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If P# =P/P-1 what is P##? P/(P-1) 1/P P 2 - P P

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Senior Manager
Joined: 14 Jun 2007
Posts: 397

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If P# =P/P-1 what is P##? P/(P-1) 1/P P 2 - P P [#permalink]

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23 Jun 2007, 14:29
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If P# =P/P-1 what is P##?

P/(P-1)
1/P
P
2 - P
P - 1

Kudos [?]: 15 [0], given: 0

Senior Manager
Joined: 04 Jun 2007
Posts: 345

Kudos [?]: 33 [0], given: 0

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23 Jun 2007, 22:27
anonymousegmat wrote:
If P# =P/P-1 what is P##?

P/(P-1)
1/P
P
2 - P
P - 1

p## = (p/(p-1))# = [p/(p-1)]/[[p/(p-1)]-1]=[p/(p-1)]/[1/(p-1)]=p

Kudos [?]: 33 [0], given: 0

Senior Manager
Joined: 14 Jun 2007
Posts: 397

Kudos [?]: 15 [0], given: 0

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24 Jun 2007, 09:31
sumande wrote:
anonymousegmat wrote:
If P# =P/P-1 what is P##?

P/(P-1)
1/P
P
2 - P
P - 1

p## = (p/(p-1))# = [p/(p-1)]/[[p/(p-1)]-1]=[p/(p-1)]/[1/(p-1)]=p

sumande,

thanks. i was trying to factor and do all sorts of weirdness; essentially we just replace each p with p/p-1

thats not so tough

i ended up with:

(p/p-1)# = p/p(1-(1/p))# = 1/(1-1/p)# 1- 1/(p/p-1) = -1/p

Kudos [?]: 15 [0], given: 0

Director
Joined: 06 Sep 2006
Posts: 734

Kudos [?]: 49 [0], given: 0

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24 Jun 2007, 14:42
P## = [ P#/(P#-1) ] ....X

P# = P/(P-1) -----A
So, P# - 1 = P/(P-1) - 1 = P - P + 1/(P-1)
P#-1 = 1/(P-1) .....B

Plugging A & B in X.

P## = [P/(P-1)] / [1/(P-1)]

= P/(P-1) * (P-1)

= P.

C.

Kudos [?]: 49 [0], given: 0

24 Jun 2007, 14:42
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