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If s^4*v^3*x^7 < 0 , is svx < 0? 1) v < 0 2) x >

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If s^4*v^3*x^7 < 0 , is svx < 0? 1) v < 0 2) x > [#permalink]

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New post 22 Jan 2010, 04:33
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If \(s^4*v^3*x^7 < 0\), is \(svx < 0?\)

1) \(v < 0\)
2) \(x > 0\)
[Reveal] Spoiler: OA

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Re: Power [#permalink]

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New post 22 Jan 2010, 05:40
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Rules to consider:
1. A negative number to the power of an odd number is always negative.
2 .A negative number to the power of an even number is always positive.
3. A negative times a negative is always positive.
4. A negative times a positive is always negative.

You have s^4 which is an even power and will always be positive.
V^3 will be negative if V is negative.
V^3 will be negative if V is negative.
X^7 will be positive if X is positive.
X^7 will be positive if X is positive.

So, solution one, v<0:
We know s^4 is positive.
We know v^3 is negative since v<0 and it is an odd power.
But, x could be either positive or negative, meaning x^7 could be either.
Since we know s^4*v^3*x^7 is negative, we have:
(+)*(-)*(?) = (-)
Therefore we can determine x^7 is positive, and thus x is positive.
HOWEVER, don't get tripped up on 'S' - just because s^4 is positive, s could be positive or negative.
s*v*x = (?)*(-)*(+) is < 0 if s is positive and > 0 if s is negative.
Therefore, one alone is not enough.

Likewise, with solution two, x>0:
We have s^4*v^3*x^7 < 0
So, (+)*(?)*(+) = (-)
Therefore, v^3 must be negative. Thus, v < 0.
HOWEVER, don't get tripped up on 'S' - just because s^4 is positive, s could be positive or negative.
s*v*x = (?)*(-)*(+) is < 0 if s is positive and > 0 if s is negative.
Therefore, two alone is not enough.


Put them together:
V<0
X>0
S^4*V^3*X^7 = (+)*(-)*(+)<0, which they already tell you, so this gives you no new information. S could still be positive or negative (since it is an even power).

So, is s*v*x < 0? We can't determine this, because s could be negative or positive.


So, your answer is e - neither solution alone nor together.

Confusing?

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Re: Power [#permalink]

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New post 22 Jan 2010, 06:09
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rahulms wrote:
If \(s^4*v^3*x^7 < 0\), is \(svx < 0?\)

1) \(v < 0\)
2) \(x > 0\)


\(s^4*v^3*x^7 < 0\) says V and X are of different sign and S could be positive or negtive.

to find \(svx < 0?\)

Yes/No question type

V*X= negative and S=positive will give answer Yes

V*X= negative and S=negative will give answer No

Question stem = find whether S is positive or negative

1. V is negative but it doesn't say anything about S; Not SUFF

2. X is positive but it doesn't say anything about S; Not SUFF

1 & 2
Again no clue whether S is positive or negative; Not SUFF

Answer E

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Re: Power [#permalink]

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New post 09 Feb 2010, 14:46
If , S^4*v^3*x^7<0, is svx<0?

1) v<0
2) x>0

sol: consider only stmt 1, where v is negative but S is always positive, becoz of its even power, but X can be postive or negative, becoz of its odd power.
so svx can greater or lesser than zero. finally stmt 1 alone is not suffcient to answer
now consider stmt 2, where again S is always positive and V can be positive or negative, becoz of its odd power, and x is always postive.
so, SVX can again greater or lesser than zero. finally stmt 2 alone is not sufficient to answer

but combining both stmt 1 and stmt 2, S is always positive, V is always negative, and X is always positive, now SVX < 0 , can be answered,
both stmts combinely are sufficient to answer.

OA is C

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Re: Power [#permalink]

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New post 09 Feb 2010, 19:40
pradeepsrilatha wrote:
If , S^4*v^3*x^7<0, is svx<0?

1) v<0
2) x>0

sol: consider only stmt 1, where v is negative but S is always positive, becoz of its even power, but X can be postive or negative, becoz of its odd power.
so svx can greater or lesser than zero. finally stmt 1 alone is not suffcient to answer
now consider stmt 2, where again S is always positive and V can be positive or negative, becoz of its odd power, and x is always postive.
so, SVX can again greater or lesser than zero. finally stmt 2 alone is not sufficient to answer

but combining both stmt 1 and stmt 2, S is always positive, V is always negative, and X is always positive, now SVX < 0 , can be answered,
both stmts combinely are sufficient to answer.

OA is C

S is not always positive but S^2 is always positive

E is the answer

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Re: Power [#permalink]

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New post 10 Feb 2010, 16:58
i do not understand what is the need in solving this

the value of s is not given , so that means s can be 0 and s^2 will also be 0

so, E
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If s^4*v^3*x^7 < 0 , is svx < 0? 1) v < 0 2) x > [#permalink]

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New post 10 Feb 2010, 17:11
ichha148 wrote:
i do not understand what is the need in solving this

the value of s is not given , so that means s can be 0 and s^2 will also be 0

so, E


Yes, the answer is E. But since \(s^4*v^3*x^7 < 0\), then s cannot be zero, so we do need some "solving".
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Re: Power [#permalink]

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New post 10 Feb 2010, 17:48
oops , yes ,missed that , we do need to solve this :)
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Re: If s^4*v^3*x^7 < 0 , is svx < 0? 1) v < 0 2) x > [#permalink]

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New post 23 Sep 2016, 09:49
Definitely agree with E being the answer.

1) v<0 --> means x>0, but we still don't know value of s, thus INSUFFICIENT

2) x>0 --> means v<0, but we still don't know value of s, thus INSUFFICIENT

A,B,D eliminated...

(1) + (2) --> Still INSUFFICIENT as we don't know value of s - it can still be either positive or negative

D eliminated...

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Re: If s^4*v^3*x^7 < 0 , is svx < 0? 1) v < 0 2) x > [#permalink]

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Re: If s^4*v^3*x^7 < 0 , is svx < 0? 1) v < 0 2) x >   [#permalink] 28 Nov 2017, 22:08
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