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Is |x - y| > |x| - |y|? (1) y < x (2) xy < 0

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

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New post 27 Mar 2016, 20:53
Only xy<1 is sufficient to answer the question.
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Re: Is |x - y| > |x| - |y|? (1) y < x (2) xy < 0 [#permalink]

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New post 13 Aug 2016, 11:38
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Is |x - y | > |x | - |y | ?
(1) y < x
(2) xy < 0

Consider 1) y < x, let \(x = 1 , y = 0\)
then | x - y | = |1 - 0| = 1
| x | - | y | = |1| - |0| = 1 and 1 > 1 ; Therefore False
let \(x = 1 , y = -1\)
then | x - y | = 2
| x | - | y | = 0 and 2 > 0 ; Therefore True
Eliminate A and D, since it's inconsistent

Consider 2) xy < 0 , let \(x = 1 , y = -1\)
then | x - y | = |1 - -1| = 2
| x | - | y | = |1| - |1| = 0 and 2 > 1 ; Therefore True
let \(x = -1 , y = 1\)
then | x - y | = 2
| x | - | y | = 0 and 2 > 0 ; Therefore True
Consistent hence B
PS: The inequality need not be True or False, it just need to be consistent.

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

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New post 13 Aug 2016, 11:46
Is |x - y | > |x | - |y | ?
(1) y < x
(2) xy < 0


1) Y < X;
when x are y are negative, x - y is greater than 0, so |x-y| is also greater than 0 whereas |x| < |y| (as the signs are flipped) and |x| - |y| is less than 0. So, |x-y| > |x| - |y|

When X and Y are positive, |x-y| = x-y. And |x| = x and |y| = y, so |x| - |y| = x - y. So, |x-y| = |x| - |y| = x-y. So, this statement is not sufficient.

2) xy< 0. This statement tells us that x and y have opposite signs. If x < 0, then y >0. for e.g. x =-3, y=4; |x-y| = |-3-4| = 7 and |x| - |y| = 3 -4 = -1
If x > 0 and y < 0, for e.g. x = 5, y =-2; |x-y| = |5 - (-2)| = 7, whereas |x| - |y| = |5| - |-2| = 3, so, |x-y| > |x| - |y|.

This statement is sufficient.

My answer is B.

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

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New post 13 Aug 2016, 22:48
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


Even if you take the fractions, we will always get the above equation satisfied.

Please Note that xy<0 means either x or y is -ve but not both.

So, as per the scenario you have listed, we if x =1/2, y needs to be a -ve integer/fraction.

Hence, try substituting the -ve and +ve combination in LHS and RHS, you will get the equation satisfied.
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Re: Is |x - y| > |x| - |y|? (1) y < x (2) xy < 0 [#permalink]

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New post 17 Aug 2016, 05:59
Some basic inequality rules
|x-y|>=|x|-|y|
|x|+|y|>=|X+Y|

st1: y<x
substitute x=1,y=-3
implies 4>-3
substitute x=3;y=1
implies 2!>2
st1 N.S
st2:xy<0
substitute x=1,y=-3
implies 4>-3
substitute x=-1,y=3
implies 4>-3
substitute x=-1,y=1
2>0
the results are consistent.
Hence B

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

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New post 30 Oct 2016, 21:26
B is the correct answer

Lets do for A. Take following values:

x= 5, y=3.......arg is FALSE
x= 3, y=-1....arg is TRUE
x=0, y=-4....arg is TRUE

As, the argument falls apart and gives two different inference, so, A is OUT

Lets do for B with following values:

x= -5, y=3.......arg is TRUE
x= 3, y=-1....arg is TRUE

So B hold true. No need to check other options

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

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New post 05 Nov 2016, 07:27
mmphf wrote:
Is |x-y|>|x|-|y| ?

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



My way is number Line way

Just look at the problem and see Is |x-y|>|x|-|y| ?

MOD difference between X and Y on a number line has to be more than the difference of MOD X and MOD Y

This is only possible when X and Y have different signs. Lets try

X = -3 and Y =5 | -3 + 5 | > |-3| - |5| => 2> -2

Put in any values of X and Y with different polarities , It will satisfy the equation.


All the best.

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

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New post 23 Nov 2016, 02:21
Squaring both sides
x^2 + y^2 - 2xy > x^2 +y^2 - 2 |x||y| => xy<|x||y|
RHS is always +ve
2) xy<0
so inequality will hold true inevery case
hence answer is B

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

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New post 18 Dec 2016, 08:29
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.


Bunuel, allow me to ask you -

1. if the Option (2) said that |x|<|y| - even then we would have our answer right? because we know that it does not meet a part of the '=' conditon for |x-y|>|x|+|y|??

2. Stepping back a little- in option (2) we get to know that the '=' condition does not hold true, however how do we conclude from that that |x-y|>|x|-|y|?

Would appreciate your help. Thank you.

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

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New post 24 Dec 2016, 01:21
study wrote:
Is |x - y| > |x| - |y|?

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


the situation |x - y| > |x| - |y| is only possible if x and y have opposite signs i.e
if x>0 then y<0 therefore |x-(-y)|= |x+y| = x+y
or if x<0 and y>0 therefore |-x-y|= |-(x+y)| = x+y

Option B because xy<0 satisfies the equation. Option A doesn't tell us whether x, y have opposite signs

put that kudos if you like the response! 8-)

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

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New post 12 Feb 2017, 16:12
Bunuel wrote:
LM wrote:
Please tell the quick approach.... it took me loner than I should have taken....


Algebraic approach is given in my first post. Below is another approach:

\(|x-y|>|x|-|y|\)?

(1) \(y<x\)

Try two positive number \(x=3>y=1\) --> is \(|3-1|>|3|-|1|\)? --> is \(2>2\)? Answer NO.

Try ANY other case but both positive: \(x=-5>y=-7\) --> is \(|-5-(-7)|>|-5|-|-7|\)? --> is \(2>-2\)? Answer YES.

Two different answers. Not sufficient.

(2) \(xy<0\), means \(x\) and \(y\) have different signs.

Now we can spot here that when \(x\) and \(y\) have different signs \(x-y\) always "contribute" to each other so that its absolute value will increase: \(x=3\), \(y=-1\) --> \(|x-y|=|3+1|=|4|=4\) or \(x=-3\), \(y=1\) --> \(|x-y|=|-3-1|=|4|=4\).

But \(|x|-|y|\) is difference (thus not "contributing") of two positive values (as neither equals to zero). \(x=3\), \(y=-1\) --> \(|x|-|y|=|3|-|-1|=2\) or \(x=-3\), \(y=1\) --> \(|x|-|y|=|-3|-|1|=2\).

So \(xy<0\) means \(|x-y|>|x|-|y|\) is always true.

Sufficient.

Answer: B.

Hope it helps.


This graphical approach will help to verify those value.
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Re: Is |x - y| > |x| - |y|? (1) y < x (2) xy < 0 [#permalink]

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New post 12 Mar 2017, 19:53
study wrote:
Is \(|x - y| > |x| - |y|\)?

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


Explanation from RON (Manhattan)

Statement (1)
let's try x = 2, y = 1
is |2 - 1| > |2| - |1| ?
no.
let's try x = -1, y = -2
is |-1 - (-2)| > |-1| - |-2| ?
yes.
Insufficient.

Statement (2)

this means we have to pick OPPOSITE SIGNS. therefore, there are basically 6 cases to try:
* x is negative "bigger", y is positive "smaller"
* x is negative, y is positive, same magnitude
* x is negative "smaller", y is positive "bigger"
* x is positive "bigger", y is negative "smaller"
* x is positive, y is negative, same magnitude
* x is positive "smaller", y is negative "bigger"

If you try all of these --

-2, 1 --> is |-3| > 2 - 1? YES
-1, 1 --> is |-2| > 0? YES
-1, 2 --> is |-3| > 1 - 2? YES
2, -1 --> is |3| > 2 - 1? YES
1, -1 --> is |2| > 0? YES
1, -2 --> is |3| > 1 - 2? YES

the pattern is pretty clear: in each of these cases, the two numbers' magnitudes are working together on the left, but against each other on the right. therefore, the left-hand side is always going to be bigger.
Sufficient.
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Is |x - y| > |x| - |y|? (1) y < x (2) xy < 0 [#permalink]

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New post 15 Dec 2017, 21:52
Bunuel wrote:
Basically the question asks whether the distance between the two points x and y on the line is greater than the difference between the individual distances of x and y from 0.

\(|x-y|>|x|-|y|\)?

(1) \(y<x\), 3 possible cases for \(|x-y|>|x|-|y|\):

A. ---------------\(0\)---\(y\)---\(x\)---, \(0<y<x\) --> in this case \(|x-y|>|x|-|y|\) becomes: \(x-y>x-y\) --> \(0>0\). Which is wrong;
B. ---------\(y\)---\(0\)---------\(x\)---, \(y<0<x\) --> in this case \(|x-y|>|x|-|y|\) becomes: \(x-y>x+y\) --> \(y<0\). Which is right, as we consider the range \(y<0<x\);
C. ---\(y\)---\(x\)---\(0\)--------------, \(y<x<0\) --> in this case \(|x-y|>|x|-|y|\) becomes: \(x-y>-x+y\) --> \(x>y\). Which is right, as we consider the range \(y<x<0\).

Two different answers. Not sufficient.

(2) \(xy<0\), means \(x\) and \(y\) have different signs, hence 2 cases for \(|x-y|>|x|-|y|\):

A. ----\(y\)-----\(0\)-------\(x\)---, \(y<0<x\) --> in this case \(|x-y|>|x|-|y|\) becomes: \(x-y>x+y\) --> \(y<0\). Which is right, as we consider the range \(y<0<x\);
B. ----\(x\)-----\(0\)-------\(y\)---, \(x<0<y\) --> in this case \(|x-y|>|x|-|y|\) becomes: \(-x+y>-x-y\) --> \(y>0\). Which is right, as we consider the range \(x<0<y\).

In both cases inequality holds true. Sufficient.

Answer: B.

Hi Bunuel,

Tried hard to understand the following related to the red part above but at the end no success. Would you please clear my doubt?

In 1 B & C, you have changed the sign of x and y in RHS, and not modified the LHS(red part) according to the sign of x and y.---- Understood the explanation of st. 1
But in 2 B, You have modified also the LHS part according to the sign of x and y. (red part) ------- Why have you not followed the process that you have followed in st 1 ?

What did this difference in procedure depend on?
Thank You in advance.
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Re: Is |x - y| > |x| - |y|? (1) y < x (2) xy < 0 [#permalink]

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New post 15 Dec 2017, 23:05
TaN1213 wrote:
Bunuel wrote:
Basically the question asks whether the distance between the two points x and y on the line is greater than the difference between the individual distances of x and y from 0.

\(|x-y|>|x|-|y|\)?

(1) \(y<x\), 3 possible cases for \(|x-y|>|x|-|y|\):

A. ---------------\(0\)---\(y\)---\(x\)---, \(0<y<x\) --> in this case \(|x-y|>|x|-|y|\) becomes: \(x-y>x-y\) --> \(0>0\). Which is wrong;
B. ---------\(y\)---\(0\)---------\(x\)---, \(y<0<x\) --> in this case \(|x-y|>|x|-|y|\) becomes: \(x-y>x+y\) --> \(y<0\). Which is right, as we consider the range \(y<0<x\);
C. ---\(y\)---\(x\)---\(0\)--------------, \(y<x<0\) --> in this case \(|x-y|>|x|-|y|\) becomes: \(x-y>-x+y\) --> \(x>y\). Which is right, as we consider the range \(y<x<0\).

Two different answers. Not sufficient.

(2) \(xy<0\), means \(x\) and \(y\) have different signs, hence 2 cases for \(|x-y|>|x|-|y|\):

A. ----\(y\)-----\(0\)-------\(x\)---, \(y<0<x\) --> in this case \(|x-y|>|x|-|y|\) becomes: \(x-y>x+y\) --> \(y<0\). Which is right, as we consider the range \(y<0<x\);
B. ----\(x\)-----\(0\)-------\(y\)---, \(x<0<y\) --> in this case \(|x-y|>|x|-|y|\) becomes: \(-x+y>-x-y\) --> \(y>0\). Which is right, as we consider the range \(x<0<y\).

In both cases inequality holds true. Sufficient.

Answer: B.

Hi Bunuel,

Tried hard to understand the following related to the red part above but at the end no success. Would you please clear my doubt?

In 1 B & C, you have changed the sign of x and y in RHS, and not modified the LHS(red part) according to the sign of x and y.---- Understood the explanation of st. 1
But in 2 B, You have modified also the LHS part according to the sign of x and y. (red part) ------- Why have you not followed the process that you have followed in st 1 ?

What did this difference in procedure depend on?
Thank You in advance.


When \(x \le 0\) then \(|x|=-x\), or more generally when \(\text{some expression} \le 0\) then \(|\text{some expression}| = -(\text{some expression})\). For example: \(|-5|=5=-(-5)\);

When \(x \ge 0\) then \(|x|=x\), or more generally when \(\text{some expression} \ge 0\) then \(|\text{some expression}| = \text{some expression}\). For example: \(|5|=5\)



(2) \(xy<0\), means \(x\) and \(y\) have different signs, hence 2 cases for \(|x-y|>|x|-|y|\):

B. ----\(x\)-----\(0\)-------\(y\)---.

This means that \(x<0<y\). So, \(x - y < 0\), \(x<0\) and \(y>0\). According to the properties above if \(x - y < 0\), then \(|x-y|=-(x-y)\), if \(x<0\), then \(|x|=-x\) and if \(y>0\), then \(|y|=y\).

Hope it's clear.
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Re: Is |x - y| > |x| - |y|? (1) y < x (2) xy < 0 [#permalink]

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New post 16 Dec 2017, 09:56
Bunuel wrote:
When \(x \le 0\) then \(|x|=-x\), or more generally when \(\text{some expression} \le 0\) then \(|\text{some expression}| = -(\text{some expression})\). For example: \(|-5|=5=-(-5)\);

When \(x \ge 0\) then \(|x|=x\), or more generally when \(\text{some expression} \ge 0\) then \(|\text{some expression}| = \text{some expression}\). For example: \(|5|=5\)



(2) \(xy<0\), means \(x\) and \(y\) have different signs, hence 2 cases for \(|x-y|>|x|-|y|\):

B. ----\(x\)-----\(0\)-------\(y\)---.

This means that \(x<0<y\). So, \(x - y < 0\), \(x<0\) and \(y>0\). According to the properties above if \(x - y < 0\), then \(|x-y|=-(x-y)\), if \(x<0\), then \(|x|=-x\) and if \(y>0\), then \(|y|=y\).

Hope it's clear.

Following this same approach, shouldn't the following part become (-x-y)?
Quote:
1C. ---\(y\)---\(x\)---\(0\)--------------, \(y<x<0\) --> in this case \(|x-y|>|x|-|y|\) becomes: \(x-y>-x+y\) --> \(x>y\). Which is right, as we consider the range \(y<x<0\).


in 2 B you have modified the \(x-y\) to -(x-y) because x is negative and y is positive. So in the same way the 1C should take (-x-y) as per the sign of x.
What am I missing?
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Re: Is |x - y| > |x| - |y|? (1) y < x (2) xy < 0 [#permalink]

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TaN1213 wrote:
Bunuel wrote:
When \(x \le 0\) then \(|x|=-x\), or more generally when \(\text{some expression} \le 0\) then \(|\text{some expression}| = -(\text{some expression})\). For example: \(|-5|=5=-(-5)\);

When \(x \ge 0\) then \(|x|=x\), or more generally when \(\text{some expression} \ge 0\) then \(|\text{some expression}| = \text{some expression}\). For example: \(|5|=5\)



(2) \(xy<0\), means \(x\) and \(y\) have different signs, hence 2 cases for \(|x-y|>|x|-|y|\):

B. ----\(x\)-----\(0\)-------\(y\)---.

This means that \(x<0<y\). So, \(x - y < 0\), \(x<0\) and \(y>0\). According to the properties above if \(x - y < 0\), then \(|x-y|=-(x-y)\), if \(x<0\), then \(|x|=-x\) and if \(y>0\), then \(|y|=y\).

Hope it's clear.

Following this same approach, shouldn't the following part become (-x-y)?
Quote:
1C. ---\(y\)---\(x\)---\(0\)--------------, \(y<x<0\) --> in this case \(|x-y|>|x|-|y|\) becomes: \(x-y>-x+y\) --> \(x>y\). Which is right, as we consider the range \(y<x<0\).


in 2 B you have modified the \(x-y\) to -(x-y) because x is negative and y is positive. So in the same way the 1C should take (-x-y) as per the sign of x.
What am I missing?


C. ---\(y\)---\(x\)---\(0\)--------------, \(y<x<0\):

x - y > 0, so |x - y| = x - y;
y < 0, so |y| = y;
x < 0, so |x| = x.

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

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New post 16 Dec 2017, 19:31
Bunuel wrote:
TaN1213 wrote:
Bunuel wrote:
When \(x \le 0\) then \(|x|=-x\), or more generally when \(\text{some expression} \le 0\) then \(|\text{some expression}| = -(\text{some expression})\). For example: \(|-5|=5=-(-5)\);

When \(x \ge 0\) then \(|x|=x\), or more generally when \(\text{some expression} \ge 0\) then \(|\text{some expression}| = \text{some expression}\). For example: \(|5|=5\)



(2) \(xy<0\), means \(x\) and \(y\) have different signs, hence 2 cases for \(|x-y|>|x|-|y|\):

B. ----\(x\)-----\(0\)-------\(y\)---.

This means that \(x<0<y\). So, \(x - y < 0\), \(x<0\) and \(y>0\). According to the properties above if \(x - y < 0\), then \(|x-y|=-(x-y)\), if \(x<0\), then \(|x|=-x\) and if \(y>0\), then \(|y|=y\).

Hope it's clear.

Following this same approach, shouldn't the following part become (-x-y)?
Quote:
1C. ---\(y\)---\(x\)---\(0\)--------------, \(y<x<0\) --> in this case \(|x-y|>|x|-|y|\) becomes: \(x-y>-x+y\) --> \(x>y\). Which is right, as we consider the range \(y<x<0\).


in 2 B you have modified the \(x-y\) to -(x-y) because x is negative and y is positive. So in the same way the 1C should take (-x-y) as per the sign of x.
What am I missing?


C. ---\(y\)---\(x\)---\(0\)--------------, \(y<x<0\):

x - y > 0, so |x - y| = x - y;
y < 0, so |y| = y;
x < 0, so |x| = x.

\(|x-y|>|x|-|y|\) --> x - y > -x - (-y) --> x - y > -x + y

Finally got it. Thanks a lot for being so patient with me.
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Re: Is |x - y| > |x| - |y|? (1) y < x (2) xy < 0 [#permalink]

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New post 17 Dec 2017, 01:16
I am a little bit concerned here.
If the question states x and y, can we make an assumption that x ≠ y?
For example here, without this assumption, i can let x = y and (2) is NS?

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

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New post 17 Dec 2017, 01:22
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khiemchii wrote:
I am a little bit concerned here.
If the question states x and y, can we make an assumption that x ≠ y?
For example here, without this assumption, i can let x = y and (2) is NS?


Unless it is explicitly stated otherwise, different variables CAN represent the same number.

For (2) though x = y is not possible because in this case we'd get x^2 < 0, which is not true for any real number, so x = y does not satisfy xy < 0 and therefore should not be considered for the second statement. Similarly, for (1) it's give that y < x, so x = y cannot be true.
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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|? (1) y < x (2) xy < 0 [#permalink]

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New post 23 Dec 2017, 11:08
study wrote:
Is |x - y| > |x| - |y|?

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


[Reveal] Spoiler:
Attachment:
3_DS_Absolute_B.JPG


Is |x - y| > |x| - |y|
if we do some work on this statement then we find that
1)IF X=2 Y=6 , THEN LHS= 4 AND RHS =-4 ((( X,Y ARE POSITIVE ---- SAME SIGN AND X<Y)))
CASE HOLDS HERE

2)IF X=-2 , Y=-6 THEN LHS=4 AND RHS =-4 AGAIN ((((X,Y ARE BOTH NEGATIVE ---- SAME SIGN
AND X>Y )))
CASE HOLDS HERE AGAIN.

3) IF X AND Y ARE OPPOSITE SIGNS , LET X=-2 AND Y=6 THEN LHS=8 AND RHS =-4 CASE HOLDS

AND LET X=6 AND Y=-2 THEN LHS =8 AND RHS 4 CASE HOLDS AGAIN
therefore for following three cases the given above inequality Is |x - y| > |x| - |y|
holds

ST 1
y<x we dont know whether x and y have same signs or not insufficient
st 2 xy<0
here both have opposite signs case 3 holds hence sufficient
ANS =b

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Re: Is |x - y| > |x| - |y|? (1) y < x (2) xy < 0   [#permalink] 23 Dec 2017, 11:08

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