Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

|x - y| will be greater when "x is greater than 0 and y is less than 0" or vice versa.
For xy to be < 0 the previous statement has to hold true.
Hence Choice B.

The simplest way is to use substitution for X & Y.

Statement I
Use 2 sets of values for X&Y i.e (5,3) and (-3,-5)...Insufficient

Statement II
Again use 2 sets of Values for X & Y (with one being negative)
(3,-5) and (-5,3)... using either of these, the solution can be deduced that |X-Y|>|X| - |Y|.

(1)
If y < x, and y is 1, x is 2, then |x-y| = 1 and |x|-|y|. |x-y| !> |x| - |y|
If y < x, and y is -1, x is 2, then |x-y| = 3 and |x|-|y| = 1. |x-y| > |x| - |y|

Insufficient as we can't tell if the inequality holds.

(2) xy < 0 -> x can be negative or y can be negative.

Consider x negative, y positive -> x=-1, y=2
|x-y| = 3, |x|-|y| -1, so |x-y| > |x| - |y|

Consider y negative, x positive -> x=2, y=1
|x-y| = 1, |x|-|y| = 1, so |x-y| = |x| - |y|

Again, we can't tell which inequality holds. Insufficient.

Using (1) and (2), we know x must be positive and y must be negaitve and so |x-y| = |x| - |y|. Thus, ans = C

(1) If y < x, and y is 1, x is 2, then |x-y| = 1 and |x|-|y|. |x-y| !> |x| - |y| If y < x, and y is -1, x is 2, then |x-y| = 3 and |x|-|y| = 1. |x-y| > |x| - |y|

Insufficient as we can't tell if the inequality holds.

(2) xy < 0 -> x can be negative or y can be negative.

Consider x negative, y positive -> x=-1, y=2 |x-y| = 3, |x|-|y| -1, so |x-y| > |x| - |y|

Consider y negative, x positive -> x=2, y=1 |x-y| = 1, |x|-|y| = 1, so |x-y| = |x| - |y|

Again, we can't tell which inequality holds. Insufficient.

Using (1) and (2), we know x must be positive and y must be negaitve and so |x-y| = |x| - |y|. Thus, ans = C

ywilfred, your values x=2, y=1 are incorrect, I guess u meant x=2 and y=-1 , once u take these values your answer shuld be B

NOTE: |x-y|= x-y if x>y or y-x if y>x |x|= x if x > 0 or -x if x < 0 |y|= y if y>0 or -y if y<0

1. x=1, y= 0
----> |x-y| = 1
|x|-|y|= 1
---> |x-y|=|x|-|y| ---> answer to the question : NO

x=1, y= -1
|x-y| = 2
|x|-|y| = 0
--->answer to the question : YES
---> insuff

2. xy< 0 --> one is <0 and the other is >0
case 1: y>0> x
----> |x-y| = y-x
|x|-|y|= -x - y
we have -x+y > -x-y (coz y >0) ---> |x-y|> |x|-|y|
--->answer to the question: YES

case 2: x>0>y
---> |x-y| = x-y
|x|-|y|= x - (-y) = x+y
we have x-y> x + y (coz y <0 )
----> |x-y|> |x|-|y|
----> answer to the question: YES

(1) If y < x, and y is 1, x is 2, then |x-y| = 1 and |x|-|y|. |x-y| !> |x| - |y| If y < x, and y is -1, x is 2, then |x-y| = 3 and |x|-|y| = 1. |x-y| > |x| - |y|

Insufficient as we can't tell if the inequality holds.

(2) xy < 0 -> x can be negative or y can be negative.

Consider x negative, y positive -> x=-1, y=2 |x-y| = 3, |x|-|y| -1, so |x-y| > |x| - |y|

Consider y negative, x positive -> x=2, y=1 |x-y| = 1, |x|-|y| = 1, so |x-y| = |x| - |y|

Again, we can't tell which inequality holds. Insufficient.

Using (1) and (2), we know x must be positive and y must be negaitve and so |x-y| = |x| - |y|. Thus, ans = C

Wrong substitution for X & Y.........in statement II