Set 1 : Q5 Number Properties

[Gmater-1]

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

I got A but Official Answer is

SPOILER: D

I think option B also confirms that m is positive

[nazar]

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

My approach:
Here, the things that should be taken into consideration:
1. v^2 is positive whatever may be the value of v.
2. t^-4=1/t^4 is positive.
3. m^3 may be positive or negative.

Actually, whether m is negative or positive has been asked?

Stem1:
m>v^2, meaning that m is +ve. SUFF.

Stem2:
m>t^-4 meaning that m is +ve. SUFF.

Answer is D.

[mannu08]

I was stuck with the question.
I guess Nazar exp makes sense.

[larbgai]

Any number to an even exponent is positive.

Don't even think about what the number is, because it is totally irrelevant.

m>v^2 =>
m > some positive number. If it is greater than a positive number, it is positive! 1 is sufficient.

2) m>t^(-4)
=> m > 1/(t^4)
=> m> 1/(some positive number)
..... 1/(positive number) = positive number

=> m> positive number

=> m is positive

Statement 2 is sufficient.

Answer is D.

[winna.7]

M > v2 means M is +ve M > 0.SUFF
M > 1/t64 = >M is +ve M > 0 again SUFF

SO answer is D!

[Mba_asp]

Good question. Answer is D. Agree with above explanations.