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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.
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.
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
Re: If s^4*v^3*x^7 < 0 , is svx < 0? 1) v < 0 2) x >
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03 Feb 2019, 04:59
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63% (01:27) correct 
