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655-705 (Hard)|   Absolute Values|         
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GMAT Club World Cup 2026 (Day 2): If |x - y| = y - x, x y, and y < 0 08 Jul 2026, 05:27
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IMO : Option B
As per the question |x-y| =y-x and y<0
and since mod of anything is always positive, hence |x-y| >0 , which means y-x >0
but since y <0 then for y-x to be positive x has to be negative since -(-x) = +x and also that |x| >|y|
now the question, in order to solve it lets break it into four different parts :
1. x/|x| = -1 since x<0
2. |-y|/-y = +1 since y<0 and |y| >0 and -y >0
3. x-y/|y-x| = -1 since x-y <0 and |y-x|>0
4. |xy| /xy = 1 since both |xy| and xy will be >0
now getting the total all the 4 parts ===> -1+1-1+1 = 0
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GMAT Club World Cup 2026 (Day 2): If |x - y| = y - x, x y, and y < 0 08 Jul 2026, 05:35
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|x-y| = y-x --> x-y<0 --> x<y
Also given, y<0 & x = y
So we get x<y<0 and |x|>|y|

Now,
x/|x| + |-y|/-y + x-y/|y-x| + |xy|/xy = -1 + 1 + -1 + 1 = 0

Option B


Bunuel
If \(|x-y|=y-x\), \(x \ne y\), and \(y<0\), what is the value of \(\frac{x}{|x|}+\frac{|-y|}{-y}+\frac{x-y}{|y-x|}+\frac{|xy|}{xy}\)?

A. -2
B. 0
C. 1
D. 2
E. 4


 


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GMAT Club World Cup 2026 (Day 2): If |x - y| = y - x, x y, and y < 0 08 Jul 2026, 06:35
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Taking the expression as sum of 4 values. The first and the last values sum to +2 or -2 depending on x>0 or x<0. The second term value is 1 so it comes out to be (2/-2) + 1 - (y-x)/|y-x|

If x>0 => 2 + 1 + 1 => 4
if x<0 => -2 + 1 -1 => -2

Not sure between 4 and -2 so I would choose -2 scenario with x<0 no specific reason.
Bunuel
If \(|x-y|=y-x\), \(x \ne y\), and \(y<0\), what is the value of \(\frac{x}{|x|}+\frac{|-y|}{-y}+\frac{x-y}{|y-x|}+\frac{|xy|}{xy}\)?

A. -2
B. 0
C. 1
D. 2
E. 4


 


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GMAT Club World Cup 2026 (Day 2): If |x - y| = y - x, x y, and y < 0 08 Jul 2026, 06:37
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Bunuel
If \(|x-y|=y-x\), \(x \ne y\), and \(y<0\), what is the value of \(\frac{x}{|x|}+\frac{|-y|}{-y}+\frac{x-y}{|y-x|}+\frac{|xy|}{xy}\)?

A. -2
B. 0
C. 1
D. 2
E. 4


 


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Since both x and y < 0, the equation simplifies to: 1-1+1+1=2. D.
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akvirginkar
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GMAT Club World Cup 2026 (Day 2): If |x - y| = y - x, x y, and y < 0 08 Jul 2026, 06:44
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|x-y|=y-x; from that we can tell that x<y (|x|=x if x>=0, and -x if x<0; from that logic, substituting 'x' as 'x-y' we get x<y). Now, let's split the expression into its constituent parts.
i) x/|x| - we know that x is negative, since it's given that y is negative and x is less than y. The mod expression always returns a positive number (or zero if x = 0, which is not the case here). So, this will give us the answer -1.
ii) |-y|/-y - since y is negative, following the same logic as above, we get 1 as the answer.
iii) x-y/|y-x| - since we've established x<y, the numerator is negative and the denominator is positive, giving us -1.
iv) |xy|/xy - product of two negatives is positive, so denominator and numerator both positive, giving us 1.
Adding all four taken together we get -1+1-1+1 which is zero.

The correct answer is B. 0
Bunuel
If \(|x-y|=y-x\), \(x \ne y\), and \(y<0\), what is the value of \(\frac{x}{|x|}+\frac{|-y|}{-y}+\frac{x-y}{|y-x|}+\frac{|xy|}{xy}\)?

A. -2
B. 0
C. 1
D. 2
E. 4


 


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GMAT Club World Cup 2026 (Day 2): If |x - y| = y - x, x y, and y < 0 08 Jul 2026, 06:47
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It is easier to check with some numbers in these type of questions. Lets consider x=-2, y=-1 which satisfies the given condition.

Putting in values, we get
first term -2/2 = -1
second term 1/1 = 1
third term -2 - (-1) / 1 = -1
fourth term 2/2 = 1

Adding these, -1+1+-1+1 = 0

Therefore the answer is 0
Bunuel
If \(|x-y|=y-x\), \(x \ne y\), and \(y<0\), what is the value of \(\frac{x}{|x|}+\frac{|-y|}{-y}+\frac{x-y}{|y-x|}+\frac{|xy|}{xy}\)?

A. -2
B. 0
C. 1
D. 2
E. 4


 


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GMAT Club World Cup 2026 (Day 2): If |x - y| = y - x, x y, and y < 0 08 Jul 2026, 07:11
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Taking any two values which satisfy the equation, we can take {x,y} = {-1, -2}, for the first equation it becomes | -2 - (-1)| = -1 - (-2)....

Substitute the above values in the second equation,

(-2/2) + (1/1) + (-2 - (-1)) / | -1 - (-2)|) + | (-2)(-1) / (-1)(-2)

Term 1 = -1, term 2 = 1, term 3 = -1, term 4 = 1, all cancel out each other.

we will get 0.

IMO, it's C
Bunuel
If \(|x-y|=y-x\), \(x \ne y\), and \(y<0\), what is the value of \(\frac{x}{|x|}+\frac{|-y|}{-y}+\frac{x-y}{|y-x|}+\frac{|xy|}{xy}\)?

A. -2
B. 0
C. 1
D. 2
E. 4


 


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GMAT Club World Cup 2026 (Day 2): If |x - y| = y - x, x y, and y < 0 08 Jul 2026, 07:45
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|x−y|=y−x, so x-y<0, x<y and y<0
So both x and y are negative numbers

Approach 1: algebre

\(x/|x|\)+\(|−y|/−y\)+\(x−y/|y−x|\)+\(|xy|/xy\)
\(x/-x\) + \(-y/-y\) + \(x-y/y-x\) + \(xy/xy\)
-1+1-1+1=0

Approach 2: take x= -2 y= -1

\(-2/2\) + \(1/1\) + \(-1/1\) + \(2/2 \)= -1+1-1+1 = 0

ans B
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GMAT Club World Cup 2026 (Day 2): If |x - y| = y - x, x y, and y < 0 08 Jul 2026, 07:51
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If you are good doing experiment with numbers, then Modulus becomes easy.
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patnaik1988
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GMAT Club World Cup 2026 (Day 2): If |x - y| = y - x, x y, and y < 0 08 Jul 2026, 07:54
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Hi, If Mod(x-y)=y-x it implies x-y<0 or x<y and since y<0, x should be further negative than y. Its better to assume -ve integers of x=-2 and y=-1 in the equation to get -2/2 + 1/1 + -1/1 + 2/2 = 0 (Answer choice B)
Bunuel
If \(|x-y|=y-x\), \(x \ne y\), and \(y<0\), what is the value of \(\frac{x}{|x|}+\frac{|-y|}{-y}+\frac{x-y}{|y-x|}+\frac{|xy|}{xy}\)?

A. -2
B. 0
C. 1
D. 2
E. 4


 


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GMAT Club World Cup 2026 (Day 2): If |x - y| = y - x, x y, and y < 0 08 Jul 2026, 07:58
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We are given mod of X minus Y is equal to Y minus X .
So X minus Y is negative
and also it is given that Y is negative
So the value of X must be negative. Let us say if X is equal to -3 and Y is equal to -2, then we can solve this question.
X divided by mod X is -1, mode of minus Y divided by minus Y is +1.
X minus Y divided by Mod of Y minus X is -1, mod XY divided by XY equal +1
Adding all will give zero option, correct B is correct
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GMAT Club World Cup 2026 (Day 2): If |x - y| = y - x, x y, and y < 0 08 Jul 2026, 07:59
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We need to find out
x/|x| + (|-y|/-y) + |x-y|/|y-x| + |xy|/xy

We are given,
|x-y| = y-x
x!=y and y<0

=> |x - y| = - (x - y)
This is possible when x - y <0
=> x < y
=> x < y < 0

x/|x| = x/-x = -1

-y/|-y|
=> -y > 0 since y< 0
=> -y/|-y| = 1

x-y / |y-x|
Since y - x > 0
=> |y - x| = y -x
=> x -y/|y-x| = -1

xy/|xy|
Since x and y <0
=> xy > 0
=> |xy| = xy

=> xy/|xy| = 1

Assembling all the 4 terms
x/|x| + (|-y|/-y) + ((x-y)/|x-y|) + |xy|/xy = -1 + 1 - 1 + 1 = 0

B. 0
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