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Debatable OA

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Debatable OA [#permalink] New post 21 Jun 2011, 10:51
00:00
A
B
C
D
E

Difficulty:

  5% (low)

Question Stats:

50% (02:15) correct 50% (00:39) wrong based on 6 sessions
***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|
[Reveal] Spoiler: OA

Last edited by wizardsasha on 13 Nov 2011, 09:40, edited 2 times in total.
Added the warning & m tag
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Re: Inequality Help! [#permalink] New post 21 Jun 2011, 11:08
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] New post 21 Jun 2011, 11:32
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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, 11:45, edited 1 time in total.
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Re: Inequality Help! [#permalink] New post 21 Jun 2011, 11:34
Hmm, this calls for an expert :D
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Re: Inequality Help! [#permalink] New post 27 Jun 2011, 09:56
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] New post 27 Jun 2011, 10:10
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] New post 27 Jun 2011, 10:33
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I interpreted the question wrongly, A only can fulfill the condition.
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Re: Inequality Help! [#permalink] New post 28 Jun 2011, 18:50
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] New post 28 Jun 2011, 22:18
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] New post 28 Jun 2011, 23:27
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] New post 29 Jun 2011, 03:47
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] New post 04 Jul 2011, 00:31
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] New post 04 Jul 2011, 00:48
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.
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Re: Inequality Help! [#permalink] New post 04 Jul 2011, 01:41
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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 = 2|z|
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] 04 Jul 2011, 01:41
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