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Functions - use of pound signs?

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Functions - use of pound signs?

by Krysia0101 » Tue Jan 24, 2012 4:18 am
Hi everyone,
I'm not really familiar with the use of pound sign in functions such as the following:

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

Thank you for any pointers!
~krysia
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by Anurag@Gurome » Tue Jan 24, 2012 4:35 am
Krysia0101 wrote:Hi everyone,
I'm not really familiar with the use of pound sign in functions such as the following:

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

Thank you for any pointers!
~krysia
#p# = ap^3+ bp - 1
#-5# = 3 implies a(-5)^3 + b(-5) - 1 = 3 or 125a + 5b = -4
#5# = a(5)^3 + b(5) - 1 = 125a + 5b - 1 = -4 - 1 = -5

The correct answer is E.
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by GMATGuruNY » Tue Jan 24, 2012 4:41 am
Krysia0101 wrote:Hi everyone,
I'm not really familiar with the use of pound sign in functions such as the following:

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

Thank you for any pointers!
~krysia
FOLLOW THE DIRECTIONS.

#p# = ap³ + bp - 1

implies

#-5# = a(-5)³ + b(-5) - 1 = -125a - 5b - 1.

Since #-5# = 3, we get:
-125a - 5b - 1 = 3.
-125a - 5b = 4
125a + 5b = -4.

#p# = ap³ + bp - 1

implies

#5# = a(5)³ + b(5) - 1 = 125a + 5b - 1.

Since 125a + 5b = -4, 125a + 5b - 1 = -4-1 = -5.

The correct answer is E.
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by Krysia0101 » Tue Jan 24, 2012 6:55 am
Thank you for the replies! That was very helpful :D
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