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

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Intern
Joined: 29 Apr 2005
Posts: 5
Location: INDIA
Is t x u positive ? (1) (t/u)(t-u) > 0 (2) (t^2/u^3) > [#permalink]

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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: 1480
Location: Germany

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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: 1480
Location: Germany

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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: 545
Location: Canuckland

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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: 669

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24 May 2005, 10:12
E is the answer

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: 1480
Location: Germany

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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

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27 May 2005, 02:43
E.

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

---> answer is E.
27 May 2005, 02:43
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# Is t x u positive ? (1) (t/u)(t-u) > 0 (2) (t^2/u^3) >

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