Is y greater than x? (1) |y|<1 and |x| <2 (2) y>0

From statement 1)

|y|<1 & |x|<2

|y|<1: This means two things: y<1 or -y<1.
-y<1 can be rewritten as y>-1 (when multiplied by -ve, the sign changes. In this case lesser than becomes greater than)
We can summarize as: -1<y<1

Similarly, we can summarize -2<x<2

This is all we know from statement 1. If you draw a number line, x and y could lie anywhere as long as they satisfy respective conditions.
For ex: Y=0 and X=1; In this case Y is less than X.
or Y=1 and X = -1; In this case Y is greater than X.

Hence statement 1 is insufficient. AD BCE

From statement 2)

y>0 and x is not 0

Again, we know nothing about the number except that y is positive and x can be positive or negative. Using simple numbers for X= -1 and Y =1, we can see that Y is indeed greater than X.
But if X= 100 and Y = 1, Y is less than X.

Hence statement 1 is insufficient. BCE

Now your chance of getting this question right is 50%. Taking 1 and 2 together:

We know Y > 0 and range of Y is -1<Y<1, combining these two we can write 0<Y<1; i.e Y is a positive fraction.
Similarly -2<x<2 but x cannot be 0.

Again, take X=-1/2 and Y=1/2; Y is greater than X.
X=1/2 and Y=1/4; Y is less than X.

Hence answer is E.

Statement 1, we cant do anything from this, as the modulus can be + or -, explained below
|y|<1
when y < 0, y > - 1
when y > 0, y < 1

|x| <2
when x < 0, x > -2
when x > 0, x < 2

Statement 2, y>0 and x is not 0, here x can be +ive or -ive

Combine both the statements, y < 1 and x will be still +ive or -ive

Is y greater than x,
it can be a Yes when y = 0 and x = -1
it can be a No when y = -1 and x = -1

Answer E.

One can visualize the range of x and y from statement 1 and 2 on the real number line. The reasoning will remain the same as that of whiplash's.

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