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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 ? (1) x + y < 0 (2) y < x < w [#permalink]
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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: y-w <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 ? (1) x + y < 0 (2) y < x < w [#permalink]
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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 w - y > 0 ? (1) x + y < 0 (2) y < x < w [#permalink]
Eshika wrote:
Is plugging numbers the best way to solve inequality probs or their is any other best method as this method takes lot of time

Well it REALLY depends on the problem. Algebraic approach will work in many cases, though sometimes other approaches might be faster or/and simpler, also there are certain GMAT questions which are pretty much only solvable with plug-in or trial and error methods (well at leas in 2-3 minutes).
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Re: If w + x < 0, is w - y > 0 ? (1) x + y < 0 (2) y < x < w [#permalink]
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
_________
w-y < 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 w - y > 0 ? (1) x + y < 0 (2) y < x < w [#permalink]
Simba2012 wrote:
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
_________
w-y < 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?

If you subtract -y -x > 0 from w+ x < 0 you'll get w+x-(-y-x)<0 --> w+x+y+x<0, not w-y < 0.

Hope it's clear.
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Re: If w + x < 0, is w - y > 0 ? (1) x + y < 0 (2) y < x < w [#permalink]
Bunuel chetan2u
When 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 - (-X-Y) < 0
W +2X + Y < 0

So, 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 w - y > 0 ? (1) x + y < 0 (2) y < x < w [#permalink]
Keats wrote:
Bunuel chetan2u
When 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 - (-X-Y) < 0
W +2X + Y < 0

So, 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)-x-y>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 w - y > 0 ? (1) x + y < 0 (2) y < x < w [#permalink]
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I have a question. You told we can add/subtract inequalities. If we subtract the inequalities given in question x+y+1<0 and w-y>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 w - y > 0 ? (1) x + y < 0 (2) y < x < w [#permalink]
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harshitasahu wrote:
I have a question. You told we can add/subtract inequalities. If we subtract the inequalities given in question x+y+1<0 and w-y>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?

w-y>0 is not given. The question is whether w-y>0.
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Re: If w + x < 0, is w - y > 0 ? (1) x + y < 0 (2) y < x < w [#permalink]
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)

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.

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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Re: If w + x < 0, is w - y > 0 ? (1) x + y < 0 (2) y < x < w [#permalink]
ganesh999 wrote:
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)

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.

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?

Hello

See, the thing is that you are assuming w-y to be > 0.
The catch here is that IF w+x < 0 and also IF w-y > 0, THEN x+y < 0.

BUT it does NOT mean that IF w+x < 0 and IF x+y < 0, then w-y will be > 0.
Eg, consider w=-4, x=-5, y=-3. Here w+x is negative and x+y is negative, but w-y is not greater than 0, rather w is less than y here.
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Re: If w + x < 0, is w - y > 0 ? (1) x + y < 0 (2) y < x < w [#permalink]
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gmatbull wrote:
If w + x < 0, is w - y > 0 ?

(1) x + y < 0
(2) y < x < w

Given: w + x < 0

Target question: Is w - y > 0 ?

Statement 1: x + y < 0
Since we're given the inequality w + x < 0, and since the inequality symbols are facing the same direction, we can ADD the inequalities to get: 2x + w + y < 0
This does not provide enough information to determine whether w - y > 0.
So, statement 1 is NOT SUFFICIENT

If you're not convinced, consider these two conflicting cases (that satisfy the given information):
Case a: w = 0, x = -1 and y = -1. In this case, the answer to the target question is YES, w - y > 0
Case b: w = 0, x = -1 and y = 0.5. In this case, the answer to the target question is NO, w - y is not greater than 0
Since we can’t answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: y < x < w
In this case, we should recognize that we can answer the target question without even using the given information (w + x < 0)
If y < x < w, then we can also conclude that y < w
From here, if we subtract y from both sides of the inequality we get: 0 < w - y
In other words, the answer to the target question is YES, w - y > 0
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Cheers,
Brent
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Re: If w + x < 0, is w - y > 0 ? (1) x + y < 0 (2) y < x < w [#permalink]
given
If w + x < 0
means either w>x or x>w and larger value is -ve
target is w - y > 0

#1
x + y < 0
either both x & y are -ve or x>y or y>x and larger value is -ve
but no info of w insufficient
#2

y < x < w
since given w+x<0 so we can say that y,w,x are all -ve such that w=-1,x=-2 and y=-3 or w= 5 , x=-6 , y= -9

so
w-y will always be >0
sufficient
option B

gmatbull wrote:
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 ? (1) x + y < 0 (2) y < x < w [#permalink]
If w + x < 0, is w - y > 0 ?

(1) x + y < 0
or
x < -y
or
w + x < w - y
which means
w - y can be zero or greater than 0

(2) y < x < w
or
y - w < x - w < 0
or
w - y > w - x > 0 . . . . . . . . . . . . . . . . . . . [multiplying all sides by -1]
Sufficient
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Re: If w + x < 0, is w - y > 0 ? (1) x + y < 0 (2) y < x < w [#permalink]
The key to this question is that you cannot subtract inequalities you can only add inequalities.

I subtracted 2 inequalities and got statement 1 sufficient, but when you do the subtraction otherway round it doesn't hold true.
Re: If w + x < 0, is w - y > 0 ? (1) x + y < 0 (2) y < x < w [#permalink]
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