y? [#permalink]
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If x + y >0, is x > y? (1) x > y (2) y < 0 Please help me with approach. I am having tough time with inequality. What is the better way to solve such questions? Picking number or doing algebraically? Here how i attempted this: Given: x+y > 0 , x > y To prove: x > y Stmt1: x > y. Now y = +y when y > 0 and y = y when y < 0. As can be seen, x > +y and x > y. Hence x > y. Sufficient. Stmt2: y < 0. y = y . So is x > y? From what is given, x > y. Sufficient. Ans: D.
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Re: If x + y > 0, is x > y? [#permalink]
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08 Apr 2011, 06:43
If x+y > 0 then x and y are both +ve  A or x is ve and y is +ve with y > x  B or x is +ve and y is ve with x > y  C (1) is sufficient as if x > y, then x > y too in this case (because x + y > 0 it could be case B or case C ) => x > y (2) y < 0, so x is +ve and x > y Answer  D
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Re: If x + y > 0, is x > y? [#permalink]
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08 Apr 2011, 10:19
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First strategy  You can throw in the values of x and y to be sure of the behavior of the inequality x > y 1. x > y and x + y > 0 x = 4 y = 3 the answer is yes x = 4 y = 3 the answer is yes sufficient
2. y < 0 and x + y > 0 y = 4 x = 4.5 the answer is yes sufficient
Answer D.
Second strategy : Doing algebra will lead to same inference.  Is x > y ? The question can be rephrased as  Is x > y and x > y ? Adding the two we have x + x > y  y or 2x > 0 or x > 0 So the question becomes Is x > 0?
1. x > y x + y >0 Adding the above two. 2x +y > y or x > 0 sufficient
2. 0>y x + y > 0 Adding the above two. x + y > y or x > 0 sufficient
Hence D.



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Re: If x + y > 0, is x > y? [#permalink]
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08 Apr 2011, 18:42
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Good solution there buddy. gmat1220 wrote: First strategy  You can throw in the values of x and y to be sure of the behavior of the inequality x > y 1. x > y and x + y > 0 x = 4 y = 3 the answer is yes x = 4 y = 3 the answer is yes sufficient
2. y < 0 and x + y > 0 y = 4 x = 4.5 the answer is yes sufficient
Answer D.
Second strategy : Doing algebra will lead to same inference.  Is x > y ? The question can be rephrased as  Is x > y and x > y ? Adding the two we have x + x > y  y or 2x > 0 or x > 0 So the question becomes Is x > 0?
1. x > y x + y >0 Adding the above two. 2x +y > y or x > 0 sufficient
2. 0>y x + y > 0 Adding the above two. x + y > y or x > 0 sufficient
Hence D.



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Re: If x + y > 0, is x > y? [#permalink]
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08 Apr 2011, 20:25
Try with 1)x=0, y=0 2)x=1, y=1 3)x=1, y=1 a) eliminates 1 and 3 above, so 2 can help find out the ans b) eliminates 1 and 3 above, so 2 can help find out the ans Either i.e. D
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Re: If x + y > 0, is x > y? [#permalink]
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08 Apr 2011, 21:26
x+y> 0 this statement implies that either both are +ve or one is +ve and other negative , but with a condition that the absolute value of +ve number > than that of ve. 1 1. x>y => x must be +ve. y can be either ve or +ve. when x,y are +ve is x > y? is true. when x +ve and y ve then also is x > y? is true using 1 2. y<0 => x must be +ve and its absolute value > than that of Y. which is the same thing asked => is x > y? thus D
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Re: If x + y > 0, is x > y? [#permalink]
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09 Apr 2011, 03:32
Thanks everyone. Different approaches gave me better idea on how to approach such problems. All i need now is some more inequality practice.
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Re: If x + y > 0, is x > y? [#permalink]
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20 May 2011, 04:02
for x+y > 0 x has to be > y the statement itself is sufficient. meaning D is clean.
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Re: If x + y > 0, is x > y? [#permalink]
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20 May 2011, 06:41
jamifahad wrote: If x + y >0, is x > y? (1) x > y (2) y < 0
Please help me with approach. I am having tough time with inequality. What is the better way to solve such questions? Picking number or doing algebraically?
Here how i attempted this: Given: x+y > 0 , x > y To prove: x > y
Stmt1: x > y. Now y = +y when y > 0 and y = y when y < 0. As can be seen, x > +y and x > y. Hence x > y. Sufficient.
Stmt2: y < 0. y = y . So is x > y? From what is given, x > y. Sufficient.
Ans: D. X+Y > 0 tells (Both are positive or X>Y or vice versa) For X >Y? True when X is positive and > Y, So first two cases Stmt 1 tells X> Y so x has to be positive to satisfy X+Y > 0  true Stmt2 : Y<0 , so x has to be positive to satisfy X+Y > 0  true D



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Re: If x + y > 0, is x > y? [#permalink]
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14 May 2013, 13:39
If x+y > 0, is x> y i. x>y ii.y<0
given
x>y or x<y..... question is x<y<x ?? stem provides first part of this inequality (x<y)
from 1
x>y...suff therefore sure x<y<x .
from 2
y is ve and question becomes is x>y i.e. is x<y ... provided in the stem



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Re: If x + y > 0, is x > y? [#permalink]
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28 Nov 2013, 02:18
Bunuel wrote: bibha wrote: If x+y > 0, is x> y i. x>y ii.y<0 Thanks If x + y >0, is x > y? \(x>y\) means: A. \(x>y\), if \(y\leq{0}\); B. \(x>y\), if \(y>{0}\). So we should check whether above two inequalities are true. First inequality is given to be true in the stem (x>y), so we should check whether \(x>y\) is true. (1) x > y. Sufficient. (2) y < 0 > \(y=y\). Question becomes is \(x>y\). This given to be true in the stem. Sufficient. Answer: D. Hi Bunuel, Can you please explain this step.I could not follow from your answer explanation. (2) y < 0 > y=y. Question becomes is x>y. This given to be true in the stem. Sufficient. Thanks in advance
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Re: If x + y > 0, is x > y? [#permalink]
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28 Nov 2013, 06:42
sunita123 wrote: Bunuel wrote: bibha wrote: If x+y > 0, is x> y i. x>y ii.y<0 Thanks If x + y >0, is x > y? \(x>y\) means: A. \(x>y\), if \(y\leq{0}\); B. \(x>y\), if \(y>{0}\). So we should check whether above two inequalities are true. First inequality is given to be true in the stem (x>y), so we should check whether \(x>y\) is true. (1) x > y. Sufficient. (2) y < 0 > \(y=y\). Question becomes is \(x>y\). This given to be true in the stem. Sufficient. Answer: D. Hi Bunuel, Can you please explain this step.I could not follow from your answer explanation. (2) y < 0 > y=y. Question becomes is x>y. This given to be true in the stem. Sufficient. Thanks in advance From (2) since y<0, then y=y. Thus the question becomes: is x>y? or is x+y>0? The stem says that this is true. Therefore the second statement is sufficient. Hope it's clear.
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Re: If x + y > 0, is x > y? [#permalink]
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24 Jan 2014, 15:34
x + y > 0 => x > y. Given x > y, is x > y?
1. x >y. solving x > y and x > y (adding), x > 0 => y has to be less than x (for the sum to be greater than zero) 2. y < 0. Since x + y > 0, x has to be larger than y (which is negative) and y.
D.



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Re: If x + y > 0, is x > y? [#permalink]
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03 Feb 2015, 22:33
This is how i approached it. Can someone please confirm if i m correct in my reasoning. x+y>0 => x>y If Y<0 look at blue & Y>0 look at red(y) [x>y]0 [x>y](y) [x>y][x>y] Now , i. x>y means "blue" marked range ii. y<0 "blue" marked range Hence D.
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Re: If x + y > 0, is x > y? [#permalink]
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