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GMAT Club World Cup 2026 (Day 2): If |x - y| = y - x, x y, and y < 0

1 2 3 4 5 6 7

ChaoChiKit

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Answer: B

Explanation:

From the given information, we can deduce that x<y<0. So when solving the equation:

x/|x|=-1,
|-y|/-y=1,
(x-y)/|y-x|=-1
|xy|/xy=1,

Hence the answer equals to 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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07 Jul 2026, 23:07

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We have |x-y|=y-x
y<0
x!=y

|x-y| (+ve) output will be y-x
so, x<y
so, x<y<0

Now split the elements.
x/|x|
x is -ve
so result will be -1

|-y|/-y
y itself is -ve, so -y is +ve
-y is +ve, absolute value will turn out to be +ve
so result will be 1

(x-y)/|y-x|)
already given, |x-y|=y=x, it is also =|y-x|
and numerator can be written as -(y-x)
solving the fraction will lead to -1

|xy|/(xy)
x and y both -ve, so xy=+ve
this will lead to 1

Add up all the elements no,
-1+1-1+1=0

Answer 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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07 Jul 2026, 23:08

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Given |x-y|=y-x , and x!=y (so x-y cannot be 0), y<0 (if y is -ve and |x-y| = y-x, we get that x is also -ve and even smaller than y)

putting values in the given equation

x/(-x) + (-y)/(-y) + x-y/|(-1)/(x-y)| + xy/xy

(-1) + 1 + (-1) + 1

==>0

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07 Jul 2026, 23:41

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from the question we get some info 1)y>x , so if we take x=-5 and y=-4 as given y<0 then we just have to put value and solve
and answer will be 0

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If mod of x-y = y-x. This implies x-y<0 =>x<y
And y<0
Therefore both x and y are negative

If we solve the equation, we get -1+1-1+1 =0
Hence B is the answer

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08 Jul 2026, 00:29

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we know x is not equal to y. also y< 0

|x-y| = y-x

assume y = -a

|x- (-a)| = (-a) -x
|x+a| = - (x+a)

now we know that |p| = +/- p

and |p| = -p that means value inside modulus is negative.

so in our case, x+a must be negative

x+a < 0
we know y= -a

x - y < 0
so, x<y

means x is negative and smaller than y.
now use this to find value of our equation.

x/|x| = -1

|-y| / -y. we know y is already negative. so |- (-y)| / y =1

x-y / |y-x|= -x +y/ |-y+x| = -(x-y) / x-y = -1

|xy| / xy = 1

-1+1-1+1= 0

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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08 Jul 2026, 00:42

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|x-y| = y-x, y< 0. This implies that x< 0 and y>x because the output of modulus is always positive, and since y is negative, for (y-x) to be positive x has to be negative as well and also greater than y in magnitude.

x/|x| will give -1
|-y|/-y will give +1
(x-y)/|y-x| will give -1
|xy|/xy will give +1

-1+1-1+1 = 0 Answer B

Jay29

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08 Jul 2026, 00:58

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Rule :
|a| = -a

Either a = 0 or a<0.

Given :
|X - Y| = -(X - Y)
X not equal to Y
Y < 0

so,
X - Y < 0
Y < 0

Adding gives us X < 0

putting the values in the four terms-

(+ve/ -ve) + (+ve/ +ve) + (-ve/ +ve) + (+ve/ +ve) = -1 +1 -1 +1 = 0

Answer = B

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08 Jul 2026, 01:34

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So ,, Here i got option B as answer .
--- First , | x - y | >= 0 then RHS should also be , y - x > = 0 so y > x ( as Y not = X) ,,, 0 > Y >X
-- i now did each one Separately - x / x ( as | - x | always x ) = -1 ,, next |- y| / - y = 1 ( both will be posstevie as multiplied with - and y itself Negative)
now appling the basic logic as 0 > Y > X for all the other once x - y / | y - x | here = -1 next | xy | = xy here -x * -y is xy so here is 1 so ,,
--- -1 +1 -1 +1 = 0
----- i solved it like that if anybody has any better and short way feel free to tag or share !!

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08 Jul 2026, 02:19

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Since we are given |x-y| = y-x; x is not equal to y, and y<0
here y-x = |-(x-y)| which means than x-y <= 0, and x < y, since it is given that x is not equal to y.

So we get x < y < 0
|x| = -x
|y| = -y
|-y| = - (-y) = y
|xy| = -x * -y = xy
|y-x| = y-x = -(x-y)

So x /|x| = -1; |-y| / -y = -1; (x-y) / |y-x| = -1; and |xy| / xy = 1
addition of all = -1-1-1+1 = -2

So as per me option A is the correct answer.

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08 Jul 2026, 02:53

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Given that | x - y | = y - x implies that x - y is a negative number. If x - y is -ve and y is already given as -ve then, x also needs to be a -ve number and x < y for the given condition to hold true.
Now we know that both x and y are negative and x < y. Assuming x as -2 and y as -1 and validating if the given condition holds true-
LHS
| -2 - (-1) | = | - 1| = 1
RHS
-1 - (-2) = -1 + 2 = 1

Since LHS = RHS, this is a valid case. Now assuming the same numbers for the equation we will arrive at-
(-1) + 1 + (-1) + 1
=0

Hence B is the answer

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Here is my solution.

Attachments

IMG_7626 (1).jpeg [ 1.61 MiB | Viewed 117 times ]

doloribusnobis

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08 Jul 2026, 03:18

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|x - y| = y - x and y < 0,
i.e., y = -y
substituting y = -y:

|x+y| = -(y+x)
|x+y| / (x+y) = -1

there are 2 cases that follow.

x/|x| + |-y|/-y + (x-y)/|y-x| + |xy|/xy

when x>0
1 + -1 + -1 + -1 = -2

when x<0
-1 + -1 + -1 + 1 = -2

JW555

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08 Jul 2026, 03:20

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|x-y| = y - x, an absolute value is at least zero and because y is given negative, for (y - x) to be at least zero x must be non-positive- zero or negative. Because, in the question x is given in the denominator, it cannot be zero, so it must be negative.
So, |x - y| is positive, because X is given different than Y, then y - x is positive and the absolute value of x is greater than y's. Finally, $|x|=|-x| = -x, |y|=|-y| =-y, |x-y|=|y-x|= -(x-y), $
Therefore, $\frac{x}{|x|}=frac{x}{-x}=-1,\frac{|-y|}{-y}=\frac{-y}{-y} = 1, -1+1=0,\frac{x-y}{|y-x|}=\frac{x-y}{-(x-y)}= -1,0-1=-1,\frac{|xy|}{xy}=\frac{xy}{xy}= 1, -1+1=0$

So, the final 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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08 Jul 2026, 03:41

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The answer is B.0
Because |x-y| = y-x means y>x, and since y<0, both x and y are negative.
x/|x|=-1 (x is negative)
|-y|/(-y) = +1 (since -y is positive)
(x-y)/|y-x| = (x-y)/(y-x) = -1
|xy|/xy = +1 (xy is positive since both are negative)
Total: -1+1-1+1 = 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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08 Jul 2026, 04:27

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let x=-5 and y =-3 then |-5+3| =-3+5=2 so substitute in the fractions

-5/|-5|+|3|/3+(-5+3)/|-3+5|+|-15|/15= -1+1-1+1=0 Ans 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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08 Jul 2026, 04:32

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|x-y| = y-x tells us y-x is positive, so y is bigger than x
y is negative (given)

so x is negative and y is negative and x is smaller (more negative) than y
lets pick easy numbers= x :-2 and y :-1

Adding values into equation
x/ |x|= -2/2 = -1

|-y|/(-y) = -y = 1 so its 1/1 =1

(x-y)/|y-x|: x-y= -2-(-1)= -1 and y-x = 1

|xy|/ xy: xy = (-2)(-1)= 2

Adding : -1+1-1+1= 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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08 Jul 2026, 04:37

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IMO - ANS is D: 2 But I could be wrong. Because I'm unable to solve for real value of x & y some conceptual gap in my understanding when I remove the MOD I get the value x=y which is already stated not equal and then later I have just one thing in my mind y<0 Meaning negative from that I get that x/mod x = 1 and then Mod -y / -y = 1 and x-y /mod y-x =1 and mod xy/xy = -1 so total 3 -1 = 2
But I could be wrong, thanks team for such a thoughtful constraint question.

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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08 Jul 2026, 04:58

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Absolute value ix-yi=y-x means y-x=>0 or y>x and since y is negative then even x is negative
Now evaluate the first part of the equation x/ixi= -1
Second part IyI/-y= 1
Third part x-y/Iy-xI= -1
Final part IxyI/xy = 1
-1+1-1+1= 0
Ans 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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08 Jul 2026, 05:13

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|x-y|= y-x
so, y-x >0 and y<0 (given) then, x must be <0 n |x|>y or x<y
so, let x=-2 & y=-1 in given equation
-2/2+1/1+(-3/1)+2/2 = -2

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

This question was provided by GMAT Club
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