imo A

1,is sufficient alone but 2is not sufficient

neoreaves wrote:If xy ≠ 0, is x > y?

(1) 4x = 3y
(2) |y - x| = x - y

St 1 :

x= 0.75 Y

X can greater or lesser than Y depending on Y. If Y is positive x is less than Y. If Y is negative then X can greater than Y.

Inconsistent- Insufficient.

St 2:

|y - x| = x - y

y-x =x-y

: 2y =2x

y=x
from this we can answer whether x>y as we have deduced x=y. The answer may be a YES or a NO. But it is definite.

Going further for nailing it, try some plugging in values for X & Y with X>Y & X<Y

Case 1: X >Y
Let X =2 Y= 1

|1-2| =2-1

|-1| = 1
1=1
YES X > Y

Case 2: X= 1 Y=2
|2-1| = 1-2

|1| = -1
LHS NOT equal to RHS . So X is not less than Y.

St 2 is sufficient to say X >Y

Pick B

Plz correct me if my method is wrong.

neoreaves wrote:If xy ≠ 0, is x > y?

(1) 4x = 3y
(2) |y - x| = x - y

xy≠ 0. So, x≠ 0, y≠ 0

from 1:
x=3y/4
If y is negative, x>y
If y is positive, x<y
Insufficient.

from 2:
|y - x| = x - y
if y>x, y-x=x-y, y=x, but y>x. So, this can't be.
if x>y, y-x=-(x-y)=y-x So, this is possible.
if x=y, y-x=x-y, So, again this is possible.
Insufficient

Combining 1 and 2:
For x and y to satisfy both 1 and 2, x≠y, and x>y

IMO answer C

Seer bhai,

can u explain using some plugged values?

one question to all :

if I say, |Z| = ?

it will be Z if Z is positive.

and it will be -Z if Z is negative.


now, just put y-x in place of Z and look at statement 2 again.

does anyone want to change his answer?

harshavardhanc wrote:one question to all :

if I say, |Z| = ?

it will be Z if Z is positive.

and it will be -Z if Z is negative.


now, just put y-x in place of Z and look at statement 2 again.

does anyone want to change his answer?

harsha bhai...

plz plz no surprise..i want toc ur answer, ahahhaha!!

gmatmachoman wrote:

harshavardhanc wrote:one question to all :

if I say, |Z| = ?

it will be Z if Z is positive.

and it will be -Z if Z is negative.


now, just put y-x in place of Z and look at statement 2 again.

does anyone want to change his answer?

harsha bhai...

plz plz no surprise..i want toc ur answer, ahahhaha!!

macho Anna,

all I'm saying is if you put y-x in place of Z and look at st2 again, you'll find that

y-x should be -ve to satisfy it.

implies, y-x < 0 OR y<x. Sufficient to answer the question. hence, B

harshavardhanc wrote:

gmatmachoman wrote:

harshavardhanc wrote:one question to all :

if I say, |Z| = ?

it will be Z if Z is positive.

and it will be -Z if Z is negative.


now, just put y-x in place of Z and look at statement 2 again.

does anyone want to change his answer?

harsha bhai...

plz plz no surprise..i want toc ur answer, ahahhaha!!

macho Anna,

all I'm saying is if you put y-x in place of Z and look at st2 again, you'll find that

y-x should be -ve to satisfy it.

implies, y-x < 0 OR y<x. Sufficient to answer the question. hence, B

Harsha..that was awesome...!! B wins!!

@Harsha,
you are right. But aren't we ignoring the possibility that x can be equal to y also.

If x=y, statement 2 still holds true.

@gmatmachoman

here are the numbers to plug in

statement 2:
x,y,y-x,|y-x|,x-y
-3,-2,1,1,-1 -fails
-3,-4,-1,1,1 - passes x>y
-3,2,5,5,-5 - fails
-3,-3,0,0,0 - passes x=y
3,4,1,1,-1 - fails
3,2,-1,1,1 - passes x>y
3,-4,-7,-7,7 - passes x>y

So, x is either greater or equal to y to satisfy the condition given in statement 2[/list]

iamseer wrote:@Harsha,
you are right. But aren't we ignoring the possibility that x can be equal to y also.

If x=y, statement 2 still holds true.

@gmatmachoman

here are the numbers to plug in

statement 2:
x,y,y-x,|y-x|,x-y
-3,-2,1,1,-1 -fails
-3,-4,-1,1,1 - passes x>y
-3,2,5,5,-5 - fails
-3,-3,0,0,0 - passes x=y
3,4,1,1,-1 - fails
3,2,-1,1,1 - passes x>y
3,-4,-7,-7,7 - passes x>y

So, x is either greater or equal to y to satisfy the condition given in statement 2[/list]

agree! thanks mate!