Min y-x = min y - max x = 3-2 = 1
Max y-x = max y - min x = 8-(-2) = 10

So range seems to be 1 through 10.

I can't make out the difference between options D and E though.

here's the OA:

you can easily solve this by subtracting the two inequalities. To do this they need to be in the opposite direction; when you subtract them preserve the sign of the inequality from which you are subtracting.

3 < y < 8
multiply the second one by (-1) to reverse the sign
2 > x > -2
Subtract them to get
3 - 2 < y - x < 8 - (-2)
1 < y - x < 10

Hey can anyone tell me wht is this difficult maths doc?

gmatguru wrote:Hey can anyone tell me wht is this difficult maths doc?

hi guru, you can find it on the gmat resource wiki. it's just a word doc with the toughest gmat practice problems

if -2=<x=<2 and 3<=y<=8
then x = {-2, -1, 0, 1, 2}
y= {3, 4, 5, 6, 7, 8}
now to get the range for y-x,
for a case study, let's consider a smaller set of numbers
given p= {2,3}
q= {4,5}
p-q = {(4-2), (4-3), (5-2), (5-3)
={2, 1, 3, 2}
={1, 2, 3}
we can see that the first term '1' is the difference between the least number in set 'q' - the ' maximum' number in set 'p'
Similarly, we can see that the last term '3' is the difference between the max. number in set 'q' - the ' min.' number in set 'p'

therefore, applying this into the question given the range is thus,
= (min 'y' - max 'x') =< y-x =< (max 'y' - min 'x')
=(3-2) =< y-x=<(8-(-2))
= 1=< y-x=< 10
{1=< y-x=< 10}