IS m+z>0 1) m-3z>0 2) 4z-m<0 1) m>3z For z<0: m+z can be +ve as well as -ve Insufficient 2) m>4z Again For z<0: m+z can be +ve as well as -ve Insufficient Combining the 2 conditions we have m>3z & m>4z. These conditions will be true simultaneously only when m>4z. Which essentially gives us the s...

Sorry Warlock,

I did a mistake while combining the two conditions.
When we combine the 2 conditions we get
y-2<x<2y-6
Please note that it can only be true when y>4

Also for every value of y>4, x>2
Thus xy>0 always.

So C should be the answer.

Apologize for the mistake.

1. x-y>-2 => x>y-2 For y=1 we get x=-1 thus xy<0 For y=3 we get x=1 thus xy>0 Insufficient 2. x-2y<-6 => x<2y-6 For y=1 we get x=-4 thus xy<0 For y=4 we get x=2 thus xy>0 Insufficient Combining 1 and 2 gives x>y-2 and x<2y-6 For y=1 we get -1<x and x<-4 thus for -ve value of x, xy<0 For y=4 we get x...

The solution can be generalized as: 1. y<0 For the given condition to be true z>0 but x<0 or x>0. If x<0 then xyz>0 else if x>0 then xyz<0 2. z>0 Implies y<0 from the given condition. Again x<0 or x>0. Which again gives xyz>0 or xyz<0 3. Combining the 2 cases doesn't help as x<0 or x>0 So answer is E.

Thanks that rationalizing is certainly better..
Saves time..

2^6=64
So 0.000064 = (0.2)^6
Hence the value (0.2)^6*(1/6) = 0.2

Hope this is helpful.