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Source: — Data Sufficiency |
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by limestone » Mon Sep 27, 2010 7:48 pm
Hi,
I have a long approach to this problem. Hope anyone can give out a more concise method.

the rule for breaking the absolute syntax: [x] = x when x>=0; or [x] = -x when x<=0.
I will use plug-in and case by case method.
Original inequation: [x-y] >[x]-[y]?

1. y<x, there are three cases:

0>=x>y then (breaking the abs. syntax) x-y > -x - (-y)
or x-y>-x+y . This is correct as x>y then x-y>0 and -x+y<0

x>y>=0 then x-y>x-y. Impossible! You can stop here as this case will prove 1. to be insuff.

x>=0>=y ( I should stop at the second case, but test this to illustrate the full set of cases) then x-y>x-(-y)
or x-y > x+y. This is true as y is a negative number.

Hence 1 is insuff

2.xy<0, there are two cases:

x<=0<=y and x>=0>=y ( x & y must have different signs)

x<=0<=y then y-x>-x -y or y>-y ( eliminate "-x" in both sides) --> this is true as y>=0 then y>=-y

x>=0>=y then x-y> x - (-y) or x -y > x +y this is true as y is negative.

So 2 is suff.

IMO: B
"There is nothing either good or bad - but thinking makes it so" - Shakespeare.
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by reply2spg » Mon Sep 27, 2010 8:05 pm
Thanks, but I am completely lost. Can you plug in some values?

I can do that, but I am not following you. Sorry :(
limestone wrote:Hi,
I have a long approach to this problem. Hope anyone can give out a more concise method.

the rule for breaking the absolute syntax: [x] = x when x>=0; or [x] = -x when x<=0.
I will use plug-in and case by case method.
Original inequation: [x-y] >[x]-[y]?

1. y<x, there are three cases:

0>=x>y then (breaking the abs. syntax) x-y > -x - (-y)
or x-y>-x+y . This is correct as x>y then x-y>0 and -x+y<0

x>y>=0 then x-y>x-y. Impossible! You can stop here as this case will prove 1. to be insuff.

x>=0>=y ( I should stop at the second case, but test this to illustrate the full set of cases) then x-y>x-(-y)
or x-y > x+y. This is true as y is a negative number.

Hence 1 is insuff

2.xy<0, there are two cases:

x<=0<=y and x>=0>=y ( x & y must have different signs)

x<=0<=y then y-x>-x -y or y>-y ( eliminate "-x" in both sides) --> this is true as y>=0 then y>=-y

x>=0>=y then x-y> x - (-y) or x -y > x +y this is true as y is negative.

So 2 is suff.

IMO: B
Sudhanshu
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by limestone » Mon Sep 27, 2010 8:24 pm
Sure,

1. x>y

0>=x>y : x= -1, y = -2 ( and -1>-2)
Then [-1 -(-2)] = [-1+2] = 1
[-1] - [-2] = 1-2 = -1
1>-1 then the inequation is confirmed.

x>y>=0 : x= 2, y =1
Then [2-1] = 1
[2] -[1] = 2-1 = 1
1 = 1 hence the inequation is not confirmed.

x>=0>=y : x= 1, y=-1
then [1-(-1)] = [2] = 2
[1] - [-1] = 1-1 = 0
2>0, correct.

3 cases: 2 say yes, 1 says no ==> cannot define => insuff

2. xy<0

x<=0<=y : x= -1, y =1
then [-1 -1] = 2
[-1] - [1] = 0
2>0, true

x>=0>=y: x =1, y=-1
then [1-(-1)] = [2] = 2
[1] - [-1] = 0
2>0, true

2 cases: both say yes => suff.

Hence pick B.
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by reply2spg » Mon Sep 27, 2010 8:29 pm
Thanks Buddy, I guess I need to go 2-3 times for A.

Btw....how can you solve this question in 2 minutes? Any simple method?
limestone wrote:Sure,

1. x>y

0>=x>y : x= -1, y = -2 ( and -1>-2)
Then [-1 -(-2)] = [-1+2] = 1
[-1] - [-2] = 1-2 = -1
1>-1 then the inequation is confirmed.

x>y>=0 : x= 2, y =1
Then [2-1] = 1
[2] -[1] = 2-1 = 1
1 = 1 hence the inequation is not confirmed.

x>=0>=y : x= 1, y=-1
then [1-(-1)] = [2] = 2
[1] - [-1] = 1-1 = 0
2>0, correct.

3 cases: 2 say yes, 1 says no ==> cannot define => insuff

2. xy<0

x<=0<=y : x= -1, y =1
then [-1 -1] = 2
[-1] - [1] = 0
2>0, true

x>=0>=y: x =1, y=-1
then [1-(-1)] = [2] = 2
[1] - [-1] = 0
2>0, true

2 cases: both say yes => suff.

Hence pick B.
Sudhanshu
(have lot of things to learn from all of you)
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by limestone » Mon Sep 27, 2010 8:55 pm
In absolute value question, I care about the sign. Here are the signs of x, y, and (x-y)

For 1. x>y, I think of how many cases can happen. There are three items to mix : x,y, and zero

x is always larger than y as given above. Then I place zero in front, after x then after y to give out 3 cases:
0>=x>y
x>=0>y
x>y>=0

In each case, plug in some very simple value such as 0,1,-1,2 ...etc. and calculate as fast as possible.

For 2. I do the same
xy<0 then x>0>y or y>0>x ( I'm sorry that I used x>=0>=y or y>=0>=x in my above post, x and y cannot be zero as xy<0)
Then plug-in then test.
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by reply2spg » Mon Sep 27, 2010 9:01 pm
Big Thanks,

I was struggling with A. I think B is pretty easy, Don't know I messed it up in exam. While analyzing I came to know that B is easier than A. Since xy < 0 means either x or y must be -ve. If you put values, it will make sense.

However, I was struggling with A. Your approach helped me.

Thanks again
limestone wrote:In absolute value question, I care about the sign. Here are the signs of x, y, and (x-y)

For 1. x>y, I think of how many cases can happen. There are three items to mix : x,y, and zero

x is always larger than y as given above. Then I place zero in front, after x then after y to give out 3 cases:
0>=x>y
x>=0>y
x>y>=0

In each case, plug in some very simple value such as 0,1,-1,2 ...etc. and calculate as fast as possible.

For 2. I do the same
xy<0 then x>0>y or y>0>x ( I'm sorry that I used x>=0>=y or y>=0>=x in my above post, x and y cannot be zero as xy<0)
Then plug-in then test.
Sudhanshu
(have lot of things to learn from all of you)
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by hi.itz.mani » Mon Sep 27, 2010 9:19 pm
My approach:

we have the inequality | x - y | > |x| - |y|

sqaring both sided | x- y|^2 > (|x| - |y| )^2 or | x- y|^2 < (|x| - |y| )^2

|x^2 + y^2 - 2xy | > |x|^2 + |y|^2 - 2|x||y| or |x^2 + y^2 - 2xy | < |x|^2 + |y|^2 - 2|x||y|

since squaring will make a number positive , hence I can remove the mod sign from left and mod signs from right where I see squares
x^2 + y^2 - 2xy > x^2 + y^2 - 2|x| |y| or x^2 + y^2 - 2xy < x^2 + y^2 - 2|x| |y|

2xy > - 2 |x| |y| or 2xy < - 2 |x| |y|
hence
we need to finally find if xy > |x| |y| or xy < |x| |y|

Statement 1:
y < x , insufficient as both y and x can be positive or negative or one of them positive and other regative

Statement 2:
xy < 0 Now we know that one of the x or Y is negative and hence xy will always be less that |x| |y|

Hence Statement 2 is sufficient

IMO B
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