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(p+5+p^3(-p-5))/(-p-5)=

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(p+5+p^3(-p-5))/(-p-5)=  [#permalink]

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New post 30 May 2010, 04:54
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\(\frac{p+5+p^3(-p-5)}{-p-5}=\)

A. p+5+p^3
B. P^3+5
C. p^3
D. p^3-1
E. p^3-5
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Re: Princeton Review Test problem : ID4019  [#permalink]

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New post 30 May 2010, 05:49
5
3
undernet wrote:
\(p+5+p^3(-p-5)/-p-5\)=

A. p+5+p^3
B. P^3+5
C. p^3
D. p^3-1
E. p^3-5

OA

I could solve it by plugging numbers. But I am looking for the algebraic way to solve this equation.


Hi, and welcome to the Gmat Club. Below is algebraic solution for you question:

Guess the question must be \(\frac{p+5+p^3(-p-5)}{-p-5}=?\)

If yes than: \(\frac{p+5+p^3(-p-5)}{-p-5}=\frac{-(-p-5)+p^3(-p-5)}{-p-5}\) --> factor out \((-p-5)\) --> \(\frac{-(-p-5)+p^3(-p-5)}{-p-5}=\frac{(-p-5)(-1+p^3)}{-p-5}=-1+p^3=p^3-1\).

Answer: D.

Hope it helps.
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Re: Princeton Review Test problem : ID4019  [#permalink]

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New post 30 May 2010, 05:54
Thank you Bunnel... negating the first term didn't struck me....
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Re: (p+5+p^3(-p-5))/(-p-5)=  [#permalink]

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New post 26 Jan 2015, 11:18
1
Hi All,

This question can be solved by TESTing VALUES.

IF.....
P = 2
Then the calculation becomes....

[2 + 5 + 8(-7)] / (-7)

[7 - 56]/(-7)
[-49]/(-7) = 7

So we're looking for an answer that = 7 when P = 2

Answer A: 2 + 5 + 8 = 15 NOT a match
Answer B: 8 + 5 = 13 NOT a match
Answer C: 8 NOT a match
Answer D: 8 - 1 = 7 This IS a MATCH
Answer E: 8 - 5 = 3 NOT a match

Final Answer:

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Re: (p+5+p^3(-p-5))/(-p-5)=  [#permalink]

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New post 02 Feb 2015, 20:27
1
\(\frac{p+5+p^3(-p-5)}{-p-5}=\)

Multiply numerator/ denominator by -1

\(\frac{p^3(p+5) - 1(p+5)}{p+5} = p^3 - 1\)

Answer = D
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Re: (p+5+p^3(-p-5))/(-p-5)=  [#permalink]

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New post 03 Aug 2016, 08:46
i think its easier to work without negatives, so i just did the following:

[p+5 + p^3(-p-5)] / (-p-5) => [(p+5)(1-p^3)/-1(p+5)] => (1-p^3)/-1 => p^3 -1
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Re: (p+5+p^3(-p-5))/(-p-5)=  [#permalink]

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New post 06 Dec 2016, 08:52
Would you mind explaining the negative sign part? Is this a rule that I am missing?
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Re: (p+5+p^3(-p-5))/(-p-5)=  [#permalink]

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New post 06 Dec 2016, 11:28
1
Hi Nasahtahir,

You're going to face some GMAT questions on Test Day that 'test' you on concepts that you know, but in ways that you're probably not used to thinking about. If you choose to approach this question algebraically (which is an approach that you do not have to use), then you would find that factoring the given equation can help to simplify it. You probably already know how to factor...

For example: 2X + 4 can be factored down to 2(X +2).

In that same way, you can factor out other common 'pieces':

-2X - 10 can be factored down to -2(X + 5).

The concept of 'factoring out a negative' is what mcdude123 used:

-P + 5 can be factored down to -1(P+5). At that point, you can then factor out (P+5) out of the numerator of the fraction.

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Re: (p+5+p^3(-p-5))/(-p-5)=  [#permalink]

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New post 02 Sep 2017, 09:16
Quote:
Hi, and welcome to the Gmat Club. Below is algebraic solution for you question:

Guess the question must be p+5+p3(−p−5)−p−5=?p+5+p3(−p−5)−p−5=?

If yes than: p+5+p3(−p−5)−p−5=−(−p−5)+p3(−p−5)−p−5p+5+p3(−p−5)−p−5=−(−p−5)+p3(−p−5)−p−5 --> factor out (−p−5)(−p−5) --> −(−p−5)+p3(−p−5)−p−5=(−p−5)(−1+p3)−p−5=−1+p3=p3−1−(−p−5)+p3(−p−5)−p−5=(−p−5)(−1+p3)−p−5=−1+p3=p3−1.

Answer: D.

Hope it helps.


Hi Brent,

Can you explain what you did after factoring (-p-5)?
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Re: (p+5+p^3(-p-5))/(-p-5)=  [#permalink]

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New post 02 Sep 2017, 20:48
Zoser wrote:
Quote:
Hi, and welcome to the Gmat Club. Below is algebraic solution for you question:

Guess the question must be p+5+p3(−p−5)−p−5=?p+5+p3(−p−5)−p−5=?

If yes than: p+5+p3(−p−5)−p−5=−(−p−5)+p3(−p−5)−p−5p+5+p3(−p−5)−p−5=−(−p−5)+p3(−p−5)−p−5 --> factor out (−p−5)(−p−5) --> −(−p−5)+p3(−p−5)−p−5=(−p−5)(−1+p3)−p−5=−1+p3=p3−1−(−p−5)+p3(−p−5)−p−5=(−p−5)(−1+p3)−p−5=−1+p3=p3−1.

Answer: D.

Hope it helps.


Hi Brent,

Can you explain what you did after factoring (-p-5)?


cancelled it from numerator and denominator to get -1+p^3.
Maybe the absence of parenthesis in the denominator confused you? -p-5 in denominator is the same as (-p-5)
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Re: (p+5+p^3(-p-5))/(-p-5)=  [#permalink]

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New post 06 Sep 2017, 19:18
Sorry guy bust I still dont get it. Anyone can put a video for it?
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Re: (p+5+p^3(-p-5))/(-p-5)=  [#permalink]

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New post 06 Sep 2017, 20:02
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Hi Zoser,

You're going to find that most GMAT questions can be approached in more than one way. As such, if you don't understand one particular approach to a question, then there's a pretty good chance that there will be another approach that you mind find easier to deal with. In my approach (above), I chose to TEST VALUES:

IF.....
P = 2
Then the calculation becomes....

[2 + 5 + 8(-7)] / (-7)

[7 - 56]/(-7)
[-49]/(-7) = 7

So we're looking for an answer that = 7 when P = 2

Answer A: 2 + 5 + 8 = 15 NOT a match
Answer B: 8 + 5 = 13 NOT a match
Answer C: 8 NOT a match
Answer D: 8 - 1 = 7 This IS a MATCH
Answer E: 8 - 5 = 3 NOT a match

Final Answer:

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Re: (p+5+p^3(-p-5))/(-p-5)=  [#permalink]

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New post 06 Sep 2017, 21:05
p+5+p^3(−p−5)/−p−5
=(p+5) -p^3(p+5)/ (-1)*(p+5)
Divide Numerator and Denominator by (p+5)
=1-p^3/-1
=p^3-1
Hence Answer is D.

Kudos if it helps.
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Re: (p+5+p^3(-p-5))/(-p-5)=  [#permalink]

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New post 30 Nov 2017, 10:10
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undernet wrote:
\(\frac{p+5+p^3(-p-5)}{-p-5}=\)

A. p+5 + p^3
B. P^3 + 5
C. p^3
D. p^3 - 1
E. p^3 - 5


Let's just focus on the NUMERATOR for a second.
Given: p + 5 + p³(-p - 5)
Factor -1 from the first part to get: -1(-p - 5) + p³(-p - 5)
So, we now have: -1(-p - 5) + p³(-p - 5)
Combine terms to get: (-1 + p³)(-p - 5)
Rearrange terms to get: (p³ - 1)(-p - 5)

Now replace ORIGINAL numerator with (p³ -1)(-p - 5)
We get: (p³ - 1)(-p - 5)/(-p - 5)
Simplify to get: (p³ - 1)
Answer: D

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Re: (p+5+p^3(-p-5))/(-p-5)=  [#permalink]

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New post 30 Nov 2017, 10:19
Top Contributor
undernet wrote:
\(\frac{p+5+p^3(-p-5)}{-p-5}=\)

A. p+5+p^3
B. P^3+5
C. p^3
D. p^3-1
E. p^3-5


We're looking for an expression that is equivalent to the original expression.
So if we evaluate the original expression for a particular value of p, then the equivalent expression should also yield the same value when we plug in the same value of p.
Let's test p = 1

Take: [p + 5 + p³(-p - 5)]/[-p - 5]
Replace p with 1 to get: [1 + 5 + 1³(-1 - 5)]/[-1 - 5]
Evaluate to get: 0/-6, which equals 0

So, when p = 1, the original express evaluates to be 0
Now let's plug p = 1 into the answer choices....
A. 1 + 5 + 1^3 = 7. No good, we want 0. ELIMINATE.
B. 1^3 + 5 = 6. No good, we want 0. ELIMINATE.
C. 1^3 = 1. No good, we want 0. ELIMINATE.
D. 1^3 - 1 = 0. Great - KEEP
E. 1^3 - 5 = -4. No good, we want 0. ELIMINATE.

Answer: D

Cheers,
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Re: (p+5+p^3(-p-5))/(-p-5)=  [#permalink]

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New post 04 Dec 2017, 10:52
undernet wrote:
\(\frac{p+5+p^3(-p-5)}{-p-5}=\)

A. p+5+p^3
B. P^3+5
C. p^3
D. p^3-1
E. p^3-5


Let’s use the distributive property over division:

(p + 5)/(-p - 5) + p^3(-p - 5)/(-p - 5)

(p + 5)/-(p + 5) + p^3

-1 + p^3

p^3 - 1

Answer: D
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Re: (p+5+p^3(-p-5))/(-p-5)=  [#permalink]

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New post 25 Feb 2018, 22:34
I think people are are confused with the format as it's a bit foreign and people are wrongly eliminating the −p−5 from both the top and bottom.

Essentially the question is \(\frac{1+2+3(x)}{x}\) in which simplified would be \(\frac{1}{x}\)+ \(\frac{2}{x}\)+ \(\frac{3}{1}\) (cancelling the x's in 3x/x.

And not \(\frac{(1+2+3)x}{x}\) Here, you can eliminate the x's and get 1+2+3
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(p+5+p^3(-p-5))/(-p-5)=  [#permalink]

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New post 25 Feb 2018, 23:54
There are a few ways to do this question. I chose the simplification/elimination (however you call it) method since I could see some symmetry in the question. Else I would have also gone for value substitution.

\(\frac{(p+5+p^3(-p-5))}{-p-5} = \frac{p+5}{-p-5} + \frac{p^3(-p-5)}{-p-5} = \frac{p+5}{-(p+5)}+ p^3 = -1 + p^3\)

Ans: D
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Re: (p+5+p^3(-p-5))/(-p-5)=  [#permalink]

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New post 13 Sep 2019, 07:06
undernet wrote:
\(\frac{p+5+p^3(-p-5)}{-p-5}=\)

A. p+5+p^3
B. P^3+5
C. p^3
D. p^3-1
E. p^3-5


\(\frac{p+5+p^3(-p-5)}{-p-5}\)

Or, \(\frac{-1 (- p - 5 ) +p^3 ( -p - 5 )}{(-p -5 )}\)

Or, \(\frac{(- p - 5 )( p^3 - 1)}{(-p -5 )}\)

Or, \(( p^3 - 1)\) , Answer must be (D)
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Re: (p+5+p^3(-p-5))/(-p-5)=   [#permalink] 13 Sep 2019, 07:06
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