LoyalWater wrote:

If vmt ≠ 0, is \((v^2)*(m^3)*(t^(-4)) > 0\)

(1)\(m > v^2\)

(2) \(m > t^(-4)\)

If vmt ≠ 0, is \(v^2*m^3*t^{-4} > 0\)

Lets analyse the question quickly

V*M*T is not zero ==> none of the digit is zero. But they can be -ve or +ve

IS \(v^2*m^3*t^{-4} > 0\)? ---------------> The expression \(>0\) means

positive; so the question is asking is the expression

positive ?

ok...\(v^2\) cannot be ever negative; squaring kills the polarity and always yield a positive number.

\(t^{-4}\) cannot be ever negative . \(t^{-4}\) is actually \(\frac{1}{t^4}\) ; Diving 1 by a positive number will yield a positive number (fraction or otherwise but definitely positive)

\(m^3\) can be positive or negative depending upon whether M is +ve or -ve

SO THE REAL QUESTION IS :- IS M POSITIVE?

(1)\(m > v^2\)

m = positive since squaring will give a positive number

SUFFICIENT

(2) \(m > t^{-4}\)

m will again be positive\(t^-4\) means \(\frac{1}{t^4}\); again positive fraction or integer

SUFFICIENT

ANSWER IS D

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