Algebra: test cases

[SoumyaV658]

If y > 0, what is the value of y?
(1) y2 < y
(2) y is an integer.

In above question from (1) we can consider the values between 0 and 1 for y; from combining both 1&2 there is no integer between 0 and 1, so we can conclude 'E' as solution since we cannot solve by using both 1&2.......but below is the solution mentioned
''.......soution mentioned in the book
(2) INSUFFICIENT: The statement indicates that y is an integer. The value
of y could be 1, 2, 14, 192, or any other positive integer.
(1) INSUFFICIENT: What numbers make y2 ≤ y true?
Case 1: If y = 1, then 1 ≤ 1. Therefore, 1 is a possible value for y.
Case 2: If , then . Therefore, is a possible value for y.
There are at least two possible values for y.....'''

where they considered y^2 <= y....here how can we consider equality?

[Sage Pearce-Higgins]

As our forum rules state, please can you cite the source of this problem? It's hard for me to comment precisely without seeing the text of the problem as we present it.

You're right that, as you've written them, the two statements don't allow for any values of y. This is non-nonsensical: in DS problems the 2 statements are always consistent, so that there should be at least one possible value of y that satisfies both statements. This makes me suspect that either you've transcribed the problem inaccurately - the solution suggests that statement (1) should be "y2 ≤ y" - or there's an error in the way the problem's written. Please let me know where this problem's from and hopefully I can help clear up the confusion.

[SoumyaV658]

This problem is from Algebra- manhattan(6th edition), 1st problem set in chapter 10.

[Sage Pearce-Higgins]

Thanks for letting me know. Unfortunately there's an error in the ebook (I presume that what you're working from). That explains why the problem and the explanation have different statements. Take a look here: https://www.manhattanprep.com/gmat/errata/errata-algebra-6ed/