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m05 Q5 simple math help please [#permalink]
 17 Sep 2008, 19:00
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 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]
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}{(p-1)-1} or #2 Option => \frac{p}{(p-1)}-1Generally, 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, 22:09
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!
answer is P
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Re: m05 Q5 simple math help please [#permalink]
21 Oct 2008, 11:41
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, 02:41
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, 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}{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}Now we simplify it gradually: \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}=PHope this clears the doubts.
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Wrong answer for this problem [#permalink]
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/-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: Wrong answer for this problem [#permalink]
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}{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, 06:53
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 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, 06:57
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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, 07:15
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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
there fore answer is C
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Re: m05 Q5 simple math help please [#permalink]
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]
17 Jan 2011, 11:47
C p# = p/(p-1) p##=( p/(p-1))/(p/(p-1) -1) = p/(p-1) * (p-1)/(p-p+1) = p Posted from my mobile device 
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Re: m05 Q5 simple math help please [#permalink]
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]
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]
11 Jun 2011, 18:50
P## = (P/(P-1)) #
= P/(P-1) / ((P /(P-1))-1)
= P
Answer is C.
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Re: m05 Q5 simple math help please [#permalink]
19 Jan 2012, 08:05
Why do you subtract a 1 for the second part of the equation? Posted from my mobile device
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Re: m05 Q5 simple math help please [#permalink]
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]
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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Re: m05 Q5 simple math help please
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22 Jan 2013, 10:46
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