I'll give this a shot. In order for 1/x to be > 1, x must be 0<x<1. So we can rephrase the stem by saying is 0<x<1?

A) Insufficient because we know nothing about y

if Y = 2, x could be 1
if Y = -2, x could be -1

B) Sufficient

In order for x^3 to be > x^2, X must be > 1

for example

if x = -1, (-1)^3 = -1 which is not greater than (-1)^2 =1
if x = 1/2, (1/2)^3 = 1/8 which is not greater than (1/2)^2 = 1/4
So x must be >1

So B

we need to determine if the statement is sufficient to answer
1/x > 1?


(1) y/x > y .

=>1/x >1
so (1) is sufficient

(2) x^3 > x^2 .

x^3 - x^2 >0
=>x^2 (x-1) > 0
=>x^2 > 0 or x-1 > 0

so (2) is insufficient

I think A should be the answer

[email protected] wrote:
(1) y/x > y .

=>1/x >1
so (1) is sufficient

I don't believe that this is accurate. We don't know the sign of y so we can't divide y on both sides of the equation. If y were negative we would have to flip the sign.

I would be interested in seeing the OA

IMO B is right here

St1 Here the value of x depends upon the value of y. Let's see three scenarios below.

a. y = 0; x can be anything other than 0
b. y is -ve; x can be +ve or -ve (y= -2; x = 2; -1 > -2. OR y = -3; x = -3; 1>-3)
c. y is +ve; x must be greater than 0 and less than 1 (y=2; x=0.5; 4>2)

St 2 x^3 > x^2. If x is -ve then x^3 must be less than x^2 so given condition will be wrong. It is not correct. if x is greater than 0 and less than 1 then x^3 will be less than x^2. If x = 1 then x^3 will be equal to x^2. In all these three scenarios given sentence fails. Therefore, to satisfy given condition or statement x must be greater than 1, and this is the correct answer

What is OA?

rockeyb wrote:If x is not equal to zero , is 1/x > 1 ?

(1) y/x > y .

(2) x^3 > x^2 .

source : Kaplan.

btw I am new here...what does "ve" mean?

I guess you mean to say what is -ve or +ve means right....it is short form of negative or of positive

PeterFaulkner wrote:btw I am new here...what does "ve" mean?

PeterFaulkner wrote:btw I am new here...what does "ve" mean?

Peter welcome to this forum . I am sorry I should have typed in the complete positive instead of just +ve . It never occurred to me that some one might not realize what +ve stands for ? Sorry for that once again .

Your explanation above is perfect and [spoiler]OA = B.[/spoiler]

rockeyb wrote:

PeterFaulkner wrote:btw I am new here...what does "ve" mean?

Peter welcome to this forum . I am sorry I should have typed in the complete positive instead of just +ve . It never occurred to me that some one might not realize what +ve stands for ? Sorry for that once again .

Your explanation above is perfect and [spoiler]OA = B.[/spoiler]

Thanks for the warm welcome!

This is how i got started :

1) y\x>y
If you divide both sides by 1\y =>1\x>1 , wont that be sufficient as well?? - correct me if i am wrong

2)also sufficient

So the answer would be D??[/list][/i]

nisha.menon294 wrote:This is how i got started :

1) y\x>y
If you divide both sides by 1\y =>1\x>1 , wont that be sufficient as well?? - correct me if i am wrong

If You divide the inequality by a negative number the direction of the inequality Changes

For example, if -5x > -5y then x<y (direction of the inequality changes)

In this case you do not y is negative or positive so, we can't be sure about the sign

So A is not sufficient

nisha.menon294 wrote:This is how i got started :
1) y\x>y
If you divide both sides by 1\y =>1\x>1 , wont that be sufficient as well?? - correct me if i am wrong2)also sufficient
So the answer would be D??[/list][/i]

Fell for the trap. Never multiply or divide both sides on an inequality by an unknown. x and y are unkown till we solve the eqs.
If you have -2x < 6 y can divide by -2 , but there is more to it.
PeterFaulkner post is enough to rule out 1.)

@ [email protected] x^3 > x^2 x^3 - x^2 >0
=>x^2 (x-1) > 0 =>x^2 > 0 or x-1 > 0 so (2) is insufficient

why x² is always +ve then (x-1) has to be +ve otherwise x^2 (x-1) < 0
so x-1 > 0 or x > 1 so 1/x < 1 Sufficient
I like you method but interpreted it differently. Thanks to all for helping me solve this.