It is currently 23 Sep 2017, 06:09

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# Q: Is x > y ? (1) x(x-1) > xy (2) (x-y)^3 > 0

Author Message
Senior Manager
Joined: 20 Feb 2007
Posts: 256

Kudos [?]: 61 [0], given: 0

Q: Is x > y ? (1) x(x-1) > xy (2) (x-y)^3 > 0 [#permalink]

### Show Tags

09 Mar 2007, 19:48
1
This post was
BOOKMARKED
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Q: Is x > y ?

(1) x(x-1) > xy
(2) (x-y)^3 > 0

Solution:
(1) x(x-1) > xy
(x-1) > xy / x
x-1 > y
x > y+1 so x > y (A is SUFF)

(2) (x-y)^3 > 0

now x-y can not be negative because cube-root of negative is -ve. So x is either more than y or equal to y. x>=y

If x = y then e.g. (3-3)^3 > 0 => 0 > 0 which is NOT TRUE so B is NOT SUFF.

My answer is A alone SUFF.

Kudos [?]: 61 [0], given: 0

Director
Affiliations: FRM Charter holder
Joined: 02 Dec 2006
Posts: 727

Kudos [?]: 94 [0], given: 4

Schools: Stanford, Chicago Booth, Babson College

### Show Tags

09 Mar 2007, 22:57
Summer3 wrote:
Q: Is x > y ?

(1) x(x-1) > xy
(2) (x-y)^3 > 0

Solution:
(1) x(x-1) > xy
(x-1) > xy / x
x-1 > y
x > y+1 so x > y (A is SUFF)

(2) (x-y)^3 > 0

now x-y can not be negative because cube-root of negative is -ve. So x is either more than y or equal to y. x>=y

If x = y then e.g. (3-3)^3 > 0 => 0 > 0 which is NOT TRUE so B is NOT SUFF.

My answer is A alone SUFF.

Hey Summer, why don't u please post just the question and then compare responses with your answer later? That way people will respond. If you post answers, how will we have fun in solving??

Kudos [?]: 94 [0], given: 4

Intern
Joined: 13 Jun 2005
Posts: 28

Kudos [?]: 42 [0], given: 0

### Show Tags

10 Mar 2007, 01:22
I'd go with B.

(1)
x(x - 1) > x.y
x [ x - y - 1] > 0

if x > 0, x > y + 1 => x > y
if x < 0, x - y- 1 < 0 => x < y + 1 => x < y
(insuff)

(2)
(x - y) ^ 3 > 0
=> x - y needs to be +ve => x > y

Kudos [?]: 42 [0], given: 0

Intern
Joined: 02 Mar 2007
Posts: 30

Kudos [?]: [0], given: 0

### Show Tags

10 Mar 2007, 01:45
Summer3 wrote:
Q: Is x > y ?

(1) x(x-1) > xy
(2) (x-y)^3 > 0

Solution:
(1) x(x-1) > xy
(x-1) > xy / x
x-1 > y

dred is right. Summer3, u divide by x without knowing whether it is positive or negative. If it is negative, the sign of the inequality will change.
I go for B)

Kudos [?]: [0], given: 0

SVP
Joined: 01 May 2006
Posts: 1794

Kudos [?]: 165 [0], given: 0

### Show Tags

10 Mar 2007, 01:58
Agree with (B).... Same approach as Dred

Kudos [?]: 165 [0], given: 0

10 Mar 2007, 01:58
Display posts from previous: Sort by