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IrinaOK
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DS_number properties 03 Sep 2007, 01:31
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Fig
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DS_number properties 03 Sep 2007, 02:15
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(D) for me :)

We have to be sure that A != 0. That gives y = (C-B)/A.

From (1)
C > B implies that C != B.

That said, we could not have A = 0 otherwise B=C, contradicting C > B.

SUFF.

From (2)
A > 1 implies that A != 0.

SUFF.
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IrinaOK
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DS_number properties 03 Sep 2007, 02:20
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Fig
(D) for me :)

We have to be sure that A != 0. That gives y = (C-B)/A.

From (1)
C > B implies that C != B.

That said, we could not have A = 0 otherwise B=C, contradicting C > B.

SUFF.

From (2)
A > 1 implies that A != 0.

SUFF.
Show more


But look at the question, why do you think it is asking if A is not equal to zero? plzz explain your reasoning..
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Fig
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DS_number properties 03 Sep 2007, 03:20
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IrinaOK
Fig
(D) for me :)

We have to be sure that A != 0. That gives y = (C-B)/A.

From (1)
C > B implies that C != B.

That said, we could not have A = 0 otherwise B=C, contradicting C > B.

SUFF.

From (2)
A > 1 implies that A != 0.

SUFF.

But look at the question, why do you think it is asking if A is not equal to zero? plzz explain your reasoning..
Show more


It's because A, B and C are constants, not variables like Y :)... The trap of this question is here :).... A, B or C is considered like a fixed number. :)

So, there are 2 possible case:
> Inifinite number of Y
> One Y only
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ywilfred
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DS_number properties 03 Sep 2007, 06:25
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Fig
(D) for me :)

We have to be sure that A != 0. That gives y = (C-B)/A.

From (1)
C > B implies that C != B.

That said, we could not have A = 0 otherwise B=C, contradicting C > B.

SUFF.

From (2)
A > 1 implies that A != 0.

SUFF.
Show more


Nice one. Constants and variables are different things.
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mastergmat1
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DS_number properties 03 Sep 2007, 07:56
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Good question.

I got trapped in with B ..

C> b tells you that y > 0
and the only value for y when A,B, C are constants is Y = 1

A> 1 means that Y is again 1

So D.
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IrinaOK
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DS_number properties 03 Sep 2007, 10:25
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Fig
IrinaOK
Fig
(D) for me :)

We have to be sure that A != 0. That gives y = (C-B)/A.

From (1)
C > B implies that C != B.

That said, we could not have A = 0 otherwise B=C, contradicting C > B.

SUFF.

From (2)
A > 1 implies that A != 0.

SUFF.

But look at the question, why do you think it is asking if A is not equal to zero? plzz explain your reasoning..

It's because A, B and C are constants, not variables like Y :)... The trap of this question is here :).... A, B or C is considered like a fixed number. :)

So, there are 2 possible case:
> Inifinite number of Y
> One Y only
Show more


So in this case Y has only one possible value, but if A was equal to zero Y would have infinite bumber of values. Cuz to find Y we would have to devide by A=zero.....?
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Fig
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DS_number properties 03 Sep 2007, 14:59
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IrinaOK
Fig
IrinaOK
[quote="Fig"](D) for me :)

We have to be sure that A != 0. That gives y = (C-B)/A.

From (1)
C > B implies that C != B.

That said, we could not have A = 0 otherwise B=C, contradicting C > B.

SUFF.

From (2)
A > 1 implies that A != 0.

SUFF.

But look at the question, why do you think it is asking if A is not equal to zero? plzz explain your reasoning..

It's because A, B and C are constants, not variables like Y :)... The trap of this question is here :).... A, B or C is considered like a fixed number. :)

So, there are 2 possible case:
> Inifinite number of Y
> One Y only

So in this case Y has only one possible value, but if A was equal to zero Y would have infinite bumber of values. Cuz to find Y we would have to devide by A=zero.....?[/quote]

No :)... We cannot divid by 0, of course.... But, u will have an equation such as 0*Y + 1 = 1, where B=C=1 and A=0... and thus, 1=1. Y is independant of the truthness of this equation and have all existing numbers as possible values. :)



Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Data Sufficiency (DS) Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
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