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# Is X>Y 1. X/3Y > 1/3 2. -X+P < -Y + P

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Manager
Joined: 17 Aug 2009
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Is X>Y 1. X/3Y > 1/3 2. -X+P < -Y + P [#permalink]

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23 Nov 2009, 11:05
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67% (02:05) correct 33% (00:33) wrong based on 24 sessions

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Is X>Y

1. X/3Y > 1/3
2. -X+P < -Y + P
Senior Manager
Joined: 21 Jul 2009
Posts: 364
Schools: LBS, INSEAD, IMD, ISB - Anything with just 1 yr program.

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23 Nov 2009, 11:39
zaarathelab wrote:
Is X>Y

1. X/3Y > 1/3
2. -X+P < -Y + P

Stmt 1- X/Y > 3/3 implies that X/Y > 1. Multiplying Y on both sides, X > Y - Suff
Stmt 2- -X+P < -Y+P. Subtracting P on both sides -X < -Y. Multiplying with -1 on both sides, note, when you multiply with -1 on both sides the comparison side changes. So if -X < -Y then X > Y. - Suff.

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Manager
Joined: 07 Jul 2009
Posts: 221

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23 Nov 2009, 11:59
It's B.

Stmt 1: X/3Y>1/3
Multiply both side by 3 and we get X/Y>1
We have two conditions here (X and Y can be both positive or both negative):
1) If both X and Y are positive then X > Y
2) If both X and Y are negative, then X < Y

So, stmt 1 is NOT sufficient

Stmt 2: -X+P < -Y + P
Subtract P from both sides and we get –X < -Y
Multiply both sides by -1 and we get X>Y

So stmt 2 is sufficient.

VP
Joined: 05 Mar 2008
Posts: 1469

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23 Nov 2009, 12:00
zaarathelab wrote:
Is X>Y

1. X/3Y > 1/3
2. -X+P < -Y + P

1. Insufficient: x and y < 0 or x and y > o
Intern
Joined: 18 Nov 2009
Posts: 46

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23 Nov 2009, 13:08
SensibleGuy wrote:
zaarathelab wrote:
Is X>Y

1. X/3Y > 1/3
2. -X+P < -Y + P

Stmt 1- X/Y > 3/3 implies that X/Y > 1. Multiplying Y on both sides, X > Y - Suff
Stmt 2- -X+P < -Y+P. Subtracting P on both sides -X < -Y. Multiplying with -1 on both sides, note, when you multiply with -1 on both sides the comparison side changes. So if -X < -Y then X > Y. - Suff.

It is B
YYou cannot go from X/Y > 1 to X > Y (Multiplying Y on both sides) without knowing the sign of Y.
Intern
Affiliations: University of Florida Alumni
Joined: 25 Oct 2009
Posts: 36
Schools: Wharton, Booth, Stanford, HBS

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23 Nov 2009, 13:11
2
KUDOS
Cancelling out "common terms" on both sides of an equation

You need to be very careful when you do algebra derivations. One of the common mistake is to divide both side by "a common term". Remember you can only do this safely if the "common term" is a constant. However you CAN't do it if it contains a variable.

Example:

x(x-2)=x
You can't cancel out the x on both side and say x=3 is the solution. You must move the x on the right side to the left side.
x(x-2)-x=0
x(x-2-1)=0
The solutions are: x=0 and x=3
The reason why you can't divided both sides by x is that when x is zero, you can't divide anything by zero.

Equally important if not more, is that you CAN'T multiple or divide a "common term" that includes a variable from both side of an inequality. Not only it could be zero, but it could also be negative in which case you would need to flip the sign.

Example:

x^2>x
You CAN'T divided both sides by x and say x>1. What you have to do is to move the right side to the left:
x^2-x>0
x(x-1)>0
Solution would be either both x and x-1 are greater than zero, or both x and x-1 are smaller than zero. So your solution is: x>1 or x<0

Example:

x>1/x
Again you CAN'T multiply both sides by x because you don't know if x is positive or negative. What you have to do is to move the right side to the left:
x-1/x>0
(x^2-1)/x>0
If x>0 then x^2-1>0 =>x>1
If x<0 then x^2-1<0 =>x>-1
Therefore your solution is x>1 or 0>x>-1.
You could also break the original question to two branches from the beginning:
x>1/x
if x>0 then x^2>1 =>x>1
if x<0 then x^2<1 => x>-1
Therefore your solution is x>1 or 0>x>-1.
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Kudos are greatly appreciated and I'll always return the favor on one of your posts.

Thanks!

Senior Manager
Joined: 21 Jul 2009
Posts: 364
Schools: LBS, INSEAD, IMD, ISB - Anything with just 1 yr program.

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23 Nov 2009, 14:17
arrrrggghhhhh!!! The nightmares of number theory will continue to haunt me!!!!
_________________

I am AWESOME and it's gonna be LEGENDARY!!!

Manager
Joined: 22 Jul 2009
Posts: 196
Location: Manchester UK

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17 Dec 2009, 03:04
This was awesome.....
Re: Is x>y ?   [#permalink] 17 Dec 2009, 03:04
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