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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
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Is t x u positive ? (1) (t/u)(t-u) > 0 (2) (t^2/u^3) > [#permalink]  24 May 2005, 05:12
Is t x u positive ?

(1) (t/u)(t-u) > 0
(2) (t^2/u^3) > 0
VP
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[#permalink]  24 May 2005, 05:49
C)...will explain if correct
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VP
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[#permalink]  24 May 2005, 06:51
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.
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Director
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[#permalink]  24 May 2005, 06:57
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
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[#permalink]  24 May 2005, 09:12
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
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[#permalink]  25 May 2005, 00:57
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.
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Intern
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[#permalink]  27 May 2005, 01:43
E.

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

---> answer is E.
[#permalink] 27 May 2005, 01:43
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