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p/q<1

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Joined: 24 Jul 2010
Posts: 88

Kudos [?]: 13 [0], given: 43

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05 Apr 2013, 09:15
Can I algebraically prove that if p/q<1 then q/(p^2) is not necessarily greater than 1? Picking numbers is easy, I am just wondering if I can algebraically solve it.

Kudos [?]: 13 [0], given: 43

Manager
Status: Looking to improve
Joined: 15 Jan 2013
Posts: 174

Kudos [?]: 74 [0], given: 65

GMAT 1: 530 Q43 V20
GMAT 2: 560 Q42 V25
GMAT 3: 650 Q48 V31

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05 Apr 2013, 09:47
Not very clear on the problem without specifics about whether p & q are positive integers

Because if p > 0 and q > 0 and integers then p/q < 1 => q/p > 1 and q/p2 can be >=< 1
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Kudos [?]: 74 [0], given: 65

Manager
Joined: 24 Jul 2010
Posts: 88

Kudos [?]: 13 [0], given: 43

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05 Apr 2013, 12:01
nt2010 wrote:
Not very clear on the problem without specifics about whether p & q are positive integers

Because if p > 0 and q > 0 and integers then p/q < 1 => q/p > 1 and q/p2 can be >=< 1

Yes sorry, p and q are positive integers. When you say can q/p2 be greater or less than 1, is there a way we can reach that algebraically.

Kudos [?]: 13 [0], given: 43

Manager
Status: Looking to improve
Joined: 15 Jan 2013
Posts: 174

Kudos [?]: 74 [1], given: 65

GMAT 1: 530 Q43 V20
GMAT 2: 560 Q42 V25
GMAT 3: 650 Q48 V31

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05 Apr 2013, 13:32
1
KUDOS
q/p > 1 so divide both side by p since P is a positive integer P > 0

q/p2 > 1/p. IFF q > p

//kudos, please if this explanation is good
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KUDOS is a way to say Thank You

Kudos [?]: 74 [1], given: 65

Manager
Joined: 24 Jul 2010
Posts: 88

Kudos [?]: 13 [0], given: 43

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05 Apr 2013, 13:39
nt2010 wrote:
q/p > 1 so divide both side by p since P is a positive integer P > 0

q/p2 > 1/p. IFF q > p

//kudos, please if this explanation is good

One last question, why do you say q/p2 > 1/p if q>p? How did you get to q>p?

Kudos [?]: 13 [0], given: 43

Manager
Status: Looking to improve
Joined: 15 Jan 2013
Posts: 174

Kudos [?]: 74 [0], given: 65

GMAT 1: 530 Q43 V20
GMAT 2: 560 Q42 V25
GMAT 3: 650 Q48 V31

Show Tags

05 Apr 2013, 13:52
Remember we got to the equation q/p2 > 1/p from q/p > 1 and if you multiple both side by p you'll get the condition q > p.

Hope this is clear.
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Kudos [?]: 74 [0], given: 65

Re: p/q<1   [#permalink] 05 Apr 2013, 13:52
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