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22 Jan 2010, 05:33
Kudos
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
61% (01:23) correct
39%
(01:03)
wrong
based on 292
sessions
History
Date
Time
Result
Not Attempted Yet
22 Jan 2010, 06:40
Kudos
Bookmarks
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?
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?
General Discussion
22 Jan 2010, 07:09
Kudos
Bookmarks
rahulms
If \(s^4*v^3*x^7 < 0\), is \(svx < 0?\)
1) \(v < 0\)
2) \(x > 0\)
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
