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# is x>y?

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Manager
Joined: 31 Mar 2006
Posts: 162

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29 May 2006, 02:58
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Is x>y?
1.(x-y)(x-y)>0
2.y<0

Senior Manager
Joined: 19 Feb 2005
Posts: 486
Location: Milan Italy

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29 May 2006, 04:12
I'm sorry I misread st 1, so the explanation is not good, I still think the answer is E
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Please allow me to introduce myself: I'm a man of wealth and taste

Last edited by thearch on 29 May 2006, 05:12, edited 1 time in total.
VP
Joined: 07 Nov 2005
Posts: 1118
Location: India

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29 May 2006, 04:15
Also got E.
Same reasoning as above.
Senior Manager
Joined: 19 Feb 2005
Posts: 486
Location: Milan Italy

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29 May 2006, 05:52
amansingla4 wrote:

St 1 doesn't help much
whatever square is positive

St 2 by itself is clearly insufficient

Combining doesn't yield any valuable additional piece of information

Right?
_________________

Please allow me to introduce myself: I'm a man of wealth and taste

Director
Joined: 10 Oct 2005
Posts: 526
Location: US

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29 May 2006, 08:32
Yeah it is E....no doubt
Manager
Joined: 24 Oct 2005
Posts: 169

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29 May 2006, 09:44
I don't want to spoil the "E" party but I think is "A" (unless I am falling in a trap).

(x-y) (x-y) > 0 is equal to x>y.

so it should be sufficient.
VP
Joined: 29 Dec 2005
Posts: 1341

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29 May 2006, 09:59
positive soul wrote:
I don't want to spoil the "E" party but I think is "A" (unless I am falling in a trap).

(x-y) (x-y) > 0 is equal to x>y.

so it should be sufficient.

amansingla4 wrote:
Is x>y?
1.(x-y)(x-y)>0
2.y<0

do this way:

st 1. (x-y)(x-y)>0
(x-y)^2 >0. now put some values for x and y, you find either is > than the other.

st 2. y<0 doesnot go anywhere.
Manager
Joined: 24 Oct 2005
Posts: 169

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29 May 2006, 10:11
Thanks Prof.
29 May 2006, 10:11
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