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Debatable OA [#permalink]
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21 Jun 2011, 11:51
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50% (03:31) correct
50% (01:46) wrong based on 8 sessions
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***Warning!!!! Debatable OA*****\(If \hspace{2} xyz \ne 0, \hspace{1} is \hspace{3} x(y+z)=0?\) \(1) \hspace{4} y+z = y+z\) \(2) \hspace{4} x+y = x+y\)
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Last edited by wizardsasha on 13 Nov 2011, 10:40, edited 2 times in total.
Added the warning & m tag



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Re: Inequality Help! [#permalink]
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21 Jun 2011, 12:32
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I cannot think of scenario where statement A is not alone sufficient. Statement A should alone give the value of exp as either +ve or ve. So should be sufficient.
As per state statement 2, the value of the expression can be either 0 or ve or +ve, so it is not sufficient.
i remember this Q i read somewhere and it was required to determine  x(y+z) >=0 , in that case the both statements will be required and OA should be C.
Last edited by agdimple333 on 21 Jun 2011, 12:45, edited 1 time in total.



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Re: Inequality Help! [#permalink]
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27 Jun 2011, 11:33
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I interpreted the question wrongly, A only can fulfill the condition.



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Re: Inequality Help! [#permalink]
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04 Jul 2011, 02:41
This is what I think when I go through this problem: \(xyz \ne 0\) implies none of x, y and z is 0. Question: Is x(y+z) = 0? Since x is not 0, the question is just: Is (y + z) = 0? or Is y = z? 1)\(\hspace{4} y+z = y+z\) If y and z satisfy this relation, y cannot be equal to z if they are not 0. This is so because \(\hspace{4} z+z = z+z\) \(\hspace{4} 0 = 2z\) But z is not 0 so y is not equal to z. Hence we get a definite 'No' answer. It is sufficient. 2) \(\hspace{4} x+y = x+y\) Doesn't give us relation between y and z so not sufficient. Answer (A)
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Re: Inequality Help! [#permalink]
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21 Jun 2011, 12:08
Hmm I don't get the OA..
If xyz is not equal to zero it means neither value, x, y or z can be zero.
1) If we pick values for y and z, 2 and 0,5, then we get 1,5 = 2,5 which obviously doesn't fit. If we take 2 and 0,5 we get 2,5 = 2,5 which fits. In order to fit this y and z must therefore have the same sign. Therefore it can either be negative or positive. Still, whether x(y+z)=0, now depends on the value of x. We can see from the question that if xyz is not equal to zero it means x is also not zero. So, if x can't equal zero, x(y+z), no matter the value of y+z, can not equal zero. Why is it C then, anyone?



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Re: Inequality Help! [#permalink]
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21 Jun 2011, 12:34
Hmm, this calls for an expert :D



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Re: Inequality Help! [#permalink]
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27 Jun 2011, 10:56
my take 
1. both z and y are either +ve or ve. insufficient 2. both x and y are either +ve or ve. insufficient.
Combining both, all of x, y and z are either +ve or ve. In both cases, x(y + z) != 0. Answer is C.



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Re: Inequality Help! [#permalink]
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27 Jun 2011, 11:10
agdimple333 wrote: I cannot think of scenario where statement A is not alone sufficient. Statement A should alone give the value of exp as either +ve or ve. So should be sufficient.
As per state statement 2, the value of the expression can be either 0 or ve or +ve, so it is not sufficient.
i remember this Q i read somewhere and it was required to determine  x(y+z) >=0 , in that case the both statements will be required and OA should be C. I agree with this. Anyone thinks otherwise? Please give me an example where A alone fails. Ans: "A"
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Re: Inequality Help! [#permalink]
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28 Jun 2011, 19:50
A alone is sufficient to answer this question.
1. Sufficient
we have y>0 z>0 or y<0 z<0
also x is not equal to 0 . so x(y+z) will never be equal to 0.
2. Not sufficient
x>0 y>0 or x<0 y<0 . Also z is not equal to 0.
But if y=z then x(y+z) can be 0. if y is not equal to z , then x(y+z) cannot be 0.
Answer is A.



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Re: Inequality Help! [#permalink]
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28 Jun 2011, 23:18
This question is from Question Set 18, Q# 32. I double checked the answer there. It says C.
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Re: Inequality Help! [#permalink]
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29 Jun 2011, 00:27
I agree this is A.
Given 1 is true, x(y+z) will never equal zero. However, given 2 is true, x(y+z) can equal zero or not



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Re: Inequality Help! [#permalink]
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29 Jun 2011, 04:47
Hi,
I agree with +ve / ve explaination. Infact i also thought it the same way before even looking at the answers.
Cheers.



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Re: Inequality Help! [#permalink]
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04 Jul 2011, 01:31
jamifahad wrote: ***Warning!!!! Debatable OA*****
\(If \hspace{2} xyz \ne 0, \hspace{1} is \hspace{3} x(y+z)=0?\)
\(1) \hspace{4} y+z = y+z\)
\(2) \hspace{4} x+y = x+y\) Given : xyz not equal to 0 therefore x not equal to 0 , y not equal to 0 and z not equal to 0. from condition 1 : y+z = y+z therefore both y and z are either positive or negative but not zero(this is given), therefore (y+z) will never be zero. This condtion is not sufficient beacuse if x=0 then x(y+z) becomes '0' else x(y+z) is non zero. from condition 2 : x+y = x + y therefore x and y are either positive or negative but not zero(this is given) , therefore neither x =0 nor y = 0 . This condtion is not sufficient because what if z= y then x(y+z) = 0. From condition 1 and 2 neither x=0 nor y=z (because they both are either positive or negative). Hence we can say that x(y+z) will never be zero. Answer is C.



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Re: Inequality Help! [#permalink]
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04 Jul 2011, 01:48
zuberahmed wrote: jamifahad wrote: ***Warning!!!! Debatable OA*****
\(If \hspace{2} xyz \ne 0, \hspace{1} is \hspace{3} x(y+z)=0?\)
\(1) \hspace{4} y+z = y+z\)
\(2) \hspace{4} x+y = x+y\) Given : xyz not equal to 0 therefore x not equal to 0 ... from condition 1 : ... This condtion is not sufficient beacuse if x=0 then x(y+z) becomes '0' else x(y+z) is non zero. But, as you already said, x can't be zero.




Re: Inequality Help!
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