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Source: — Data Sufficiency |
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by abhishekg21 » Wed Oct 20, 2010 4:49 am
it is A.
as per 1 q(q-r)=0 ..so either q is 0 or q-r =0.
since it is given q !=r so q-r cannot be 0 and so q=0

2) doesnt tell us anything..
so answer is A
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by novel » Wed Oct 20, 2010 6:17 am
IMO both the statements are needed to answer the question

Statement 1-if q=1 and r=1 ,then the equation is satisfied
but if r is not equal to one ,q has to be 0 to satisfy the equation.
Satement 2-r=5 does not ell us anything about q

however using both the statements we can knoe whether the value of Q is 0 or not
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by abhishekg21 » Wed Oct 20, 2010 6:32 am
Novel as per the question q!=r so your assumption that q= and r=1 is not valid.
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by fskilnik@GMATH » Wed Oct 20, 2010 9:13 am
Hi goyalsau,

My pleasure to be here!

(1) Sufficient

Please note that q^2 = qr is equivalent to q^2 - qr = 0 that is equivalent to q(q-r) = 0.

When 2 real numbers has product zero, at least one of them must be zero, therefore:

(a) q = 0 or
(b) q-r = 0, that is, q = r.

From the fact that (b) was not allowed by the question stem, we have necessarily (a), that is what we were asked, by the way.

(2) Not Sufficient

> Take r = 5 and q = 0 to answer in the affirmative ;
> Take r = 5 and q = 1 to answer in the negative.

We are done.

Regards,
Fábio.

Important: DO NOT CANCEL common factors, put them into "evidence" (then you do not fall into the trap of sttm (1)!!
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
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by novel » Wed Oct 20, 2010 9:35 pm
fskilnik wrote:Hi goyalsau,

My pleasure to be here!

(1) Sufficient

Please note that q^2 = qr is equivalent to q^2 - qr = 0 that is equivalent to q(q-r) = 0.

When 2 real numbers has product zero, at least one of them must be zero, therefore:

(a) q = 0 or
(b) q-r = 0, that is, q = r.

From the fact that (b) was not allowed by the question stem, we have necessarily (a), that is what we were asked, by the way.

(2) Not Sufficient

> Take r = 5 and q = 0 to answer in the affirmative ;
> Take r = 5 and q = 1 to answer in the negative.

We are done.

Regards,
Fábio.

Important: DO NOT CANCEL common factors, put them into "evidence" (then you do not fall into the trap of sttm (1)!!
From 1 we can conclude thar q=0 or q= r,we have been asked if q=0,thisa cannot be answered from 1.so 1 is not sufficient.I am right.
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by fskilnik@GMATH » Thu Oct 21, 2010 4:37 am
novel wrote:From 1 we can conclude thar q=0 or q= r,we have been asked if q=0,thisa cannot be answered from 1.so 1 is not sufficient.I am right.
No, you are not. Please be more humble and read the question stem and (for instance) my answer more carefully.
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