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605-655 (Medium)|   Algebra|      
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MontrealLady
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If #p# = ap^3+ bp – 1 where a and b are constants, and #-5# = 3, what 18 Apr 2010, 08:30
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

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45% (medium)

Question Stats:

70% (02:01) correct 30% (02:26) wrong based on 796 sessions

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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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The answer I found was (5) and here is how I solved the question.
for #-5# = 3, here is the equation -125a – 5b – 1 = 3
For #5#, the equation is 125a + 5b – 1 = X

then to find X, I substract the first equation from the second one

-125a – 5b – 1 = 3
-
125a + 5b – 1 = X

that gives -2 = 3-x, so x = 5

However, the correct answer is (-5) and here is how MGMAT solved it. I got their explanation, but I'm curious to know why my way didn't work!


Since we know the value of #-5# = 3, we can plug p = -5 into our formula:

(-5)3a + (-5)b – 1 = 3
-125a – 5b – 1 = 3
-125a – 5b = 4

We are asked to solve for #5#. If we plug 5 into our formula, we get:
(5)3a + (5)b – 1 = ?
125a + (5)b – 1 = ?

To figure this out, we would need to know the value of 125a + 5b.

From the first equation we know that -125a – 5b = 4. By multiplying both sides by negative one, we see that 125a + 5b = -4.

125a + 5b – 1 = ?
–4 – 1 = -5

The correct answer is E.
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E
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Bunuel
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If #p# = ap^3+ bp – 1 where a and b are constants, and #-5# = 3, what 18 Apr 2010, 08:46
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MontrealLady
If #p# = ap3+ bp – 1 where a and b are constants, and #-5# = 3, what is the value of #5#?

The answer I found was (5) and here is how I solved the question.
for #-5# = 3, here is the equation -125a – 5b – 1 = 3
For #5#, the equation is 125a + 5b – 1 = X

then to find X, I substract the first equation from the second one

-125a – 5b – 1 = 3
-
125a + 5b – 1 = X

that gives -2 = 3-x, so x = 5
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Your approach is much better than that of MGMAT's. Though there are two problems:

When you subtract equations you'll get: \(-250a-10b=3-x\) (so you subtracted incorrectly) and this doesn't work.

What you should have done is to add equations instead of subtracting:

    \(-125a-5b-1+125a+5b-1=3+x\);
    \(-2=3+x\);
    \(x=-5\).

Hope it helps.
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If #p# = ap^3+ bp – 1 where a and b are constants, and #-5# = 3, what 12 Jul 2017, 09:43
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First equation:

a*-5^3-5b-1=3

-125a-5b= 4

Multiply both sides by -1

125a+5b= -4 (Plug this value in equation two)

Second equation:

125a +5b-1= ?

-4-1= -5

The answer is E.

Bunuel is this correct or there is some error in the above method that I used?
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If #p# = ap^3+ bp – 1 where a and b are constants, and #-5# = 3, what 12 Jul 2017, 17:17
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MontrealLady
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
Show more

Let’s solve for #-5# = 3:

a(-5)^3 + b(-5) - 1 = 3

-125a - 5b = 4

-5(25a + b) = 4

(25a + b) = -⅘

Let’s now solve for #5#:

a(5^3 + b(5) - 1

125a + 5b - 1

5(25a + b) - 1

Since 25a + b = -4/5, we have:

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

Answer: E
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If #p# = ap^3+ bp – 1 where a and b are constants, and #-5# = 3, what 24 Nov 2025, 05:01
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