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Is x^2y^3w^4z^1 < 0 ?

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Is x^2y^3w^4z^1 < 0 ?  [#permalink]

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New post 20 Apr 2016, 01:09
00:00
A
B
C
D
E

Difficulty:

  65% (hard)

Question Stats:

28% (00:57) correct 72% (00:52) wrong based on 85 sessions

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Re: Is x^2y^3w^4z^1 < 0 ?  [#permalink]

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New post 20 Apr 2016, 01:36
Its really a good post. I would willingly be waited to get the answer of this question.
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Re: Is x^2y^3w^4z^1 < 0 ?  [#permalink]

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New post 20 Apr 2016, 07:35
1
IMO E

A)yz<0, but w or x can be zero so the term can be equal to zero or less so in suff, same goes for B, and A+B
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Re: Is x^2y^3w^4z^1 < 0 ?  [#permalink]

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New post 20 Apr 2016, 10:25
IMO 'A'

yz<0 means that both 'y' and 'z' have opposite signs. And since in the question stem the powers of both these are odd, they will retain their original powers. And, hence the product of the question stem will be -ve.
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Re: Is x^2y^3w^4z^1 < 0 ?  [#permalink]

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New post 20 Apr 2016, 12:12
Is x2y3w4z1<0x2y3w4z1<0 ?

(1) yz < 0
(2) z < 0

Take the question -- x, w will always be positive . since even power .

Therefore question becomes y^3 z^1 < 0

1) yz < 0
either y or z is negative . Hence sufficient

2) z<0 . Here we dont know about y .
Hence not sufficient .

Ans is A
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Re: Is x^2y^3w^4z^1 < 0 ?  [#permalink]

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New post 20 Apr 2016, 20:24
ans e
1)yz<0 hence either y<0 or z<0 but no mention of w,,x they can be 0
2 z<0 same reason
combined still we get y>0 z,0 stil x and w can zero
hence equation can = 0 or less than zero
hence E
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Re: Is x^2y^3w^4z^1 < 0 ?  [#permalink]

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New post 20 Apr 2016, 21:10
2
Bunuel wrote:
Is \(x^2y^3w^4z^1 < 0\) ?

(1) yz < 0
(2) z < 0


Hi,
there are two sets of answer-
1) A as the answer --- Not speaking about any of the variables being 0..
2) E as the answer--- Correctly taking 0 as an option for variables ..


lets see the equation first and what is required..


\(x^2y^3w^4z^1 < 0\)
1) when all are non-zero integers..
\(x^2w^4\) will be > 0, we have to check if y and z are of opposite sign..

2) if any of the integer is 0..
if any integer is zero, entire equation will be 0
AND our answer to "Is \(x^2y^3w^4z^1 < 0\) ?" will be NO..
and suff..

lets see the statements-


(1) yz < 0
this shows that y and z are of opposite sign, so ans is YES..
But say x is 0, then ans will be NO..
Insuff

HAD the q been \(x^2y^3w^4z^1 > 0\), this statement would have been suff as ans would be NO in both cases, but not NOW

(2) z < 0
Clearly insuff

Combined,
still insuff as the cases as shown in statement 1 still remain
E
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Re: Is x^2y^3w^4z^1 < 0 ?  [#permalink]

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New post 26 Aug 2018, 00:24
Statement 1: yz<0.
As per the question, x^2 & w^4 would always be positive for all values excluding 0. So, As per st. 1, the expression could be <0 or =0. Insufficient.
Statement 2: z<0
Again the same we don't know about x and w whether they are Non-zero.

Combining both the statement we don't have any information on x and w whether they are non zero.

Answer - E.
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Re: Is x^2y^3w^4z^1 < 0 ?  [#permalink]

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New post 26 Aug 2018, 00:30
Bunuel wrote:
Is \(x^2y^3w^4z^1 < 0\) ?

(1) yz < 0
(2) z < 0


It should be E, for simple reason that \(W\) can be \(0\) and neither of the statements provide the Value of \(W\).
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Re: Is x^2y^3w^4z^1 < 0 ?   [#permalink] 26 Aug 2018, 00:30
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