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# m05 Q5 simple math help please

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If P# = \frac{P}{P-1} , what is the value of P## ?

(A) \frac {P}{P-1}
(B) \frac{1}{P}
(C) P
(D) 2 - P
(E) P - 1

[Reveal] Spoiler: OA
C

Source: GMAT Club Tests - hardest GMAT questions

Can somebody walk me through this one?

Last edited by akkane on 16 Jan 2013, 16:02, edited 2 times in total.
Fixed the question stem
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Re: m05 Q5 simple math help please [#permalink]  17 Sep 2008, 20:34
1
KUDOS
You really need to use for math formatting in order for us to figure out if there are any errors in the question as written. I can't determine which of these is the proper way for this question to be displayed:

#1 Option => \frac{p}{(p-1)-1}
or
#2 Option => \frac{p}{(p-1)}-1

Generally, this is a question that you can use substituting and get the correct answer, but that doesn't seem to be working. For example, if you make p = 4 such that you have 4##.

#1 gives you division by zero because you have \frac{4}{(4-1)-1 = 2, so 2# = \frac{2}{2-1-1} = \frac{2}{0}

#2 is all other the place with results. If p = 8, so you have 8##, you get the following:

\frac{8}{(8-1)}-1 = \frac{8}{7} - \frac{7}{7} = \frac{1}{7} and you then have \frac{1}{7}#

which = \frac{\frac{1}{7}}{\frac{1}{7} - \frac{7}{7}}-1 = -1\frac{1}{6}

1/7 dvidied by -6/7 is like 1/7 * -7/6 so the 7's cancel out and you have -1/6 left. Then subtract 1 from that to give you the answer above.

If you plug in the original numbers used above for p into the answer equations, you never get the same value as you do above so it makes the question (or answers) invalid.

Someone, please show me that I'm wrong and I just didn't see the correct approach because I'm tired.

d2touge wrote:
If P# = p/(p-1) -1 , what is the value of P##?

a) p/p-1
b) 1/p
c) p
d) 2-p
e) p-1

Can somebody walk me through this one?

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Re: m05 Q5 simple math help please [#permalink]  17 Sep 2008, 21:09
Oops, in the question, forget the extra (-1). it's just p/(p-1)

Using substitution as you suggested, I was able to solve the answer. Thanks!

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Re: m05 Q5 simple math help please [#permalink]  21 Oct 2008, 10:41
As stated above the original problem was P# = p / (p-1)

Find P##

Can someone please kindly show this to me using algebra?

How do you get (p/(p-1)) / (p/(p-1) - 1) as suggested in the answer.

Thanks alot
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Re: m05 Q5 simple math help please [#permalink]  22 Oct 2008, 01:41
How do you get (p/(p-1)) / (p/(p-1) - 1) as suggested in the answer.

in your denominator pls see that its ( (p/(p-1) ) - 1 )

which will be (p -p + 1) / (p -1) [ This is just the denominator ]
so finally p / (p -1) / (1 / (p - 1) ) which is p.
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Re: m05 Q5 simple math help please [#permalink]  23 Oct 2008, 02:26
4
KUDOS
I'll try to make this clearer by using our math syntax. See the short intro here:

Given P# = \frac{P}{P-1}, need to find P## - original problem
We might simplify this problem by saying that this '#' sign is an operation that replaces each P with \frac{P}{P-1}. P## is found by replacing all the Ps with \frac{P}{P-1} in this expression:
\frac{P}{P-1}, which equals P#

This is how it looks after these replacements:

\frac{\frac{P}{P-1}}{\frac{P}{P-1}-1}

\frac{\frac{P}{P-1}}{\frac{P}{P-1}-\frac{P-1}{P-1}}=

\frac{\frac{P}{P-1}}{\frac{P-P+1}{P-1}}=

\frac{\frac{P}{P-1}}{\frac{1}{P-1}}=

\frac{P}{P-1}*\frac{P-1}{1}=

P

Hope this clears the doubts.
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For both picking number and algebra solution explanations are incorrect:
1. Example- picking number: If we choose a negative number (The problem didn’t state that P is a positive number), let’s say -2 then -2# = -2/-2-1 = 2/3. Then P# <> P
2. Example - Algebra: P# = P/(P-1) We agree here, but how did you come to P/(P-1)/P/(P-1) -1 for P##? Shouldn’t it be P/(P-1)/(P-1) . You are multiplying both numerator and denominator what is not right P/(P-1)* #/#?, 2/3 * 2 is not 2/3 * 2/2 for putting it in simple terms. So answer should be P/ (P-1) (P-1) Not P. Please verify.
CIO
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This question is pretty confusing. I'll try to describe the approach that works for me. Let's see once again what this "#" operation does to P. We see from the question stem that this operation replaces P with \frac{P}{P-1}. I've covered that more thoroughly in my above post. So, after the operation is "executed", we get \frac{P}{P-1}, but this is not the answer as we have to find P##. Now, to find P##, we need to perform this operation # (replacing all Ps with \frac{P}{P-1}) over this expression we got previously: \frac{P}{P-1}. If you replace all Ps with \frac{P}{P-1} in this expression, you'll end up with a big one (that can be simplified to P as described in more detail in my above post).

Hope this helps.
Ikowill wrote:
For both picking number and algebra solution explanations are incorrect:
1. Example- picking number: If we choose a negative number (The problem didn’t state that P is a positive number), let’s say -2 then -2# = -2/-2-1 = 2/3. Then P# <> P
2. Example - Algebra: P# = P/(P-1) We agree here, but how did you come to P/(P-1)/P/(P-1) -1 for P##? Shouldn’t it be P/(P-1)/(P-1) . You are multiplying both numerator and denominator what is not right P/(P-1)* #/#?, 2/3 * 2 is not 2/3 * 2/2 for putting it in simple terms. So answer should be P/ (P-1) (P-1) Not P. Please verify.

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Re: m05 Q5 simple math help please [#permalink]  12 Jan 2010, 05:53
d2touge wrote:
If P# = \frac{P}{P-1} , what is the value of P## ?

(A) \frac {P}{P-1}
(B) \frac{1}{P}
(C) P
(D) 2 - P
(E) P - 1

[Reveal] Spoiler: OA
C

Source: GMAT Club Tests - hardest GMAT questions

Can somebody walk me through this one?

Let X=P#=P/(P-1)
therefore, X#=X/(X-1) put the value for X.
ans P
hence option C
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Re: m05 Q5 simple math help please [#permalink]  12 Jan 2010, 05:57
1
KUDOS
given p#=p/(p-1)
then p##= p/(p-1)# ==>

numerator => p/(p-1)
denominator => p/(p-1)-1 ==> 1/(p-1)

so p##==> p/(p-1)/1/(p-1) ==> p

so ans is C
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Re: m05 Q5 simple math help please [#permalink]  12 Jan 2010, 06:15
1
KUDOS
p# = p/(p-1)

(p#)# = p/(p-1)] / [ {p/(p-1)}-1]
= p/(p-1)] / [ {( p - (p-1)}/(p-1)]
= p/(p-1)] / [ 1/(p-1)]
= p ( p-1) / (p-1)
= p

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Re: m05 Q5 simple math help please [#permalink]  17 Jan 2011, 06:46
I tried with

# = 1, Solvng we get P = 2

P## = 2

Ony option 'C' satisfies this ....
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Re: m05 Q5 simple math help please [#permalink]  17 Jan 2011, 10:47
C

p# = p/(p-1)
p##=( p/(p-1))/(p/(p-1) -1)
= p/(p-1) * (p-1)/(p-p+1)
= p

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Re: m05 Q5 simple math help please [#permalink]  17 Jan 2011, 11:11
thanks guys. can anyone tell me whats the lvl of this question?

thanks.
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Re: m05 Q5 simple math help please [#permalink]  18 Jan 2011, 02:02
I thought this was a very simple problem.

I dint think of substituting with number - But just went with (p#) #
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Re: m05 Q5 simple math help please [#permalink]  11 Jun 2011, 17:50
P## = (P/(P-1)) #

= P/(P-1) / ((P /(P-1))-1)

= P

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Re: m05 Q5 simple math help please [#permalink]  19 Jan 2012, 07:05
Why do you subtract a 1 for the second part of the equation?

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Re: m05 Q5 simple math help please [#permalink]  19 Jan 2012, 08:39
C. multiplying buy the reciprocal is a very useful thing
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Re: m05 Q5 simple math help please [#permalink]  22 Jan 2013, 09:46
atojha wrote:
I tried with

# = 1, Solvng we get P = 2

P## = 2

Ony option 'C' satisfies this ....

But,
option (1) also satisfies it.
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M05-05: Function [#permalink]  28 Jul 2013, 20:46
If P@=P1/P−1, what is the value of P@@?

(A) P/P−1

(B) 1/P

(C) P

(D) 2−P

(E) P−1

[Reveal] Spoiler:
C

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It's not just in some of us; it's in everyone. And as we let our own light shine, we unconsciously give other people permission to do the same.

As we are liberated from our own fear, our presence automatically liberates others.

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M05-05: Function   [#permalink] 28 Jul 2013, 20:46
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