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If #p# = ap3+ bp – 1............

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If #p# = ap3+ bp – 1............ [#permalink]

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New post 18 Nov 2009, 21:44
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If #p# = ap3+ bp – 1 where a and b are constants, and #-5# = 3, what is the value of #5#?
5
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-2
-3
-5
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Re: If #p# = ap3+ bp – 1............ [#permalink]

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New post 18 Nov 2009, 22:39
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.
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Re: If #p# = ap3+ bp – 1............ [#permalink]

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New post 19 Nov 2009, 04:33
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

ANSWER: E. -5
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Re: If #p# = ap3+ bp – 1............ [#permalink]

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New post 09 Nov 2015, 16:34
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Re: If #p# = ap3+ bp – 1............   [#permalink] 09 Nov 2015, 16:34
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