hi Xlogic
can you clarify st 1, it looks very poor
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Geez--Shld we expect this on d GMAT?
Source: Beat The GMAT — Data Sufficiency |
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MAAJ
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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...
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...
"There's a difference between interest and commitment. When you're interested in doing something, you do it only when circumstance permit. When you're committed to something, you accept no excuses, only results."
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?
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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manpsingh87
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statement 1) xy>=0;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
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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MAAJ
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IMO:
Best bet (E);
Second best bet (C)
Best bet (E);
Second best bet (C)
"There's a difference between interest and commitment. When you're interested in doing something, you do it only when circumstance permit. When you're committed to something, you accept no excuses, only results."
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GMATGuruNY
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We can plug in values. To speed up the process, look for values that satisfy both statements and any other conditions given.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.
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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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
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
