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Is |x-y|>|x|-|y|

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Re: Is |x-y|>|x|-|y| [#permalink] New post 08 Mar 2013, 04:56
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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Re: Is |x-y|>|x|-|y| [#permalink] New post 25 Mar 2013, 02:08
mmphf wrote:
Is |x-y|>|x|-|y| ?

(1) y < x
(2) xy < 0



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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Re: Is |x-y|>|x|-|y| [#permalink] New post 25 Mar 2013, 03:56
Expert's post
kancharana wrote:
mmphf wrote:
Is |x-y|>|x|-|y| ?

(1) y < x
(2) xy < 0



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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Re: Is |x-y|>|x|-|y| [#permalink] New post 25 Mar 2013, 04:37
Bunuel wrote:
kancharana wrote:
mmphf wrote:
Is |x-y|>|x|-|y| ?

(1) y < x
(2) xy < 0



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).


Thank you... understood that. :)

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Re: Is |x-y|>|x|-|y| [#permalink] New post 25 Mar 2013, 10:19
kancharana wrote:
mmphf wrote:
Is |x-y|>|x|-|y| ?

(1) y < x
(2) xy < 0



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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Re: Is |x-y|>|x|-|y| [#permalink] New post 25 Mar 2013, 11:39
Expert's post
kancharana wrote:
mmphf wrote:
Is |x-y|>|x|-|y| ?

(1) y < x
(2) xy < 0



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.

B.

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Last edited by mau5 on 05 Apr 2013, 04:02, edited 1 time in total.
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Re: Is |x-y|>|x|-|y| [#permalink] New post 26 Mar 2013, 01:06
Expert's post
tulsa wrote:
kancharana wrote:
mmphf wrote:
Is |x-y|>|x|-|y| ?

(1) y < x
(2) xy < 0



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.

For more check here: math-number-theory-88376.html

Hope it helps.

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COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: Is |x-y|>|x|-|y| [#permalink] New post 04 Apr 2013, 11:18
Bunuel wrote:
mmphf wrote:
Is |x-y|>|x|-|y| ?

(1) y < x
(2) xy < 0


Is |x-y|>|x|-|y|?

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 <=?
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Re: Is |x-y|>|x|-|y| [#permalink] New post 05 Apr 2013, 01:55
Expert's post
margaretgmat wrote:
Bunuel wrote:
mmphf wrote:
Is |x-y|>|x|-|y| ?

(1) y < x
(2) xy < 0


Is |x-y|>|x|-|y|?

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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Re: Is |x-y|>|x|-|y| [#permalink] New post 26 Jun 2013, 07:17
Is |x-y|>|x|-|y| ?

(1) y < x

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=-2, y=-4)
|x-y|>|x|-|y|
(x-y)>(-x)-(-y)
x-y>-x+y
2x>2y
x>y
|-2-(-4)|>|-2|-|-4|
|2|>|2|-|4|
2>-2 TRUE

(x=2, y=-4)
|x-y|>|x|-|y|
(x-y)>x-(-y)
x-y>x+y
0>2y
|x-y|>|x|-|y|
|2-(-4)|>|2|-|-4|
6>-2 TRUE

(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

(B)

(does that make sense?)
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Re: Is |x-y|>|x|-|y| [#permalink] New post 06 Aug 2013, 08:28
Is |x-y|>|x|-|y| ?

(1) y < x
(2) xy < 0

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

B
write, if it is OK
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Re: Is |x-y|>|x|-|y| [#permalink] New post 06 Aug 2013, 13:29
kancharana wrote:
mmphf wrote:
Is |x-y|>|x|-|y| ?

(1) y < x
(2) xy < 0



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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Re: Is |x-y|>|x|-|y| [#permalink] New post 24 Aug 2013, 05:11
kancharana wrote:
mmphf wrote:
Is |x-y|>|x|-|y| ?

(1) y < x
(2) xy < 0



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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Re: Inequlities [#permalink] New post 28 Jun 2014, 19:32
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.

OA (B)
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Re: Is |x-y|>|x|-|y| [#permalink] New post 08 Jul 2014, 03:53
Is this a valid approach to solve this problem?

| X –Y | > |X| - |Y|
Squaring both sides
(X-Y)^2>(|X|-|Y|)^2

X^2-2XY+Y^2>X^2-2|XY|+Y^2

-XY>|XY|

XY<|XY| --> Can be true only for XY < 0.

1 : y > X - Insufficient
2 : XY < 0 -> Sufficient.

Hence, (B).
Re: Is |x-y|>|x|-|y|   [#permalink] 08 Jul 2014, 03:53
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