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m05 Q5 simple math help please [#permalink]
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17 Sep 2008, 19:00
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If \(P# = \frac{P}{P1}\) , what is the value of \(P##\) ? (A) \(\frac {P}{P1}\) (B) \(\frac{1}{P}\) (C) \(P\) (D) \(2  P\) (E) \(P  1\) Source: GMAT Club Tests  hardest GMAT questions Can somebody walk me through this one?
Last edited by akkane on 16 Jan 2013, 17:02, edited 2 times in total.
Fixed the question stem



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Re: m05 Q5 simple math help please [#permalink]
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17 Sep 2008, 21:34
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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 => \(\frac{p}{(p1)1}\) or #2 Option => \(\frac{p}{(p1)}1\) 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 \(\frac{4}{(41)1\) = 2, so 2# = \(\frac{2}{211} = \frac{2}{0}\) #2 is all other the place with results. If p = 8, so you have 8##, you get the following: \(\frac{8}{(81)}1 = \frac{8}{7}  \frac{7}{7} = \frac{1}{7}\) 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/(p1) 1 , what is the value of P##?
a) p/p1 b) 1/p c) p d) 2p e) p1
Can somebody walk me through this one?
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Re: m05 Q5 simple math help please [#permalink]
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17 Sep 2008, 22:09
Oops, in the question, forget the extra (1). it's just p/(p1)
Using substitution as you suggested, I was able to solve the answer. Thanks!
answer is P



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Re: m05 Q5 simple math help please [#permalink]
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21 Oct 2008, 11:41
As stated above the original problem was P# = p / (p1) Find P## Can someone please kindly show this to me using algebra? How do you get (p/(p1)) / (p/(p1)  1) as suggested in the answer. Thanks alot
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Re: m05 Q5 simple math help please [#permalink]
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22 Oct 2008, 02:41
How do you get (p/(p1)) / (p/(p1)  1) as suggested in the answer.
in your denominator pls see that its ( (p/(p1) )  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]
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23 Oct 2008, 03:26
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I'll try to make this clearer by using our math syntax. See the short intro here: Given \(P# = \frac{P}{P1}\), need to find \(P##\)  original problem We might simplify this problem by saying that this '#' sign is an operation that replaces each \(P\) with \(\frac{P}{P1}\). \(P##\) is found by replacing all the \(P\)s with \(\frac{P}{P1}\) in this expression: \(\frac{P}{P1}\), which equals \(P#\) This is how it looks after these replacements: \(\frac{\frac{P}{P1}}{\frac{P}{P1}1}\) Now we simplify it gradually: \(\frac{\frac{P}{P1}}{\frac{P}{P1}\frac{P1}{P1}}=\) \(\frac{\frac{P}{P1}}{\frac{PP+1}{P1}}=\) \(\frac{\frac{P}{P1}}{\frac{1}{P1}}=\) \(\frac{P}{P1}*\frac{P1}{1}=\) \(P\) Hope this clears the doubts.
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Wrong answer for this problem [#permalink]
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10 Jan 2010, 11:52
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# = 2/21 = 2/3. Then P# <> P 2. Example  Algebra: P# = P/(P1) We agree here, but how did you come to P/(P1)/P/(P1) 1 for P##? Shouldn’t it be P/(P1)/(P1) . You are multiplying both numerator and denominator what is not right P/(P1)* #/#?, 2/3 * 2 is not 2/3 * 2/2 for putting it in simple terms. So answer should be P/ (P1) (P1) Not P. Please verify.



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Re: Wrong answer for this problem [#permalink]
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11 Jan 2010, 08:29
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 \(\frac{P}{P1}\). I've covered that more thoroughly in my above post. So, after the operation is "executed", we get \(\frac{P}{P1}\), but this is not the answer as we have to find \(P##\). Now, to find \(P##\), we need to perform this operation \(#\) (replacing all \(P\)s with \(\frac{P}{P1}\)) over this expression we got previously: \(\frac{P}{P1}\). If you replace all \(P\)s with \(\frac{P}{P1}\) 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# = 2/21 = 2/3. Then P# <> P 2. Example  Algebra: P# = P/(P1) We agree here, but how did you come to P/(P1)/P/(P1) 1 for P##? Shouldn’t it be P/(P1)/(P1) . You are multiplying both numerator and denominator what is not right P/(P1)* #/#?, 2/3 * 2 is not 2/3 * 2/2 for putting it in simple terms. So answer should be P/ (P1) (P1) Not P. Please verify.
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Re: m05 Q5 simple math help please [#permalink]
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12 Jan 2010, 06:53
d2touge wrote: If \(P# = \frac{P}{P1}\) , what is the value of \(P##\) ? (A) \(\frac {P}{P1}\) (B) \(\frac{1}{P}\) (C) \(P\) (D) \(2  P\) (E) \(P  1\) Source: GMAT Club Tests  hardest GMAT questions Can somebody walk me through this one? Let X=P#=P/(P1) therefore, X#=X/(X1) put the value for X. ans P hence option C
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Re: m05 Q5 simple math help please [#permalink]
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12 Jan 2010, 06:57
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given p#=p/(p1) then p##= p/(p1)# ==>
numerator => p/(p1) denominator => p/(p1)1 ==> 1/(p1)
so p##==> p/(p1)/1/(p1) ==> p
so ans is C



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Re: m05 Q5 simple math help please [#permalink]
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12 Jan 2010, 07:15
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p# = p/(p1)
(p#)# = p/(p1)] / [ {p/(p1)}1] = p/(p1)] / [ {( p  (p1)}/(p1)] = p/(p1)] / [ 1/(p1)] = p ( p1) / (p1) = p
there fore answer is C



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Re: m05 Q5 simple math help please [#permalink]
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17 Jan 2011, 07:46
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]
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17 Jan 2011, 11:47
C
p# = p/(p1) p##=( p/(p1))/(p/(p1) 1) = p/(p1) * (p1)/(pp+1) = p
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Re: m05 Q5 simple math help please [#permalink]
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17 Jan 2011, 12:11
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]
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18 Jan 2011, 03:02
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]
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11 Jun 2011, 18:50
P## = (P/(P1)) #
= \(P/(P1) / ((P /(P1))1)\)
= P
Answer is C.



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Re: m05 Q5 simple math help please [#permalink]
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19 Jan 2012, 08:05
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]
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19 Jan 2012, 09:39
C. multiplying buy the reciprocal is a very useful thing



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Re: m05 Q5 simple math help please [#permalink]
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22 Jan 2013, 10:46
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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M0505: Function [#permalink]
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28 Jul 2013, 21:46
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
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