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

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Manager
Status: Current MBA Student
Joined: 19 Nov 2009
Posts: 89
Concentration: Finance, General Management
GMAT 1: 720 Q49 V40
If #p# = ap^3+ bp – 1 where a and b are constants, and #-5#  [#permalink]

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06 Oct 2010, 09:55
2
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Difficulty:

55% (hard)

Question Stats:

62% (02:06) correct 38% (02:15) wrong based on 269 sessions

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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
Math Expert
Joined: 02 Sep 2009
Posts: 64246
Re: MGMAT CAT function problem  [#permalink]

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06 Oct 2010, 10:04
2
1
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 $$125a$$ --> $$#5#=(-5b-4)+5b-1=-5$$.

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Re: MGMAT CAT function problem  [#permalink]

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07 Oct 2010, 00:04
2
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 :
$$f(5)=-f(-5)-2=-(3)-2=-5$$

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Re: MGMAT CAT function problem  [#permalink]

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07 Oct 2010, 00:51
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
Director
Joined: 14 Dec 2012
Posts: 670
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Concentration: General Management, Operations
GMAT 1: 700 Q50 V34
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Re: If #p# = ap^3+ bp – 1 where a and b are constants  [#permalink]

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17 Aug 2013, 10:01
1
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: If #p# = ap^3+ bp – 1 where a and b are constants  [#permalink]

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17 Aug 2013, 10:42
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]

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13 Dec 2018, 01:12
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Re: If #p# = ap^3+ bp – 1 where a and b are constants, and #-5#   [#permalink] 13 Dec 2018, 01:12

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

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