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If w+x<0, is wy>0?
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04 Dec 2010, 06:33
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If w + x < 0, is w  y>0? (1) x + y < 0 (2) y < x < w
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Re: If w + x < 0, is w  y . 0?
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04 Dec 2010, 06:50
gmatbull wrote: If w + x < 0, is w  y . 0? (1) x + y < 0 (2) y < x < w
I don't know why i missed this question; with regards to (1) Please explain your steps. If w + x < 0 , is w  y > 0 ?Question: is \(w>y\)? (1) x + y < 0 > for this statement best way would be to pick numbers: on DS questions when plugging numbers, goal is to prove that the statement is not sufficient. So we should try to get a YES answer with one chosen number(s) and a NO with another. If \(x=0\), \(w=1\) and \(y=2\) then the answer would be YES but if \(x=0\), \(w=2\) and \(y=1\) then the answer would be. Not sufficient. (2) \(y<x<w\) > ignore \(x\) > \(y<w\), directly tells us the answer. Sufficient. Answer: B.
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Re: If w + x < 0, is w  y . 0?
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04 Dec 2010, 15:23
This one appears okay for an algebraic option, but missed out due to some careless mistakes. I combined the equations given in the stimulus as well as in the options. stem: w + x < 0..........(a) required to answer: is w > y? (1) y + x < 0...........(b) SUBTRACTING: (a)  (b) we get w  y < 0 ==> w<y..... answer to the question is NO But again SUBTRACTING: (b)  (a) we get: yw <0 ==> w>y ..... answer to the question is YES INSUFFICIENT I simply did not take into consideration this second part. (2) I have no problem resolving (2)...SUFFICIENT Bunuel: Thanks buddy for your awesome contributions.
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Re: If w + x < 0, is w  y . 0?
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04 Dec 2010, 15:38
gmatbull wrote: This one appears okay for an algebraic option, but missed out due to some careless mistakes.
I combined the equations given in the stimulus as well as in the options. stem: w + x < 0..........(a) required to answer: is w > y? (1) y + x < 0...........(b) SUBTRACTING: (a)  (b) we get w  y < 0 ==> w<y..... answer to the question is NO
But again SUBTRACTING: (b)  (a) we get: yw <0 ==> w>y ..... answer to the question is YES INSUFFICIENT I simply did not take into consideration this second part.
(2) I have no problem resolving (2)...SUFFICIENT
Bunuel: Thanks buddy for your awesome contributions. The red part is not correct. You can not subtract inequalities with signs in the same direction, you can only add them. You can only add inequalities when their signs are in the same direction:If \(a>b\) and \(c>d\) (signs in same direction: \(>\) and \(>\)) > \(a+c>b+d\). Example: \(3<4\) and \(2<5\) > \(3+2<4+5\). You can only apply subtraction when their signs are in the opposite directions:If \(a>b\) and \(c<d\) (signs in opposite direction: \(>\) and \(<\)) > \(ac>bd\) (take the sign of the inequality you subtract from). Example: \(3<4\) and \(5>1\) > \(35<41\). As for algebraic way.We can add \(w+x<0\) and \(y+x<0\) and we'll get \(w+y+2x<0\), which tells us nothing whether \(w>y\). In DS questions when after certain algebraic manipulations you don't have clear answer then you can quite safely assume that this statement is not sufficient, though in order to get definite answer you must use number plugging.
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Re: If w + x < 0, is w  y . 0?
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13 Dec 2010, 13:41



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Re: If w + x < 0, is w  y . 0?
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14 Dec 2010, 08:01



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Re: If w+x<0, is wy>0?
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17 Jan 2012, 23:22
Is plugging numbers the best way to solve inequality probs or their is any other best method as this method takes lot of time



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Re: If w+x<0, is wy>0?
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18 Jan 2012, 04:55



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Re: If w+x<0, is wy>0?
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18 Jan 2012, 05:05
Ok ..I need to practice lots of them..Thanks Bunuel



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Re: If w + x < 0, is w  y . 0?
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01 Jun 2012, 18:00
Bunuel wrote: gmatbull wrote: This one appears okay for an algebraic option, but missed out due to some careless mistakes.
I combined the equations given in the stimulus as well as in the options. stem: w + x < 0..........(a) required to answer: is w > y? (1) y + x < 0...........(b) SUBTRACTING: (a)  (b) we get w  y < 0 ==> w<y..... answer to the question is NO
But again SUBTRACTING: (b)  (a) we get: yw <0 ==> w>y ..... answer to the question is YES INSUFFICIENT I simply did not take into consideration this second part.
(2) I have no problem resolving (2)...SUFFICIENT
Bunuel: Thanks buddy for your awesome contributions. The red part is not correct. You can not subtract inequalities with signs in the same direction, you can only add them. You can only add inequalities when their signs are in the same direction:If \(a>b\) and \(c>d\) (signs in same direction: \(>\) and \(>\)) > \(a+c>b+d\). Example: \(3<4\) and \(2<5\) > \(3+2<4+5\). You can only apply subtraction when their signs are in the opposite directions:If \(a>b\) and \(c<d\) (signs in opposite direction: \(>\) and \(<\)) > \(ac>bd\) (take the sign of the inequality you subtract from). Example: \(3<4\) and \(5>1\) > \(35<41\). As for algebraic way.We can add \(w+x<0\) and \(y+x<0\) and we'll get \(w+y+2x<0\), which tells us nothing whether \(w>y\). In DS questions when after certain algebraic manipulations you don't have clear answer then you can quite safely assume that this statement is not sufficient, though in order to get definite answer you must use number plugging. Thanks, Bunuel! Didn't know that property about inequalities. If inequalities facing the same direction, you can only add them, never subtract. Nice!



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Re: If w+x<0, is wy>0?
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08 Dec 2012, 19:01
Bunuel, I understand we cannot subtract inequalities with same sign but I can subtract if the signs are different, right? if so is this correct Given w+ x < 0 Statement (1) x+ y < 0 now multiply with 1 it results x y > 0
so now we combine the given information with statement 1 info we get
w+ x < 0 y x > 0 _________ wy < 0
I know this contradicts what we get in statement 2 but I don't understand why we can't multiply with 1 and subtract statement 1 from given info? can you help?



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Re: If w+x<0, is wy>0?
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09 Dec 2012, 07:00



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Re: If w+x<0, is wy>0?
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09 Dec 2012, 09:23
I Totally missed that part to assign the  sign. Thanks Bunuel!



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Re: If w+x<0, is wy>0?
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24 Aug 2015, 15:05
gmatbull wrote: If w + x < 0, is w  y>0?
(1) x + y < 0 (2) y < x < w Given, w+x<0 Per statement 1, x+y<0 > this could mean w=y and hence a "no" for w>y but can also be w<y or w>y. Thus giving an insufficient answer. Per statement 2, y<x<w > directly gives y<w . B is the correct answer.



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If w+x<0, is wy>0?
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05 Dec 2016, 03:55
Bunuel chetan2uWhen we subtract two inequalities that are having its sign in opposite direction, why do we retain the sign of the inequality that we "subtract from"? For example: 1: W + X < 0 2: X  Y > 0 1 2 W+X  (XY) < 0W +2X + Y < 0So, my question is why are we retaining the sign of inequality (1) and not of inequality (2)?



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Re: If w+x<0, is wy>0?
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07 Dec 2016, 03:50
Keats wrote: Bunuel chetan2uWhen we subtract two inequalities that are having its sign in opposite direction, why do we retain the sign of the inequality that we "subtract from"? For example: 1: W + X < 0 2: X  Y > 0 1 2 W+X  (XY) < 0W +2X + Y < 0So, my question is why are we retaining the sign of inequality (1) and not of inequality (2)? Hi You should take the Q as.. 1) w+x<0 2)xy>0.. First get the inequality signs same for two.. So 2) x+y<0.. If you want to change the inequality sign, multiply both sides with ''.. Now the two equations are 1) w+x<0 2)x+y<0.. Now you can add the two as the signs are same.. Remember only add not subtract as you do not know the values.. So w+x+x+y<0....w+2x+y<0
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Re: If w+x<0, is wy>0?
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27 Dec 2016, 11:50
I have a question. You told we can add/subtract inequalities. If we subtract the inequalities given in question x+y+1<0 and wy>0. The output will come as x+y<0. And the first option x+y<0, totally satisfies it. So why isn't this sufficient?



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Re: If w+x<0, is wy>0?
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27 Dec 2016, 11:55



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Re: If w+x<0, is wy>0?
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13 Mar 2017, 10:37
Question says w+x < 0, while (1) says y + x < 0.
So, both (w+x) and (y+x) are < 0. Clearly from this, we cannot find out relatively whether w > y. Could be either way.



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Re: If w+x<0, is wy>0?
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06 Jun 2018, 21:03
Bunuel wrote: gmatbull wrote: If w + x < 0, is w  y . 0? (1) x + y < 0 (2) y < x < w
I don't know why i missed this question; with regards to (1) Please explain your steps. If w + x < 0 , is w  y > 0 ?Question: is \(w>y\)? (1) x + y < 0 > for this statement best way would be to pick numbers: on DS questions when plugging numbers, goal is to prove that the statement is not sufficient. So we should try to get a YES answer with one chosen number(s) and a NO with another. If \(x=0\), \(w=1\) and \(y=2\) then the answer would be YES but if \(x=0\), \(w=2\) and \(y=1\) then the answer would be. Not sufficient. (2) \(y<x<w\) > ignore \(x\) > \(y<w\), directly tells us the answer. Sufficient. Answer: B. Hello Buneul, I'm struck with the following approach. For the condition w > y to hold, I undertook the following approach. w + x < 0 , w  y > 0 I subtracted both of them I got x + y < 0. So i deduced that for w  y > 0 to hold, we need the condition x + y < 0, condition that is present in the answer choice. and hence I wrote D as the answer. Am I missing something?




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