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

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13 Aug 2016, 11:38

1

This post received KUDOS

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.

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

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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|.

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

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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.
_________________

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

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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

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

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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|?

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

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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

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

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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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_________________

"Be challenged at EVERY MOMENT."

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"Each stage of the journey is crucial to attaining new heights of knowledge."

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

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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.
_________________

"Be challenged at EVERY MOMENT."

“Strength doesn’t come from what you can do. It comes from overcoming the things you once thought you couldn’t.”

"Each stage of the journey is crucial to attaining new heights of knowledge."

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

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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 \(x<0<y\).

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.
_________________

------------------------------ "Trust the timing of your life" Hit Kudus if this has helped you get closer to your goal, and also to assist others save time. Tq

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 \(x<0<y\).

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 \(|y|=y\).

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

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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?
_________________

------------------------------ "Trust the timing of your life" Hit Kudus if this has helped you get closer to your goal, and also to assist others save time. Tq

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
_________________

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

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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.
_________________

------------------------------ "Trust the timing of your life" Hit Kudus if this has helped you get closer to your goal, and also to assist others save time. Tq

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

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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?

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.
_________________

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