Geez--Shld we expect this on d GMAT?

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Source: — Data Sufficiency |

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by clock60 » Thu Mar 31, 2011 1:39 pm
hi Xlogic
can you clarify st 1, it looks very poor

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by MAAJ » Thu Mar 31, 2011 2:58 pm
I arrived to (C)... :B but I'm not quite sure...

What I'm sure is that from STMT 1, y = 0, and that it's not sufficient.

xy = 0 -> y = 0 because |x|>|y|

x|y| < x-y ???
x|0| < x-0 ???
0 < x ???
x > 0 ???
Not sufficient

STMT 2 tell us that x < 0, heres where I use my imagination/creativity to solve this problem :p

x|y| < x-y ???
|y| > (x-y)/x ??? flip sign because x < 0
|y| > 1 - y/x ???
Not sufficient

Combined...??? WTH:

Because STMT1: y = 0 and STMT2: |y| > 1 - y/x ???
|0| > 1 - 0/x???
0 > 1 - 0???
0> 1???
NO!, so I picked (C)..... please don't judge me after this lol...
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by XLogic » Thu Mar 31, 2011 4:23 pm
@ Maaj

Sorry I goofed. Here you go.

x and y are integers, and |x|>|y|
is x*|y|< x-y?

(1) xy >= 0
(2) x < 0

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by rohu27 » Thu Mar 31, 2011 7:25 pm
is x*|y|< x-y?

1)xy>=0, implies y=0 or x and y both are negative. (remeber |x|>|y| so x cannot be zero)
now case 1: y=0 x is x is +ve.
x*|y|=0 which will be less than x-y
case 2 : y=0 x is x is -ve
x*|y|=0 which will be greater than x-y
2 contradicting answers, so A is not suff.
2)X<0 so y can also be <0 or y=0 or y>0
x<0,y<0
x=-2,y=-1
x*mod y=-2, x-y=-1 so given condition answer is YES
but if x=-2 and y=0, x*mody=0 which is not less than x-y(-2). again contradicting, so not suff

both statements together,
xy>=0, x<0 implies either y=0 or y<0
if y=0
x*|y|>x-y.
if y<0, exmaple, x=-2,y=-1
x*|y|=-2, x-y=-1 so x*|y|< x-y

still no clear answeer

i wil go wth E..whts the OA?

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by manpsingh87 » Thu Mar 31, 2011 9:46 pm
XLogic wrote:@ Maaj

Sorry I goofed. Here you go.

x and y are integers, and |x|>|y|
is x*|y|< x-y?

(1) xy >= 0
(2) x < 0
statement 1) xy>=0;

now this is possible if x>0,y>0; x<0, y<0; also x>0, y=0; x<0 , y=0;

if x>0,y>0; then as per the question |x|>|y|; i.e. x>y; let x = 5, y=4; then we have 5*4<5-4;
which is false.
x<0,y<0, assume x=-4, y=-3; -4|-3|<-4-(-3); -12<-1; which is true.
x>0, y=0; x=4 y=0, 4*0<4-0; which is true.
x<0, y=0; x=-4 y=0, -4*0<-4-0; which is false.

as here different results are possible for different values of x hence 1 alone is not sufficient to answer the question.

2) x<0; x<0, again here different cases are possible. hence 2 alone is not sufficient to answer the question.

now combining 1 and 2;

we have two cases left x<0,y<0, and x<0, y=0;

consider x<0, y<0; let x=-4, y=-3,-4|-3|<-4-(-3)=-12<-1; which is true,

consider x<0, y=0; let x=-4 y=0 -4*0<-4-0; 0<-4; which is false.

as even after combining 1 and 2 we are not able to get the unique solution, hence answer should be E
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by MAAJ » Fri Apr 01, 2011 5:18 am
IMO:

Best bet (E);
Second best bet (C)
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by GMATGuruNY » Fri Apr 01, 2011 5:39 am
XLogic wrote:@ Maaj

Sorry I goofed. Here you go.

x and y are integers, and |x|>|y|
is x*|y|< x-y?

(1) xy ≥ 0
(2) x < 0
We can plug in values. To speed up the process, look for values that satisfy both statements and any other conditions given.

Statement 1: xy ≥ 0.
Let x = -2, y = 0.
All the conditions are satisfied:
|x|>|y| --> |-2| > |0|
xy ≥ 0 --> (-2)(0) ≥ 0
x < 0 --> -2 < 0

Plug x=-2 and y=0 into the question:
Is (-2)*|0| < -2-0?
0 < -2. No.

Let x = -2, y = -1.
All the conditions are satisfied:
|x|>|y| --> |-2| > |-1|
xy ≥ 0 --> (-2)(-1) ≥ 0
x < 0 --> -2 < 0

Plug x=-2 and y= -1 into the question:
Is -2*|-1| < -2 -(-1)?
-2 < -1. Yes.

Since in the first case the answer is No and in the second case the answer is Yes, insufficient.

Statement 2: x<0.
In each case tried under statement 1, x<0.
Since in the first case the answer is No and in the second case the answer is Yes, insufficient.

Statements 1 and 2 combined:
The two cases tried satisfy both statements.
Since in the first case the answer is No and in the second case the answer is Yes, insufficient.

The correct answer is E.
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by umesh321 » Fri Apr 01, 2011 7:48 am
My approach:-

x and y are integers, and |x|>|y|
is x*|y|< x-y?
(1) xy ≥ 0
(2) x < 0

(1)1) xy ≥ 0
if both x and y are positive integers
==> x*y<x-y
==> y<1-y/x
which is not valid
if both x and y are negative integers

==> which is not always
x*|y|< x-y
-IxI*IyI<-IxI+IyI
IxI< IyI(1+IxI)
which is also not valid
(2) when x<0
no solution
1 in conjunction with 2
IxI< IyI(1+IxI)
it also does not give any solution
Answer E