>0?

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by eagleeye » Thu Jun 28, 2012 5:53 pm
alex.gellatly wrote:If vmt does not equal 0, is v^2m^3t^-4 > 0?

1) m>v^2
2) m>t^-4

Thanks
Hi Alex:

Here's a fast way of doing this:

We are given vmt is not 0. This means that none of v,m, or t equal 0.
Second we are asked if (v^2)*(m^3)*(t^-4) > 0.

We can rewrite (v^2)*(m^3)*(t^-4) as (v^2)*(m^2)*(t^-4)*m

When numbers have even powers, they are always positive (provided they are not equal to 0, which we already established).
So v^2, m^2, t^-4 are all positive in this case.
Therefore, the sign of the whole expression is determined by whether m is greater than or smaller than 0. That's what we will look for.

With that in mind, let's look at the statements:

1) m>v^2
Since m > v^2 and v^2 is positive, m>0. Hence the expression is +ve. Sufficient.

2) m>t^-4
Since m > t^-4 and t^-4 is positive, m is positive as well. Sufficient.

Hence the correct answer is D.

Let me know if this helps :)

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by sandeep_thaparianz » Sat Jun 30, 2012 5:27 am
+1 ForD