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# If #p# = ap3+ bp – 1............

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Manager
Joined: 11 Aug 2009
Posts: 129
Followers: 1

Kudos [?]: 36 [0], given: 3

If #p# = ap3+ bp – 1............ [#permalink]  18 Nov 2009, 20:44
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Question Stats:

100% (01:42) correct 0% (00:00) wrong based on 10 sessions
If #p# = ap3+ bp – 1 where a and b are constants, and #-5# = 3, what is the value of #5#?
5
0
-2
-3
-5
Manager
Joined: 28 Jul 2009
Posts: 127
Followers: 2

Kudos [?]: 11 [0], given: 12

Re: If #p# = ap3+ bp – 1............ [#permalink]  18 Nov 2009, 21:39
From the given equation,
-125a-5b=4
==> 125a+25b= -4 ....................... (1)

Now we have to find #5#

#5#= 125a+25b -1
From (1)
#5# = -4 -1 = -5

I hope I did it right.
Manager
Joined: 13 Aug 2009
Posts: 203
Schools: Sloan '14 (S)
Followers: 3

Kudos [?]: 79 [0], given: 16

Re: If #p# = ap3+ bp – 1............ [#permalink]  19 Nov 2009, 03:33
If #p# = ap3+ bp – 1 where a and b are constants, then we can simply this equation further:

#p# = ap3+ bp – 1
#p# = p*(a3+ b) – 1
And let (a3+b) equal some constant C:
#p# = p*(C) – 1

#-5# = 3 = -5*(C)-1
C=-4/5

#5# = 5*(-4/5)-1
#5# -5

Re: If #p# = ap3+ bp – 1............   [#permalink] 19 Nov 2009, 03:33
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