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If #p# = ap^3+ bp – 1 where a and b are constants, and #-5#

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tonebeeze
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If #p# = ap^3+ bp – 1 where a and b are constants, and #-5# [#permalink] New post   06 Oct 2010, 10:55
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Can you please walk me through how to best approach this problem? Thanks

If #p# = ap^3+ bp – 1 where a and b are constants, and #-5# = 3, what is the value of #5#?

A. 5
B. 0
C. -2
D. -3
E. -5
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E
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Re: MGMAT CAT function problem [#permalink] New post   06 Oct 2010, 11:04
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tonebeeze wrote:
Can you please walk me through how to best approach this problem? Thanks

If #p# = ap3+ bp – 1 where a and b are constants, and #-5# = 3, what is the value of #5#?

a. 5

b. 0

c. -2

d. -3

e. -5


Given: \(#p#=ap^3+bp-1\) and \(#-5#=a(-5)^3+b(-5)-1=-125a-5b-1=3\) --> \(125a=-5b-4\).
Question: \(#5#=a5^3+b*5-1=125a+5b-1=?\) --> substitute \(125a\) --> \(#5#=(-5b-4)+5b-1=-5\).

Answer: E.
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Re: MGMAT CAT function problem [#permalink] New post   07 Oct 2010, 01:04
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tonebeeze wrote:
Can you please walk me through how to best approach this problem? Thanks

If #p# = ap3+ bp – 1 where a and b are constants, and #-5# = 3, what is the value of #5#?

a. 5

b. 0

c. -2

d. -3

e. -5


\(f(p)=ap^3+bp-1\)
\(f(-p)=a(-p)^3+b(-p)-1=-ap^3-bp+1-2=-(ap^3+bp-1)-2=-f(p)-2\)

Using the above simplification :
\(f(5)=-f(-5)-2=-(3)-2=-5\)

Answer is (e)
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Re: If #p# = ap^3+ bp – 1 where a and b are constants [#permalink] New post   17 Aug 2013, 11:01
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Stiv wrote:
If \(#p# = ap^3+ bp - 1\)where a and b are constants, and #-5# = 3, what is the value of #5#?
A 5
B 0
C -2
D -3
E -5


#p# = ap^3 + bp - 1

#-5# = 3
putting p = -5 in above equation
-125a -(5b +1) = 3 or
#-5# = (125a+5b+1) = -3
therefore 125a+5b = -4 .....(1

now putting p = 5
#5# = 125 a+5b - 1
using equation 1(125a+5b = -4)
#5# = -4-1 = -5

hence E
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Re: MGMAT CAT function problem [#permalink] New post   07 Oct 2010, 01:51
Just see these funny signs (there are many others) as a function, so #p# = f(p) = ap3+ bp – 1

They can use any other sign: xy = x+y
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Re: If #p# = ap^3+ bp – 1 where a and b are constants [#permalink] New post   17 Aug 2013, 11:42
Stiv wrote:
If \(#p# = ap^3+ bp - 1\)where a and b are constants, and #-5# = 3, what is the value of #5#?
A 5
B 0
C -2
D -3
E -5


#p# = ap^3+ bp - 1
or , #5# = 125a+ 5b - 1

#-5# = -125a -5b - 1
or, 3 = -125a-5b-1 or, 125a+5b= -4

so #5# = -4 -1 = -5
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Re: If #p# = ap^3+ bp – 1 where a and b are constants, and #-5# [#permalink] New post   07 Jan 2022, 09:42
#-5# => -125a -5b-1 = 3 (I)
#5# => 125a +5b-1= x (II)
I+ II => 0+0-2=3+x => (solve for x) : (-3)
x= -5
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Re: If #p# = ap^3+ bp 1 where a and b are constants, and #-5# [#permalink] New post   12 Apr 2023, 08:29
I think that one can solve this one with fairly straight forward algebra:

#p# = ap^3 + bp - 1
-> #p# = p(p^2 * a + b) - 1
#-5# = -5(25a + b) - 1 = 3 -> 25a + b = (-4/5)

Now substitute that into #5#:
#5# = 5 * (-4/5) - 1 = -4 -1 = -5
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Re: If #p# = ap^3+ bp 1 where a and b are constants, and #-5# [#permalink]
12 Apr 2023, 08:29
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