Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack
GMAT Club

 It is currently 28 Mar 2017, 09:08

### 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

# Is t x u positive ? (1) (t/u)(t-u) > 0 (2) (t^2/u^3) >

Author Message
Intern
Joined: 29 Apr 2005
Posts: 5
Location: INDIA
Followers: 0

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

Is t x u positive ? (1) (t/u)(t-u) > 0 (2) (t^2/u^3) > [#permalink]

### Show Tags

24 May 2005, 06:12
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Is t x u positive ?

(1) (t/u)(t-u) > 0
(2) (t^2/u^3) > 0
VP
Joined: 30 Sep 2004
Posts: 1487
Location: Germany
Followers: 6

Kudos [?]: 336 [0], given: 0

### Show Tags

24 May 2005, 06:49
C)...will explain if correct
_________________

If your mind can conceive it and your heart can believe it, have faith that you can achieve it.

VP
Joined: 30 Sep 2004
Posts: 1487
Location: Germany
Followers: 6

Kudos [?]: 336 [0], given: 0

### Show Tags

24 May 2005, 07:51
sparky wrote:
E.

why not C) ?

1) t/u * t-u > 0 => (t(t-u))/u > 0 => t(t-u) > 0 or t(t-u) < 0 => insuffcient

2) t^2/u^3 > 0 => insufficient but u must be +ve

1)+2) u is +ve so => t(t-u) > 0 => t>0 or t>u.we know from 2) that u is +ve so u * t is +ve.

correct me if i am wrong. thx.
_________________

If your mind can conceive it and your heart can believe it, have faith that you can achieve it.

Director
Joined: 18 Apr 2005
Posts: 547
Location: Canuckland
Followers: 1

Kudos [?]: 37 [0], given: 0

### Show Tags

24 May 2005, 07:57
christoph wrote:
sparky wrote:
E.

why not C) ?

1) t/u * t-u > 0 => (t(t-u))/u > 0 => t(t-u) > 0 or t(t-u) < 0 => insuffcient

2) t^2/u^3 > 0 => insufficient but u must be +ve

1)+2) u is +ve so => t(t-u) > 0 => t>0 or t>u.we know from 2) that u is +ve so u * t is +ve.

correct me if i am wrong. thx.

what if in '1)+2)' t<0 and u>0 ?
Director
Joined: 18 Feb 2005
Posts: 672
Followers: 1

Kudos [?]: 6 [0], given: 0

### Show Tags

24 May 2005, 10:12

The question is

Is t x u positive ?

(1) (t/u)(t-u) > 0
(2) (t^2/u^3) > 0

From 1) we have nothing.....

From 2) we have (t/u)^2*(1/u) > 0

This means " u " is always positive because (t/u)^2 is always >0
Combine 1 and 2

From 2 we know that u is positive ....But t may be a -ve number and the product (t/u)(t-u) is still positive

and t may be positive also.....

So it is a straight E
VP
Joined: 30 Sep 2004
Posts: 1487
Location: Germany
Followers: 6

Kudos [?]: 336 [0], given: 0

### Show Tags

25 May 2005, 01:57
christoph wrote:
sparky wrote:
E.

why not C) ?

1) t/u * t-u > 0 => (t(t-u))/u > 0 => t(t-u) > 0 or t(t-u) < 0 => insuffcient

2) t^2/u^3 > 0 => insufficient but u must be +ve

1)+2) u is +ve so => t(t-u) > 0 => t>0 or t>u.we know from 2) that u is +ve so u * t is +ve.

correct me if i am wrong. thx.

another !?""**? mistake

1) + 2) i concluded above that t(t-u) > 0 => t>0 or t>u => that is wrong. it is either both t>0 and t>u OR both t<0 and t<u. insufficient.
_________________

If your mind can conceive it and your heart can believe it, have faith that you can achieve it.

Intern
Joined: 10 May 2005
Posts: 2
Followers: 0

Kudos [?]: 2 [0], given: 0

### Show Tags

27 May 2005, 02:43
E.

(2)---> u>0
(1) --> ut < (u/t^2) --> maybe ut < 0 or ut > 0

27 May 2005, 02:43
Display posts from previous: Sort by