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If v*m*t not = 0 , is v^2*m^3*t^{-4} > 0 ? 1) m>v^2 2) [#permalink]
 22 Jan 2010, 04:21
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If v*m*t not = 0, is v^2*m^3*t^{-4} > 0?
1) m>v^2
2) m>t^{-4}
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Sign of v^2m^3t^{-4} depends on m, because both x^2 and t^(-4) are always positive.
Each statement alone is sufficient t determine that m>0. So (D)
Last edited by shalva on 22 Jan 2010, 12:03, edited 1 time in total.
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shalva wrote: Sign of v^2m^3t^4 depends on m, because both x^2 and t^(-4) are always positive.
Each statement alone is sufficient t determine that m>0. So (D) Statement 1m > v^2So, m is positive as it's greater than v^2but, what about t? what if t's negative? couldn't t^{-4} be a negative number as it's ^{-4}? Statement 2m > t^{-4}again, t could be negative or positive. Combining the two statementswe do not know whether t is positive So, IMO it's E
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rahulms wrote: If v*m*t not = 0, is v^2*m^3*t^{-4} > 0?
1) m>v^2
2) m>t^{-4} V^2 will always be positive so V^2>0Same for T^{-4} so T^{-4} >0Question stem asks whether m is positive or negative (m>0 or m<0) Yes/No question 1. m>v^2Given V^2>0 so m is positive or m >0Ans Yes STATEMENT SUFFICIENT 2. m>t^{-4}given T^{-4}so T^{-4} >0so m is positive or m >0Ans Yes STATEMENT SUFFICIENT Answer D
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rahulms wrote: shalva wrote: Sign of v^2m^3t^4 depends on m, because both x^2 and t^(-4) are always positive.
Each statement alone is sufficient t determine that m>0. So (D) Statement 1m > v^2So, m is positive as it's greater than v^2but, what about t? what if t's negative? couldn't t^{-4} be a negative number as it's ^{-4}? Statement 2m > t^{-4}again, t could be negative or positive. Combining the two statementswe do not know whether t is positive So, IMO it's E Even power of any number is always positive... i.e. greater than zero. t^{-4} can be written as \frac{1}{t^{4}}result will be less than one but greater than zero.
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rahulms wrote: If v*m*t not = 0, is v^2*m^3*t^{-4} > 0?
1) m>v^2
2) m>t^{-4} Shalva's solution is correct. First of all: knowing that v*m*t\neq{0} implies that none of the variables equals to zero. Second of all: v^2*m^3*t^{-4}>0? --> \frac{v^2*m^3}{t^4}> 0? Now when this inequality holds true? Obviously as v^2 and t^4 are positive this inequality will hold true if and only m^3>0, or, which is the same, when m>0. (1) m>v^2 --> m is more than some positive number ( v^2), hence m is positive. Sufficient. (2) m>t^{-4} --> m>\frac{1}{t^4} --> Again m is more than some positive number ( \frac{1}{t^4} ), hence m is positive. Sufficient. Answer: D.
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agree with D
in both conditions the only variable that can change the sign of the expression is M. Since in both conditions we are told that M most be greater than some always positive number than in both condition we can answer the question if M is positive and thus the expression is positive.
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D it is!
thank you folks
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Answer is D.
A and B alone are sufficient to answer.
If m > v^2 => m has to be positive.
thus m * v^2 = +ve and t^-4 is going to be positive irrespective of value of t.
Hence A sufficient.
Similarly if m > t^-4 => m should be +ve.
Hence m * t^-4 will be +ve.
v^2 will be +ve irrespective of the value of v.
Hence B alone is also sufficient.
Thus answer is D.
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Need Solution for some DS problems from SET1 [#permalink]
23 Jun 2010, 12:30
I have problem in solving following DS questions from SET1 (I don't know the source of this). Can anyone help me in resolvign them?
DS Questions: Need help
Q5:
If vmt ≠ 0, is v2m3t-4 > 0?
(1) m > v2
(2) m > t-4
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D for Sure 
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