ps Equation : PS Archive
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# ps Equation

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Senior Manager
Joined: 17 Mar 2009
Posts: 309
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Kudos [?]: 473 [0], given: 22

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08 Aug 2009, 08:55
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50% (02:02) correct 50% (03:02) wrong based on 7 sessions

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If x and y are positive and x^2y^2 = 18 – 3xy, then x^2 =?

A) (18 – 3y )/y3
B) 18/y2
C) 18/(y2 + 3y)
D) 9/y2
E) 36/y2

OA:
[Reveal] Spoiler:
D

Senior Manager
Joined: 17 Mar 2009
Posts: 309
Followers: 9

Kudos [?]: 473 [0], given: 22

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08 Aug 2009, 17:48
can any one solve it..
Manager
Joined: 18 Jul 2009
Posts: 169
Location: India
Schools: South Asian B-schools
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Kudos [?]: 98 [2] , given: 37

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08 Aug 2009, 18:28
2
KUDOS
OA Dassume xy = a thus we have a quadratic eqn a^2+3a-18=0

roots of this eqn are (a-3)(a+6) => this gives xy = 3 or xy= -6
since x & y are +ve ....xy = -6 is negated..as x or y have to be -ve to have xy = -ve

thus xy =3 this x = 3/y...thus x^2 = 9/y^2...OA D
_________________

Bhushan S.
If you like my post....Consider it for Kudos

Senior Manager
Joined: 17 Mar 2009
Posts: 309
Followers: 9

Kudos [?]: 473 [0], given: 22

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08 Aug 2009, 18:55
Thanks for the solution... i got it..
Re: ps Equation   [#permalink] 08 Aug 2009, 18:55
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