kakz wrote:If yz does not equal zero. Is 0<y<1?
(1)y<1/y
(2)y=z^2

(1)y<1/y

0<y<1 or y<0
you can take y = 1/2 and y = -2 as examples
Insufficient

(2)y=z^2
this means y will always be positive
Insufficient


Combining the two statements

y<0 would be eliminated because of the 2nd statement
only range left is 0<y<1



Option C

+1 C
st(1) is logically deconstructed as y<-1 and 0<y<1
st(2) implies x^2 is always positive
combining st(1&2): y can be only within range (0,1)
answer Yes, Sufficient both

y=-1/2 ?
-1/2 < -2 ?

rijul007 wrote: (1)y<1/y

0<y<1 or y<0
you can take y = 1/2 and y = -2 as examples
Insufficient

(2)y=z^2
this means y will always be positive
Insufficient


Combining the two statements

y<0 would be eliminated because of the 2nd statement
only range left is 0<y<1



Option C

rijul007 wrote:

kakz wrote:If yz does not equal zero. Is 0<y<1?
(1)y<1/y
(2)y=z^2

(1)y<1/y

0<y<1 or y<0
you can take y = 1/2 and y = -2 as examples
Insufficient

Not quite.. y<-1, not y<0.

Try -1/2 as an example.


Edit: oops PEMDAS already covered it.

chieftang wrote:

rijul007 wrote:

kakz wrote:If yz does not equal zero. Is 0<y<1?
(1)y<1/y
(2)y=z^2

(1)y<1/y

0<y<1 or y<0
you can take y = 1/2 and y = -2 as examples
Insufficient

Not quite.. y<-1, not y<0.

Try -1/2 as an example.


Edit: oops PEMDAS already covered it.

Oh ya, thanks for correcting guys...

kakz wrote:If yz does not equal zero. Is 0<y<1?
(1)y<1/y
(2)y=z^2

Statement 1: y < 1/y.
The CRITICAL POINTS are where y = 1/y or where the inequality is undefined.
y = 1/y when y=1 or y=-1.
y < 1/y is undefined when y=0.
When y is any other value, y < 1/y or y > 1/y.
Thus, there are 4 ranges to consider: y < -1, -1<y<0, 0<y<1, and y>1.

To determine the range of y, test one value to the left and right of each critical point.
If y = -2, then the inequality becomes -2 < -1/2.
This works. Thus, y<-1 is part of the range.

If y = -1/2, then the inequality becomes -1/2 < -2.
Doesn't work. Thus, -1<y<0 is not part of the range.

If y = 1/2, then the inequality becomes 1/2 < 2.
This works. Thus, 0<y<1 is part of the range.

If y = 2, then the inequality becomes 2 < 1/2.
Doesn't work. Thus, y>1 is not part of the range.

Thus, it's possible that y<-1 or that 0<y<1.
INSUFFICIENT.

Statement 2: y=z².
Since the square of a number cannot be negative, and it is given that yz≠0, we know that y>0.
Thus, it's possible that y=1/2 (in which case 0<y<1) or that y=2 (in which case y>1).
INSUFFICIENT.

Statements 1 and 2 combined:
The only range that satisfies both statements is 0<y<1.
SUFFICIENT.

The correct answer is C.

An algebraic way to determine the CRITICAL POINTS of y<1/y is to multiply by y².
Since y²≥0, we don't have to worry about changing the direction of the inequality:

y²(y) < y²(1/y)
y³ < y
y³ - y < 0
y(y² - 1) < 0
y(y+1)(y-1) < 0.

The CRITICAL POINTS are where y(y+1)((y-1) = 0:
y=-1, y=0 and y=1.

@Mitch, your *analyze it and see* is ideal for math extra-curricula (sorry) but not now and GMAT. Agree?
you are prompting for graphical solution here, which is cumbersome to build and complete within 1-2 min. Thereby, i was trying to deconstruct logically (which is actually the same process as you say and others put, combined and done mentally).

BTW, when you say function y=1/y is not defined for y=0 and this is already implied from the question as x*y cannot be zero

pemdas wrote:@Mitch, your *analyze it and see* is ideal for math extra-curricula (sorry) but not now and GMAT. Agree?
your are prompting for graphical solution here, which is cumbersome to build and complete within 1-2 min. Thereby, i was trying to deconstruct logically (which is actually the same process as you say and others put, combined and done mentally).

BTW, when you say function y=1/y is not defined for y=0 and this is already implied from the question as x*y cannot be zero

I would not encourage a graphical solution here.
For most test-takers, I would recommend plugging in different types of numbers to see which satisfy y < 1/y.
Recognizing the critical points (-1,0,1) is not necessary, but it can speed up the process by clarifying which types of numbers need to be tested.

Although the question stem precludes y=0, this value is still a critical point: while POSITIVE fractions (0<y<1) satisfy y<1/y, NEGATIVE fractions (-1<y<0) do not.

nice explantion Gmatguru!!!