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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.

Funny, but I remember form university that |a-b|>||a|-|b||>|a|-|b|, therefore the above inequality is valid for all numbers a,b can somebody verify the inequality?

This is a very simple question. !x-y!>|x|-|y| can only happen if both the numbers are of different signs.

If xy<0 then these numbers are of opposite signs. Hope this clears.

X=2 y=3 then |x-y|=|x|-|y| if x=-2 and y = 3 then |x-y|>|x|-|y|

Red part is not correct \(|x-y|>{|x|-|y|}\) also holds true when \(x\) and \(y\) have the same sign and the magnitude of \(y\) is more than that of \(x\) (so for \(|y|>|x|\)). Example: \(x=2\) and \(y=3\) --> \(|x-y|=1>-1={|x|-|y|}\); \(x=-2\) and \(y=-3\) --> \(|x-y|=1>-1={|x|-|y|}\).

Actually the only case when \(|x-y|>{|x|-|y|}\) does not hold true is when \(xy>{0}\) (so when \(x\) and \(y\) have the same sign) and \(|x|>|y|\) (simultaneously). In this case \(|x-y|={|x|-|y|}\) (as shown in my previous post). Example: \(x=3\) and \(y=2\) --> \(|x-y|=1={|x|-|y|}\); \(x=-3\) and \(y=-2\) --> \(|x-y|=1={|x|-|y|}\).

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

But in both the examples, its being shown that case 2 is sufficient. Am i mistaken?

|x-y| represents the distance between x and y on the number line.

|x|-|y|, on the other hand, first takes the absolute value of both numbers - and thereby moving them both to the positive side of the number line - and THEN calculates the difference between x and y

Visually, it makes sense that if x and y are of different signs (for example, x=-5, y=5), then the difference between the two numbers on a number line is greater if measured before moving them both to the positive side of the number line.

At this point I logically deduced that it is impossible for |x-y| to be less than |x|-|y|. I also deduced at this point that if x and y have the same sign, it does not matter when the absolute value is taken because the difference between them will be the same either way.

After this thought process, the problem becomes much easier.

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

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. _________________

(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'

This is a very simple question. !x-y!>|x|-|y| can only happen if both the numbers are of different signs.

If xy<0 then these numbers are of opposite signs. Hope this clears.

X=2 y=3 then |x-y|=|x|-|y| if x=-2 and y = 3 then |x-y|>|x|-|y|

Red part is not correct \(|x-y|>{|x|-|y|}\) also holds true when \(x\) and \(y\) have the same sign and the magnitude of \(y\) is more than that of \(x\) (so for \(|y|>|x|\)). Example: \(x=2\) and \(y=3\) --> \(|x-y|=1>-1={|x|-|y|}\); \(x=-2\) and \(y=-3\) --> \(|x-y|=1>-1={|x|-|y|}\).

Actually the only case when \(|x-y|>{|x|-|y|}\) does not hold true is when \(xy>{0}\) (so when \(x\) and \(y\) have the same sign) and \(|x|>|y|\) (simultaneously). In this case \(|x-y|={|x|-|y|}\) (as shown in my previous post). Example: \(x=3\) and \(y=2\) --> \(|x-y|=1={|x|-|y|}\); \(x=-3\) and \(y=-2\) --> \(|x-y|=1={|x|-|y|}\).

This post has left me confused. Earlier for these questions I used to plug in values and got to know if the inequality is valid or not.

As per the algebraic method, this inequality stands when

1. If x and y have different signs (got this by the plug in method) 2. If x and y have same signs but y>x (above post)

So now, we need to look out for a scenario when x and y are positive and y>x. So had the second statement been xy>0 then the answer would have been C?

Yes. If the 2nd statement had mentiond that xy>0 ---> this would mean that both x and y are of the same sign but without knowing whether y>x , this statement would not have been sufficient.

With statement 1, y<x and statement 2 (new) xy>0, you would have been able to answer the question unambiguously, leading to C. _________________

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.

My Answer is E. Lets check this with pluy and play method. Consider - x= 5 and y = 2 -> 3 > 3 x= 2 and y = -2 -> 4 > 0 A not sufficient.

Consider - x= 5 and y = 2 -> 3 > 3 x= -2 and y = -5 -> 3 > -3 B not sufficient.

Answer is E. Cheers! _________________

----------------------------------------------------------------------------------------- What you do TODAY is important because you're exchanging a day of your life for it! -----------------------------------------------------------------------------------------

Funny, but I remember form university that |a-b|>||a|-|b||>|a|-|b|, therefore the above inequality is valid for all numbers a,b can somebody verify the inequality?

This is a very simple question. !x-y!>|x|-|y| can only happen if both the numbers are of different signs.

If xy<0 then these numbers are of opposite signs. Hope this clears.

X=2 y=3 then |x-y|=|x|-|y| if x=-2 and y = 3 then |x-y|>|x|-|y|

Red part is not correct \(|x-y|>{|x|-|y|}\) also holds true when \(x\) and \(y\) have the same sign and the magnitude of \(y\) is more than that of \(x\) (so for \(|y|>|x|\)). Example: \(x=2\) and \(y=3\) --> \(|x-y|=1>-1={|x|-|y|}\); \(x=-2\) and \(y=-3\) --> \(|x-y|=1>-1={|x|-|y|}\).

Actually the only case when \(|x-y|>{|x|-|y|}\) does not hold true is when \(xy>{0}\) (so when \(x\) and \(y\) have the same sign) and \(|x|>|y|\) (simultaneously). In this case \(|x-y|={|x|-|y|}\) (as shown in my previous post). Example: \(x=3\) and \(y=2\) --> \(|x-y|=1={|x|-|y|}\); \(x=-3\) and \(y=-2\) --> \(|x-y|=1={|x|-|y|}\).

Hope it's clear.

thanks. I missed that

gmatclubot

Re: Is |x-y|>|x|-|y|
[#permalink]
14 Jan 2012, 14:52

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