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How it is B? Did they mention that X and Y are integers? No right, the answer should be E. If they provide details about X and Y as integers then it will be B otherwise it will be E.

can anyone help me about the scenario whether we consider fractions or not in this case?

Scenario:

x=1/2, y=1/3 ==> |1/2-1/3|=1/6 and |1/2|-|1/3|=1/6

The point is that x = 1/2 and y = 1/3 do not satisfy xy < 0 (the second statement). _________________

How it is B? Did they mention that X and Y are integers? No right, the answer should be E. If they provide details about X and Y as integers then it will be B otherwise it will be E.

can anyone help me about the scenario whether we consider fractions or not in this case?

Scenario:

x=1/2, y=1/3 ==> |1/2-1/3|=1/6 and |1/2|-|1/3|=1/6

It is B because if you use the data of statement 2, you can say, "Yes, |x-y| is greater than |x|-|y|"

(2) xy < 0 This means that one of x and y is positive and the other is negative. You cannot take x = 1/2 and y = 1/3. It is not about fractions/integers. It is about positive/negative numbers (most mod questions are about positive/negative numbers) When xy < 0, |x-y|>|x|-|y| always holds. Only when x and y both are positive or both are negative and |x|>|y|, then |x-y|=|x|-|y| _________________

How it is B? Did they mention that X and Y are integers? No right, the answer should be E. If they provide details about X and Y as integers then it will be B otherwise it will be E.

can anyone help me about the scenario whether we consider fractions or not in this case?

Scenario:

x=1/2, y=1/3 ==> |1/2-1/3|=1/6 and |1/2|-|1/3|=1/6

The point is that x = 1/2 and y = 1/3 do not satisfy xy < 0 (the second statement).

How it is B? Did they mention that X and Y are integers? No right, the answer should be E. If they provide details about X and Y as integers then it will be B otherwise it will be E.

can anyone help me about the scenario whether we consider fractions or not in this case?

Scenario:

x=1/2, y=1/3 ==> |1/2-1/3|=1/6 and |1/2|-|1/3|=1/6

It's implied that it is integers on the GMAT? Is this perception by me correct or completely out of the blue? _________________

How it is B? Did they mention that X and Y are integers? No right, the answer should be E. If they provide details about X and Y as integers then it will be B otherwise it will be E.

can anyone help me about the scenario whether we consider fractions or not in this case?

Scenario:

x=1/2, y=1/3 ==> |1/2-1/3|=1/6 and |1/2|-|1/3|=1/6

From F.S 1, we have that x>y. Thus |x-y| = x-y. Thus, we have to answer whether x-y>|x|-|y|.

or x-|x|>y-|y|. Now for x>0, and y>0, we have is 0>0 and hence a NO. Again, for x>0 and y<0, we have a YES. Insufficient.

For F.S 2, we know that x and y are of opposite signs. Thus, x and y being on the opposite sides of the number line w.r.t the origin, the term |x-y| will always be more than the difference of the absolute distance of x and y from origin.Sufficient.

How it is B? Did they mention that X and Y are integers? No right, the answer should be E. If they provide details about X and Y as integers then it will be B otherwise it will be E.

can anyone help me about the scenario whether we consider fractions or not in this case?

Scenario:

x=1/2, y=1/3 ==> |1/2-1/3|=1/6 and |1/2|-|1/3|=1/6

It's implied that it is integers on the GMAT? Is this perception by me correct or completely out of the blue?

No, that's completely wrong, we cannot assume that x and y are integers, if this is not explicitly stated.

Generally, GMAT deals with only Real Numbers: Integers, Fractions and Irrational Numbers. So, if no limitations, then all we can say about a variable in a question that it's a real number.

Probably the best way to solve this problem is plug-in method. Though there are two properties worth to remember: 1. Always true: \(|x+y|\leq{|x|+|y|}\), note that "=" sign holds for \(xy\geq{0}\) (or simply when \(x\) and \(y\) have the same sign);

2. Always true: \(|x-y|\geq{|x|-|y|}\), note that "=" sign holds for \(xy>{0}\) (so when \(x\) and \(y\) have the same sign) and \(|x|>|y|\) (simultaneously). (Our case)

So, the question basically asks whether we can exclude "=" scenario from the second property.

(1) y < x --> we can not determine the signs of \(x\) and \(y\). Not sufficient. (2) xy < 0 --> "=" scenario is excluded from the second property, thus \(|x-y|>|x|-|y|\). Sufficient.

Answer: B.

(1) x>y x=-2,y=-4 then 2>-2 --> yes x=4,y=-2 then 6>2 --> yes can't get a no, so sufficient

(2) xy<0 x=4,y=-2 then 6>2 --> yes x=-2,y=4 then 6>-2 --> yes can't get a no, so sufficient

ans: D why is the answer B? is the question mis-written and the inequality sign should have >= or <=?

Probably the best way to solve this problem is plug-in method. Though there are two properties worth to remember: 1. Always true: \(|x+y|\leq{|x|+|y|}\), note that "=" sign holds for \(xy\geq{0}\) (or simply when \(x\) and \(y\) have the same sign);

2. Always true: \(|x-y|\geq{|x|-|y|}\), note that "=" sign holds for \(xy>{0}\) (so when \(x\) and \(y\) have the same sign) and \(|x|>|y|\) (simultaneously). (Our case)

So, the question basically asks whether we can exclude "=" scenario from the second property.

(1) y < x --> we can not determine the signs of \(x\) and \(y\). Not sufficient. (2) xy < 0 --> "=" scenario is excluded from the second property, thus \(|x-y|>|x|-|y|\). Sufficient.

Answer: B.

(1) x>y x=-2,y=-4 then 2>-2 --> yes x=4,y=-2 then 6>2 --> yes can't get a no, so sufficient

(2) xy<0 x=4,y=-2 then 6>2 --> yes x=-2,y=4 then 6>-2 --> yes can't get a no, so sufficient

ans: D why is the answer B? is the question mis-written and the inequality sign should have >= or <=?

What about the case x = 4, y = 2 in statement 1? then we get 2 > 2 --> No Hence statement 2 is not sufficient. _________________

If y is less than x then (x-y) is going to be positive, however, we don't know if x and y are positive or negative:

I. (x-y) > x -y ===> 0 > 0

II. (x-y) > -x -y ===> 2x > 0

III. (x-y) > -x +y ===> 2x > 2y

IV. (x-y) > x +y ===> 0 > 2y

The way I see it, is with case I.) 0>0 isn't true, II.) x must be some non-negative # that isn't zero, III.) x > y but we already know that, IV.) y must be some non-negative # that isn't zero. So we know that x is positive, y is negative and that x > y but we still can't get a single answer. All we know for sure is that y < x

(x=4, y=2) |x-y|>|x|-|y| (x-y)>(x)-(y) x-y>x-y 0>0 |x-y|>|x|-|y| |4-2|>|4|-|2| 2>2 FALSE (0>0 isn't possible, nor does it confirm y or x) NOT SUFFICIENT

(2) xy < 0

So either x is less than zero or y is less than zero. x & y ≠ 0.

There are two possible cases: (x is positive and y is negative) or (x is negative and y is positive)

I. (x is positive and y is negative) |x-y|>|x|-|y| (x-y)>(x)-(-y) x-y>x+y 0>2y (which holds with the premise in the first case that y is negative)

II. (x is negative and y is positive) |x-y|>|x|-|y| -(x-y)>(-x)-(y) -x+y>-x-y 2y>0 (which holds with the premise in the second case that y is positive) SUFFICIENT

what if like that (x-y)^2>x^2-y^2 so x^2-2xy+y^2>x^2-y^2 and x^2-2xy+y^2-x^2+y^2>0, and 2y^2-2xy>0 and 2y(y-x)>0 finally, y>0 and y-x>0 (y>x)

Then, 1) y < x, not sufficient, because it negates only one final condition and y may be both positive and negative 2) xy < 0, sufficient, because confirms that when y>0, y>x when x is negative

How it is B? Did they mention that X and Y are integers? No right, the answer should be E. If they provide details about X and Y as integers then it will be B otherwise it will be E.

can anyone help me about the scenario whether we consider fractions or not in this case?

Scenario:

x=1/2, y=1/3 ==> |1/2-1/3|=1/6 and |1/2|-|1/3|=1/6

.......

st(1), use x=3 , y = 2 and then x=1 , y = -1 , we will have a double case. ----insufficient st(2), use x= -5 , y = 10 and then x=10 , y = -5, we will have a single solution and its yes |x-y|>|x|-|y| .so its sufficient. you can use fractions in st(2) maintaining one positive and the other negative. st(2) will provide the same. so Answer is (B) _________________

How it is B? Did they mention that X and Y are integers? No right, the answer should be E. If they provide details about X and Y as integers then it will be B otherwise it will be E.

can anyone help me about the scenario whether we consider fractions or not in this case?

Scenario:

x=1/2, y=1/3 ==> |1/2-1/3|=1/6 and |1/2|-|1/3|=1/6

Fractions and integers have no role to play here. Check Bunuel's post above.

Whenever xy < 0, i.e. x is negative but y is positive OR x is positive but y is negative, |x-y| is greater than |x|-|y|.

e.g. x = -1/2, y = 1/3 |x-y| = |-1/2-1/3| = 5/6 |x|-|y| = 1/2 - 1/3 = 1/6

So |x - y| > |x|-|y|

Do you see the logic here? If one of x and y is positive and the other is negative, in |x - y|, absolute values of x and y get added and the sum is positive. While in |x|-|y|, the absolute values are subtracted. _________________

How it is B? Did they mention that X and Y are integers? No right, the answer should be E. If they provide details about X and Y as integers then it will be B otherwise it will be E.

can anyone help me about the scenario whether we consider fractions or not in this case?

Scenario:

x=1/2, y=1/3 ==> |1/2-1/3|=1/6 and |1/2|-|1/3|=1/6

It really does not matter; no one is saying that they are integers. The problem with your approach is that you considered invalid values for the fractions.

According to b xy<0; so either x or y must be -ve. Now, lets put the valid values as x=1/2 and y=-1/3; In LHS we get |1/2+1/3|=5/6 and in RHS we get 1/6; therefore the inequality holds, hence statement b is sufficient. _________________

--It's one thing to get defeated, but another to accept it.

First of all we need to consider different cases to solve this problem. take option 1) y<x this option can be subdivided into two blocks...when both are x,y>0 and x>y. lets take x=2, y=1 lx-yl = l2 -1l = 1 Right hand side of the equation = lxl - lyl = l2l - l1l = 1....so equation is invalid. lets take another example when x= 1 and y = -1... lx-yl = l1 - (-1)l = 2 and Rgiht hand side = 0 which make our equation valid....hence we cannot conclude anything from this option.

take option 2) xy<0 under this option there can be two cases....a) x>0 and y<0 (b) x<0 and y>0 lets take a) and use some values.... x=2 and y = -1... simplifing the equation we get...lx-yl = 3 where lxl - lyl = 1 it makes equation valid. now take b) x= -2 and y = 1...we get lx-yl = 3 and lxl - lyl = 1 its also satisfy our given equation. so this option is sufficent to answer the given question.

(II) For all real x and y, \(|x - y| \geq |x| - |y|\) \(|x - y| = |x| - |y|\) when (1) x and y have the same sign and x has greater (or equal) absolute value than y (2) y is 0 \(|x - y| > |x| - |y|\) in all other cases

Question: Is |x-y|>|x|-|y|?

We need to establish whether the "equal to" sign can hold or not.

(1) y < x Doesn't tell us whether they have the same sign or opposite. So we don't know whether the equal to sign will hold or greater than. Not sufficient.

(2) xy < 0 Tells us that one of x and y is positive and the other is negative (they do not have same sign). Also tells us that neither x nor y is 0. Hence, the "equal to" sign cannot hold. Sufficient to answer 'Yes'

I have recently started with the prep thus looking for patterns to solve the Absolute Value questions. Wanted to check if the below solution stands valid.

Square both the sides of the question stem |x-y|^2>(|x|-|y|)^2 --- |x-y|^2 = (x-y)^2 x^2+y^2-2xy>x^2+y^2-2|x||y| xy<|x||y|

From 1 y<x - In Sufficient

From 2 xy<0 (Either of them is negative i.e. x +ve y -ve or y +ve or x -ve) Sufficient

gmatclubot

Re: Is |x-y|>|x|-|y|
[#permalink]
11 Jan 2015, 06:48

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