kanusha wrote:
Can any one solve this in simpliest Way,helpful for me
Thank you.
vmt ≠ 0, so we know that v, m, and t must all be either a positive or negative number, but cannot be 0.
(1) The value of m is greater than v^2. Whether we set v at a positive or negative number, it's square will be positive.
m has to be positive because it must be greater than v. With two positive numbers multiplied together, the only way that v^2m^3t^4 can be less than zero is if t^4 was negative. Since it is multiplied by an even exponent, t^4 will be a positive number regardless if t is negative or positive. Therefore, if (1) is valid, the solution can only be positive, for a "yes" response. Sufficient.
(2) If the value of m is greater than t^4, then both numbers must be positive, and again since v^2 is an even exponent, the square will be a positive number. Again, the only valid solution is a greater than zero, resulting in only a "yes" response. This is similar to (1). Sufficient.
Both statements are sufficient alone.
Answer: D