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11. If W +x <0, is w-y>0?

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Source: — Data Sufficiency |
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by niraj_a » Tue Dec 09, 2008 5:35 am
from the stem, they are asking if w > y

statement II gives you that info directly while statement I does not.

what's the source?
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Re: 11. If W +x <0, is w-y>0?

by hwiya320 » Tue Dec 09, 2008 7:19 am
mmgmat2008 wrote:(1) x+y <0
(2) y<x<w
Answer b

I got this wrong. why not D for this question?
Stem is stating,
W + X < 0
W < -X or W > X,
asking, is W>Y?

1) x +y < 0
which means x < -Y or -X > Y
so we knew that -X is greater than W, and now we know that -X is greater than Y. That doesn't prove that W is greater than Y.

I think what you did was X > -Y then flip the sign and got rid of negative for ALL numbers, X < Y... I assume that's how you determined that W>Y>X?

This Q is from GMATPrep.
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Thanks

by mmgmat2008 » Wed Dec 10, 2008 9:19 am
This type of questions is my weakness.
Thanks for the answers. I will try to not make the same mistake tomorrow at my G day!:)
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by cramya » Wed Dec 10, 2008 7:34 pm
Think of it this way.

No info about x,y,w are given (may or may not be integers)

Since w+x<0 is given all we know is either both are negative or one of them is negative and other positive with the abolsute value of the negative number greater than the absolute value of the positive number.

Given w+x<0

w+x<0 i.e w< - x Is w-y>0 or w > y

Stmt II(always try to pick the easier of the 2 statements - just my opinion)

y<x<w

Exactly what we need w>y

SUFF

Now its down to B) or D) .

This way u have increased the probability of answering the question right (50% prob)


Given w+x<0 i.e w < -x (1)

Stmt I

x+y <0 I.E X< - Y (2)

U can pick numbers here where w > y and where w < y satisfying (1) and (2)


INSUFF


Choose B)
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by chipbmk » Mon Nov 23, 2009 4:02 pm
Why can we not do this for statement 1:

Given: W+X<0

Statement 1: X+Y<0

Given --> X+W<0
Statement 1--> X+Y<0
______ (subtract the statements)
= W-Y<0 or W<Y

Doesnt that give you a definitive "no" to the question is w-y>0? or is w>y? appears like it does to me ... let me know
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by mehravikas » Tue Nov 24, 2009 6:36 pm
you can never subtract inequalities. you can only add them.

- 1 < 0
-2 < 0
If you add both inequalities you get -3 < 0, which is correct.

If you subtract 2nd statement from 1 i.e.

-1 < 0
-2 < 0
- -
-------

1 < 0 - incorrect
chipbmk wrote:Why can we not do this for statement 1:

Given: W+X<0

Statement 1: X+Y<0

Given --> X+W<0
Statement 1--> X+Y<0
______ (subtract the statements)
= W-Y<0 or W<Y

Doesnt that give you a definitive "no" to the question is w-y>0? or is w>y? appears like it does to me ... let me know
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by missrochelle » Sun Aug 29, 2010 4:22 pm
mehravikas wrote:you can never subtract inequalities. you can only add them.

- 1 < 0
-2 < 0
If you add both inequalities you get -3 < 0, which is correct.

If you subtract 2nd statement from 1 i.e.

-1 < 0
-2 < 0
- -
-------

1 < 0 - incorrect
[/quote]


IF YOU ADD THE INEQUALITY IN STATEMENT 1, TO THE STEM THEN HOW DO YOU GET -3 <1?

W + X < 0
X+ Y < 0
W+2X+Y < 0
WHICH DOESN'T ALLOW YOU TO ISOLATE W-Y..... IS THIS INCORRECT?
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by andy123 » Sun Mar 13, 2011 9:32 am
W + X < 0
W < -X or W > X,
asking, is W>Y?


Stupid Question but please help :

Is it OK to convert:

W < -X --- >> to ---- >> W > X,

I thought if we multiply by -1 , then we have to multiply both the sides of the equations and reverse the inequality ??

so in thi case: W < -X should not it be converted to -W > X ???
:)
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by Night reader » Sun Mar 13, 2011 10:33 am
an interesting problem popped up today. I found answer which is not OA cited in the post, please contribute

... this question's title contains part of this DS, so I am copying/pasting the title of the post back into the post :)
w<-x OR x<-w. Is w>y?
st(1) x<-y --> as x<-w both -w and -y can be greater than y, BUT we don't know if if w>y Not Sufficient;
st(2) y<x<w --> as x<-w and x<w, then we can deduce x<0 and |w|<x . If x<0 then y<0 too. And if y<x then |y|>|x|. If w<0 then w lies to the right from y, i.e. -y<-w. If w>0 then w lies to the left of y again, i.e. y<w. Since we don't know the sign of w, and w can be both +ve and -ve statement(2) is Not Sufficient;
Combined st(1&2): x<-y and y<x<w --> -x>y U x>y suggest y<0 and |x|<y. This is too not sufficient, as both w>y and -w>y present (the same condition was met in statement 2 Alone) BUT we need to meet both conditions w>y AND -w<-y

IOM E
If W +x <0, is w-y>0?
(1) x+y <0
(2) y<x<w
mmgmat2008 wrote:(1) x+y <0
(2) y<x<w
Answer b

I got this wrong. why not D for this question?
My knowledge frontiers came to evolve the GMATPill's methods - the credited study means to boost the Verbal competence. I really like their videos, especially for RC, CR and SC. You do check their study methods at https://www.gmatpill.com
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by andy123 » Mon Mar 14, 2011 1:58 am
2nd equation is easy:

Question prompt requires us to prooove: w-y > 0

which means: W > y ??

Option B: already gives the olution y<x<W - thus B.
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by HSPA » Mon Mar 14, 2011 2:57 am
w-y = w+x-x-y = (w+x)-(x+y)


Now let us use A (x+y <0) : (w+x)-(x+y) = -ve - ( -ve) [INSUFFICIENT]... we dont know how much negitive each fellow is
Now let us use B :
Consider w = 1, x = -2, y = -3
then (W+X)-(X+Y) = -1+5
w= 100, x = -101 then y is more negitive

But now
w= -1, x=0 so w+x <0 then x>y so y = -1... no w>y so y = -2...

Hello All ..... B seems fine for me
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by Night reader » Mon Mar 14, 2011 3:35 am
Perhaps I got lost in inequality translations, after plug in and with condition given (w+x<0) I figured |x| could not be more than |w| - so B is Sufficient.
thanks
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by ankur.agrawal » Mon Mar 14, 2011 7:13 am
mmgmat2008 wrote:(1) x+y <0
(2) y<x<w
Answer b

W+x<0 ; W-y>0 or is W>y ?

1) Take w =-2 & x=-3 y=-4 ; These value satisfies W+x<0 & x+y <0

Is W>y. From Values taken it seems Yes it is.

Lets test another set of values : x=1 ; y=-2 ; w=-3

Is W>y . No . A yes & a No . This Statement is not sufficient.

2) Taking the same value for the first case as above:

Take w =-2 & x=-3 y=-4. These values satisfies y<x<w & W+x<0 .

Is w> y . Yes . Lets check another set of values.

w=2 ; x=-3 ; y=-4. again a yes.

If we use this method i think the answer should be D. lets check algebraically for the 2nd Statement.

W+x <0 ; y<x<w

Ist Case : Both Negative : Yes W>y
2nd Case: W + & x - (x has to be more negative than w is positive) = Yes W>y
3rd Case : W- & x+ ( W has to be more negative than x is +)= Yes W>y.

Am i missing sumthing.

I Think B
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by ankur.agrawal » Mon Mar 14, 2011 7:15 am
mmgmat2008 wrote:(1) x+y <0
(2) y<x<w
Answer b

W+x<0 ; W-y>0 or is W>y ?

1) Take w =-2 & x=-3 y=-4 ; These value satisfies W+x<0 & x+y <0

Is W>y. From Values taken it seems Yes it is.

Lets test another set of values : x=1 ; y=-2 ; w=-3

Is W>y . No . A yes & a No . This Statement is not sufficient.

2) Taking the same value for the first case as above:

Take w =-2 & x=-3 y=-4. These values satisfies y<x<w & W+x<0 .

Is w> y . Yes . Lets check another set of values.

w=2 ; x=-3 ; y=-4. again a yes.

If we use this method i think the answer should be D. lets check algebraically for the 2nd Statement.

W+x <0 ; y<x<w

Ist Case : Both Negative : Yes W>y
2nd Case: W + & x - (x has to be more negative than w is positive) = Yes W>y
3rd Case : W- & x+ ( W has to be more negative than x is +)= Yes W>y.

Am i missing sumthing.

I Think B
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