Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized for You
we will pick new questions that match your level based on your Timer History
Track Your Progress
every week, we’ll send you an estimated GMAT score based on your performance
Practice Pays
we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
It appears that you are browsing the GMAT Club forum unregistered!
Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club
Registration gives you:
Tests
Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.
Applicant Stats
View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more
Books/Downloads
Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!
Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:
Re: m05 Q5 simple math help please [#permalink]
17 Sep 2008, 20:34
1
This post received 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?
_________________
------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.
Re: m05 Q5 simple math help please [#permalink]
23 Oct 2008, 02:26
4
This post received 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 \(P\)s with \(\frac{P}{P-1}\) in this expression: \(\frac{P}{P-1}\), which equals \(P#\)
Wrong answer for this problem [#permalink]
10 Jan 2010, 10:52
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.
Re: Wrong answer for this problem [#permalink]
11 Jan 2010, 07:29
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 \(P\)s with \(\frac{P}{P-1}\)) over this expression we got previously: \(\frac{P}{P-1}\). If you replace all \(P\)s 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.
Let X=P#=P/(P-1) therefore, X#=X/(X-1) put the value for X. ans P hence option C _________________
--------------------------------------------------------------------------------------- If you think you can,you can If you think you can't,you are right.
Our deepest fear is not that we are inadequate. Our deepest fear is that we are powerful beyond measure. It is our light not our darkness that most frightens us.
Your playing small does not serve the world. There's nothing enlightened about shrinking so that other people won't feel insecure around you.
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.