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

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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 $$y<x<0$$.

(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.

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?
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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:
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 $$y<x<0$$.

(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.

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?

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

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

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1
ManishKM1 wrote:
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.

Given:
|x - y| > |x| - |y|, you cannot square it. You can square only when you know that both sides are positive. Here, the right hand side may not be positive. For example, if x is 2 and y is 5.
3 > -3
Squaring does not work here since you get 9 > 9 which doesn't hold.
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GMAT 1: 750 Q50 V42 Re: Is |x - y| > |x| - |y|? (1) y < x (2) xy < 0  [#permalink]

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VeritasPrepKarishma

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

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

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Mudit27021988 wrote:
VeritasPrepKarishma

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

Posted from my mobile device

For all real x and y, |x – y| >= |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 (i.e. y is not 0 and |x| < |y| or x and y have opposite signs)

So |x – y| > |x| – |y| is not the same as |x| > |y|

Check: https://www.veritasprep.com/blog/2014/0 ... -the-gmat/
https://www.veritasprep.com/blog/2014/0 ... t-part-ii/
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Re: Is |x - y| > |x| - |y|? (1) y < x (2) xy < 0  [#permalink]

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1
Hi Bunuel,

could you check the following approach:

Absolute value property: If $$|x|+|y| > |x+y|$$ then $$xy<0$$

Re-write the original statement: $$|x-y| > |x|-|y|$$ to $$|x-y|+|y| > |x+y-y|$$

According to the property above the final question becomes: $$(x-y)(y) < 0$$? --> Is $$xy < y^2$$ ?

Stm 2: $$xy<0$$ --> sufficient.

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

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Hi Bunuel,

could you check the following approach:

Absolute value property: If $$|x|+|y| > |x+y|$$ then $$xy<0$$

Re-write the original statement: $$|x-y| > |x|-|y|$$ to $$|x-y|+|y| > |x+y-y|$$

According to the property above the final question becomes: $$(x-y)(y) < 0$$? --> Is $$xy < y^2$$ ?

Stm 2: $$xy<0$$ --> sufficient.

Is that a valid approach?

Yes, that's a valid approach. Kudos.
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Re: Is |x - y| > |x| - |y|? (1) y < x (2) xy < 0  [#permalink]

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study wrote:
Is |x - y| > |x| - |y|?

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

For |x - y| we have

(i) if x - y > 0, x > y

Hence we have if x > 0, then y >= 0 or y < 0

if x < 0, then y < 0

&

(ii) if x - y < 0, x < y,

we have if y > 0, then x >=0 or x < 0

if y < 0, then x < 0

Statement 1: y < x , hence we have case 1, however there are multiple possibilities, since we do not know signs of x & y.

Hence Statement 1 is Not Sufficient.

Statement 2:

xy < 0

Hence either x > 0, y < 0 or x < 0 , y > 0,

lets take x > 0, y < 0 case & plug in numbers
x = 1, y = -1, we have for |x - y| > |x| - |y|
LHS = 2
RHS = 0, hence LHS > RHS

lets take the case where x < 0 , y > 0
x = -1, y = 1, we get
LHS = 2
RHS = 0, hence LHS > RHS

Statement 2 is Sufficient.

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

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Dear GMATGuruNY

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

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

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1
Mo2men wrote:
Dear GMATGuruNY

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

Thanks

Squaring both sides of an inequality is a good approach when both sides are guaranteed to be nonnegative.
Here, the right side of the question stem is NOT guaranteed to be nonnegative, so squaring the inequality is not recommended:
If x=2 and y=1, then |x| - |y| > 0
If x=1 and y=2, then |x| - |y| < 0
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Re: Is |x - y| > |x| - |y|? (1) y < x (2) xy < 0  [#permalink]

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study wrote:
Is |x - y| > |x| - |y|?

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

Attachment:
3_DS_Absolute_B.JPG

My Approach during the actual Prep exam:
Try values.
1. y as -3 and x as 2 gives us 5>-1 works.
y as 1 and x as 5 gives us 4>4. Doesn't work

2. y as 1 and x as -5 gives us 6>4. Works.
y as negative is already tested which works too.

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

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

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Top Contributor
study wrote:
Is |x - y| > |x| - |y|?

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

Attachment:
3_DS_Absolute_B.JPG

Target question: Is |x - y| > |x| - |y|?

Statement 1: y < x
Let's test some values.
There are several values of x and y that satisfy statement 1. Here are two:
Case a: x = 2 and y = 1. In this case, |x - y| = |2 - 1| = 1 and |x| - |y| = |2| - |1| = 1. So, the answer to the target question is NO, |x - y| is NOT greater than |x| - |y|
Case b: x = 2 and y = -1. In this case, |x - y| = |2 - (-1)| = 3 and |x| - |y| = |2| - |-1| = 1. So, the answer to the target question is YES, |x - y| IS greater than |x| - |y|
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: xy < 0
This tells us that one value (x or y) is positive, and the other value is negative. This sets up two possible cases:

Case a: x is positive and y is negative.
So, we're taking a positive value (x) and subtracting a negative value (y). Doing so yields a positive value that is bigger than x.
In other words, we have: 0 < x < |x - y|

Now let's examine |x| - |y|
Since x is positive, we know that |x| = x
Since y ≠ 0, we know that 0 < |y|
So, |x| - |y| = x - |y| = some number less than x
In other words, |x| - |y| < x

When we combine the inequalities we get: |x| - |y| < x < |x - y|
In this case, the answer to the target question is YES, |x - y| IS greater than |x| - |y|

Case b: x is negative and y is positive.
Here, we're taking a negative value (x) and subtracting a positive value (y). Doing so yields a negative value that is less than x.
In other words, we have: x - y < x < 0 < |x|
Important: since the MAGNITUDE of x - y is greater than the MAGNITUDE of x, we can write: |x| < |x - y|

Now let's examine |x| - |y|
Since x ≠ 0, we know that 0 < |x|
Since |x| is a positive number, we know that subtracting |y| (another positive value) will yield a number that is LESS THAN |x|
In other word, |x| - |y| < |x|

When we combine the inequalities we get: |x| - |y| < |x| < |x - y|
In this case, the answer to the target question is YES, |x - y| IS greater than |x| - |y|

In both possible cases, the answer to the target question is the same: YES, |x - y| IS greater than |x| - |y|
Since we can answer the target question with certainty, the combined statements are SUFFICIENT

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

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

|x|= the distance between x and 0 = the RED segment on the number lines below.
|y| = the distance between y and 0 = the BLUE segment on the number lines below.
|x-y| = the distance BETWEEN X AND Y.

Statement 1: y < x
Case 1: |x| - |y| = RED - BLUE.
|x-y| = RED - BLUE.
Thus, |x-y| = |x| - |y|.

Case 2: |x| - |y| = RED - BLUE.
|x-y| = RED + BLUE.
Thus, |x-y| > |x| - |y|.
INSUFFICIENT.

Statement 2: xy < 0
Since x and y have different signs, they are on OPPOSITE SIDES OF 0. In each case:
|x| - |y| = RED - BLUE.
|x-y| = RED + BLUE.
Thus, |x-y| > |x| - |y|.
SUFFICIENT.

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