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

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

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

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.

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

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

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.

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 \(y<x<0\).

Following this same approach, shouldn't the following part become (-x-y)?


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?

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?

Can anyone tell me the flaw in my logic

|x - y| > |x| - |y|
Squaring both sides
x^2 +y^2-2xy> x^2 +y^2-2|x||y|
simplifying, xy<|x||y|,

It will only be possible when product of xy is -ve. or xy<0, Hence B.

VeritasPrepKarishma

Is it incorrect to say that question is asking is |x|>|y| ?

Posted from my mobile device

Dear GMATGuruNY

In this question stem, can I square both sides? or is it wrong?

Thanks