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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 \(P\)s with \(\frac{P}{P-1}\) in this expression: \(\frac{P}{P-1}\), which equals \(P#\)

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$$.

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.

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
_________________

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