(1) if X>1 and Y>1, is X<Y?

(1) X^2/(XY+X)<1
(2) XY/Y^2-Y<1

I chose D, but OA is E ,why????

Stat. (1) X^2/(XY+X) = x^2/x(y+1) = x/y+1<1
Multiply times y+1 (you can do that because it's positive, since y>1)

x<y+1

so if x is smaller than y+1, does that mean that x is smaller than y? Not necessarily:
If y=2, then y+1=3.
x could be 1.5, or 2, or 2.5, any value that is smaller than y+1=3: These examples show that x can be smalelr, equal to or greater than y.

Stat. (2) is not so clear - is the -y in the denominator, or outside of the fraction? :

If the latter

XY/Y^2-Y = x/y - y = (x-y^2)/y < 1
Multiply times y (which is positive, since y>1)

x-y^2<y
x<y+y^2.

Since both x and y are greater than 1, x will indeed be smaller than y, and stat. (2) would be sufficient.

If the former xy/(y^2-Y) = xy/y(y-1) = x/(y-1) < 1
Multiply times y-1 (which is positive, since y>1) to get
x<y-1
And again, if x is smaller than y-1, then x is definitely smaller than y, and stat. (2) would be sufficient.

In any case, the answer seems to be B, not E.

Q2: 4 books or 3?

Q3: Stat. (1) and (2) are both insufficient because m can be 100 (and then m^3 = 10^6, with 7 digits) or 400 = 4*10^2 (in which case m^3 = 64*10^6) = 8 digits)

Q4: You need stat. (1) to fix them as integers, otherwise for stat. (2) x can be 2 and y=4.5, in which case (1+4.5)^3 is not an even integer - it is not an integer at all.

If x and y are integers, then the only possible options for xy=9 are x=3, y=3, or x=1, y=9 - bot of which will yield an even integer for (x+y)^3.