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So if we take the case x = -1, y = 1 Then |x – y| = |-2| = 2 and |x| - |y| = 1 – 1 = 0

Again, if x = 5 , y = -1 Then |x – y| = |6| = 6 and |x| - |y| = 5 – 1 = 4

So both 1 and 2 are insuff.

Combine them -> It is obvious that y < 0 and x > 0, so by adding a negative sign the magnitude increases and on the right side the magnitude will be less as the difference is between two positive numbers (i.e. the modulus values).

e.g. x = 2, y = -5

|x – y| = |7| and |x| - |y| = 2 – 5 = -3

So |x – y| > |x| - |y| Answer - C
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(1) y < x if x=3 and y=2, left hand abs(x-y) = 1, and right hand abs(x) - abs(y) = 1...No. But if x=3 and y=-2, left hand is 5 and right is 1...Yes. INSUFFICIENT.

(2) xy < 0 Let's think the following two cases.

(a) x>0 and y<0 In this case abs(x-y) > abs(x), as in the second plug-in in the discussion of (1) above. So abs(x-y) naturally is greater than abs(x) - abs(y) because abs(x) > abs(x)-abs(y)...Yes.

(b) x<0 and y>0 In this case abs(x-y) = abs(x)+abs(y). Since abs(x) + abs(y) > abs (x) - abs(y), abs(x-y) > abs(x)-abs(y)...Yes.

Re: Is |x - y | > |x | - |y | ? (1) y < x (2) xy < 0 [#permalink]

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06 Mar 2013, 22:36

Answer: B

1) Taking statement (1), if y < x, then there can be two cases - a) y is negative, it can lead to two sub cases -- (i) x is negative ==> as y < x so |y| > |x| ==> |x| - |y| will be < 0, and |x - y| > 0 ==> |x - y| > |x| - |y| (ii) x is positive ==> |x - y| would be sum of absolute value of x and y, essentially |x| + |y| ==> |x - y| > |x| - |y|

Problem statement is true.

b) y is positive ==> x can only be positive ==> |x - y| = |x| - |y|

Problem statement is false.

Since we do not know, whether y is positive or negative we can not conclude from statement 1.

2) Taking statement (2), if xy < 0 ==> two sub cases a) x < 0 and y > 0 ==> |x - y| = |x| + |y| which is greater than |x| - |y| b) x > 0 and y < 0 ==> |x - y| = |x| + |y|, which is again greater than |x| - |y|

Statement (2) is sufficient enough to conclude the problem statement.

So basically the explanation is that find the signs test in the following cases

++ - - + - - +

since statement doesn't 1 doesn't help in any way its insufficient and statement 2 either both positive or both negative when we plug examples its never true so its sufficient?
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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
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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).
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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's implied that it is integers on the GMAT? Is this perception by me correct or completely out of the blue?
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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

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.
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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)
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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

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.
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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.

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