Inequality

[colakumarfanta]

Q x, y, z, are integers Is xy <1 ?
1) xyz > 1
2) xyz ^2 > 1

[debmalya_dutta]

1) xyz > 1 = > insufficient by itself
x=1/2 ;y=4 ;z = 3 here xy => 1
x=1/2 ;y=1/3 ;z = 8 here xy <1
2) xyz ^2 > 1 => again insufficient
Together is insufficient ;

option e

[Rahul@gurome]

Solution
This is being solved assuming the second statement is xy(z^2) > 1
The question is Is xy < 1? Let xy be "a". Since both x and y are integers, xy or "a" is also an integer.
Consider (1) alone.
It says xyz > 1or (xy)z >1 or az > 1.
Since both a and z are integers this is possible only if both are positive or both are negative.
Take an example. Let a = 1 and z = 6. Then az = 1*6 = 6 > 1.
Let a = -1 and z = -6. Then az = 6 > 1
So if a >0, we have that a >= 1, because it is an integer and so xy >= 1.
If a < 0, we have that a <= 1, because it is an integer and so xy<=1.
Since nothing definite can be said about xy, (1) is not enough.
Consider (2) alone.
If xy(z^2) > 1, we have that a(z^2) > 1. Since z^2 is positive, being a perfect square, a >= 1 or xy >= 1. So xy is not less than 1.
Hence (2) is enough.

The correct answer is (B).

[debmalya_dutta]

BTW ...I was thinking that the option 2 was more in line with (XYZ)^2 and not (XY)Z^2

if its the later case and (XY)Z^2 >1
if X = 1/2 and y = 1/4 and Z= 10 would be a scenario where XY <1
and X = 10 and y = 1/2 and Z= 1/4 would be a scenario where XY >1

So, would I be incorrect in sticking to my initial choice of "e"

[Ashish8]

I also solved this like debmalya where I'm assuming option 2 is (xyz)^2.

Then I got E

If its xy(z)^2 > 1.

Then xy can be positive or both negative and Z can be positive or negative

Then X * Y > 0.

Answer: B

[vcb]

@debmalya:

The question stem states that x,y,z are integers..So you cant consider x=1/2 and y=1/4!

[debmalya_dutta]

Right .
@vcb - oversight at my end .thanks for correcting me

So, assuming that it is actually xy(z)^2 >1 and not (xyz)^2 > 1
z ^2 is always positive
so , for the product of xy(z)^2 to be greater than 1 , xy > 1 and in that case , B is sufficient to answer the question