If S is a data set of three numbers x, y and z, is the range of the nu

I think question stem should tell us whether statements can be negative or not. Otherwise it is time-consuming.

Let's analyze

Case 1
Stmtnt 1: x-y<1 If this can only be positive, I mean between 0 and 1, then x>y. Insufficient

Stmtnt 2 : y-z<2 again, if this can only be positive, between 0 and 2, then y>z. Insufficient

If we combine both conclusions we'll get x>y>z meaning that we can write it in ascending order {z,y,x}. In this case range will be x-z. And if we solve both equations:

x-y<1
y-z<2

We add them and get x-z<3 , can be less than 2 or not. Insufficient Ans E

Case 2

If both statements can also be negative.we should check negative cases

Stmtnt 1: x-y<1 Assume that this is negative,then x<y, Insufficient

Stmtnt 2 :y-z<2 again, Assume that this is negative, then y<z, Insufficient

If we combine both we get x<y<z and write it in ascending order {x,y,z}. In this case range will be z-x. If x-z<3,then we flip it z-x>-3. Again we don't know exactly if it is less than 2 or not. Insufficient Ans E

Case 3

Now, we need to check another scenario 1st statement positive, 2nd negative.

x-y<1 can only be positive: x>y Insufficient

y-z<2, is negative: y<z Insufficient

Combining them x>y ,y<z then x>z. In ascending order it will be {y,z,x}, range will be x-y which less than 1,also less than 2. [b]Sufficient. Ans C[/b]

Case 4

1st stmnt negative, 2nd positive

x-y<1 negative, x<y Insufficient

y-z<2 positive, y>z Insufficient

Combining them x<z<y In ascending order it will be {x,z,y}, range y-x. If x-y<1 then y-x>-1. Again, we don't know the exact answer. Insufficient Ans E

I don't know what I'm missing here. But it seems one can't do this within 2 mins.