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mmgmat2008
- Junior | Next Rank: 30 Posts
- Posts: 23
- Joined: Tue Nov 11, 2008 6:46 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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Stem is stating,mmgmat2008 wrote:(1) x+y <0
(2) y<x<w
Answer b
I got this wrong. why not D for this question?
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
[/quote]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
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?
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
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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