If w + x < 0 , is w  y > 0 ?
1) x + y < 0
2) y < X < w
for statement 1, why cant be minus both fractions together ( w+x<0  { x+y<0} because they are toward the same sign <0....????
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Great question diebeatsthegmat!diebeatsthegmat wrote:If w + x < 0 , is w  y > 0 ?
1) x + y < 0
2) y < X < w
for statement 1, why cant be minus both fractions together ( w+x<0  { x+y<0} because they are toward the same sign <0....????
You are correct that we can add inequalities by stacking when the signs are facing the same direction  BUT what happens when we subtract? Think about a regular old set of equalities (something we can stack and either add OR subtract). For example:
2x + y = 23
x + y = 14
Now, if I want to subtract the second equation from the first (to eliminate y), we am essentially multiplying everything in the second equation by 1
2x + y = 23
1(x + y) = 1(14)

2x + y = 23
x  y = 14

x = 9
So why isn't this allowed for the inequalities?? Well, rather than just tell you, I want you to consider. When you stack the 2 inequalities with their signs going in the same direction:
w + x < 0
x + y < 0
What happens when you try to "subtract" the second one (ie. multiply it though by a negative 1)?
Whit
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If w + x < 0 , is w  y > 0 ?

either w or X or both of them are ve.
For wy >0.

Y = +Ve then W>Y, w = +ve
or Y = ve then either w = +ve
or w = ve but [Y] > [W] ([]= numerical value)
1) x + y < 0

this means either X or Y or both are ve.
or either of X or Y could be Zero and another one ve.
2) y < X < w

This means X = ve, W = ve
Y = ve
So we know w  y > 0
Hence B

either w or X or both of them are ve.
For wy >0.

Y = +Ve then W>Y, w = +ve
or Y = ve then either w = +ve
or w = ve but [Y] > [W] ([]= numerical value)
1) x + y < 0

this means either X or Y or both are ve.
or either of X or Y could be Zero and another one ve.
2) y < X < w

This means X = ve, W = ve
Y = ve
So we know w  y > 0
Hence B
This time no looking back!!!
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thank you whit for a quick reply...Whitney Garner wrote:Great question diebeatsthegmat!diebeatsthegmat wrote:If w + x < 0 , is w  y > 0 ?
1) x + y < 0
2) y < X < w
for statement 1, why cant be minus both fractions together ( w+x<0  { x+y<0} because they are toward the same sign <0....????
You are correct that we can add inequalities by stacking when the signs are facing the same direction  BUT what happens when we subtract? Think about a regular old set of equalities (something we can stack and either add OR subtract). For example:
2x + y = 23
x + y = 14
Now, if I want to subtract the second equation from the first (to eliminate y), we am essentially multiplying everything in the second equation by 1
2x + y = 23
1(x + y) = 1(14)

2x + y = 23
x  y = 14

x = 9
So why isn't this allowed for the inequalities?? Well, rather than just tell you, I want you to consider. When you stack the 2 inequalities with their signs going in the same direction:
w + x < 0
x + y < 0
What happens when you try to "subtract" the second one (ie. multiply it though by a negative 1)?
Whit
if i multiplied the second one by a netative 1 i cant add or subtract them all together because the signs will be opposite...

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I would go with E.
We have w+x<0 in the question stem.
From A.) we have x+y<0, subtracting the the two inequations we have w  y , however we don't know
which of the two inequation has a larger negative value so we cannot conclude the sign of wy.Not sufficient
From B.) xw >0, adding to the question stem we have x<0.Also we have x>y so we can conclude 0>x>y.
Now the question is is w>y (or is w> a negative quantity) when w < x (w < a positive quantity).
Again we cannot be sure. Not sufficient.
From A we get x+y<0, From B we get 0>x>y.Combing them gives no new information. Hence E.
We have w+x<0 in the question stem.
From A.) we have x+y<0, subtracting the the two inequations we have w  y , however we don't know
which of the two inequation has a larger negative value so we cannot conclude the sign of wy.Not sufficient
From B.) xw >0, adding to the question stem we have x<0.Also we have x>y so we can conclude 0>x>y.
Now the question is is w>y (or is w> a negative quantity) when w < x (w < a positive quantity).
Again we cannot be sure. Not sufficient.
From A we get x+y<0, From B we get 0>x>y.Combing them gives no new information. Hence E.
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You got it!! Nice catchdiebeatsthegmat wrote:thank you whit for a quick reply...
if i multiplied the second one by a netative 1 i cant add or subtract them all together because the signs will be opposite...
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Be VERY careful! We cannot subtract inequalities in the same way we can subtract equalities (see my discussion with diebeatsthegmat above for the logic).AbhiJ wrote:I would go with E.
We have w+x<0 in the question stem.
From A.) we have x+y<0, subtracting the the two inequations we have w  y , however we don't know
which of the two inequation has a larger negative value so we cannot conclude the sign of wy.Not sufficient
Again, be careful when you translate the stem. Asking "Is wy>0?" is the same as asking "Is w>y?" so is W a bigger number than Y?. Statement 2 tells us that y<x<w and. So, if w is a number bigger than x (further to the right on the number line) and y is a number smaller than x (further to the left on the number line), then we KNOW that w is bigger than y (further to the right). SUFFICIENT.AbhiJ wrote:From B.) xw >0, adding to the question stem we have x<0.Also we have x>y so we can conclude 0>x>y.
Now the question is is w>y (or is w> a negative quantity) when w < x (w < a positive quantity).
Again we cannot be sure. Not sufficient.
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Hey, to me it looks like C (It is possible to tell the answer by using both the equations given but not using one of them alone).
Please let me know, if you also agree to this?
Thanks
Mayank
Please let me know, if you also agree to this?
Thanks
Mayank
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Hi Mayank!mankey wrote:Hey, to me it looks like C (It is possible to tell the answer by using both the equations given but not using one of them alone).
Please let me know, if you also agree to this?
Thanks
Mayank
See the complete discussion above, but when the stem asks "Is wy>0?", it is the same as asking "Is w>y?" so is W a bigger number than Y?. Statement 2 tells us that y<x<w. So, if w is a number bigger than x (further to the right on the number line) and y is a number smaller than x (further to the left on the number line), then we KNOW that w is bigger than y (further to the right). Statement (2) is SUFFICIENT on its own.
If we test statement (1) by adding the inequality (x+y<0) to the inequality in the question (w+x<0), we get that w+2x+y<0, but there is no way to determine the relative sizes of W and Y (we have to be careful to NOT subtract inequalities because this is not allowed). Therefore statement (1) is insufficient.
The correct answer is B.
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Its my pleasure! This is actually something that LOTS of people confuse so I'm really happy that it was asked at all!!
Whit
Whit
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Given: w + x < 0diebeatsthegmat wrote: ↑Mon Aug 29, 2011 1:39 amIf w + x < 0 , is w  y > 0 ?
1) x + y < 0
2) y < X < w
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
Answer: B
Cheers,
Brent