This is a moderately difficult, ‘could be’ kind of question. In a ‘could be’ knid of question, the strategy is always to find ONE case to make a statement true, ONCE. After all, we are not trying to prove that the statement IS true, but we are only trying to figure out if there are cases which CAN make it true.
So, the trick is to take simple values in and around Zero to try and see if the statements can be made true.
We know that y is a positive number. xz>0 means that x and z are either both positive or both negative. However, in all the three statements, x and z are shown to be greater than ZERO. So, it would not be wrong for us to consider that x and z are both positive.
But while we consider positive values, we will have to be careful to pick values from all ranges, especially from the 0 to 1 range which contains proper fractions.
Also, let’s keep in mind that |x| < |y+2| < |z| , so the values that we pick should also satisfy the above inequality.
To test statement I, let’s take y = ½, x = 1 and z = 3. For these values, xz>0 and |1| < | 2.5| < |3|. So, statement I could be true. Therefore, options B and C can be eliminated.
To test statement II, let’s take x = 1, y = 2 and z = 5. For these values, xz>0 and |1| < | 4 | < |5|. Statement II could be true. Now, options A and D can be eliminated.
The only option left is E and this has to be the answer. However, let’s test statement III by taking x = 2, y = 1/10 and z = 3. For these values, xz>0 and |2| < |2.1| < |3|. Statement III could be true.
Taking values and proving statements true, is the way forward in ‘could be’ type of questions. Also look at the statements to see if they have a few clues on offer.
Hope this helps!
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60% (02:19) correct 
