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M05-05

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M05-05  [#permalink]

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New post 16 Sep 2014, 00:24
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A
B
C
D
E

Difficulty:

  25% (medium)

Question Stats:

69% (01:08) correct 31% (02:02) wrong based on 84 sessions

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Re M05-05  [#permalink]

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New post 16 Sep 2014, 00:24
Official Solution:

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. \(P - 1\)


Solution 1. Pick numbers. If \(P = 2\). Then \(P@ = \frac{2}{2 - 1} = 2\). If \(P@ = P\), then \(P@@ = P\).

Solution 2. Algebra: \(P@ = \frac{P}{(P-1)}\). Take this value and substitute it for \(P\) to find \(P@@\). We will get \(\frac{\frac{P}{(P-1)}}{\frac{P}{(P-1)} - 1}\). Simplify to get the following expression: \(\frac{\frac{P}{(P-1)}}{\frac{1}{(P-1)}}\). Dividing eliminates \(P-1\) and we get \(\frac{P}{1}\) or P.


Answer: C
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Re: M05-05  [#permalink]

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New post 10 Dec 2014, 16:38
Hi Bunuel,

I do not quite understand your solution to this problem here. Are these two different P's ? If yes, then can you elaborate how you solved this using the Algebra ? I am not sure how you got the expression after 'We will get' in your solution.

Thanks !

S
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Re: M05-05  [#permalink]

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New post 11 Dec 2014, 06:40
SD007 wrote:
Hi Bunuel,

I do not quite understand your solution to this problem here. Are these two different P's ? If yes, then can you elaborate how you solved this using the Algebra ? I am not sure how you got the expression after 'We will get' in your solution.

Thanks !

S


No. \(@\) defines some functions for P: \(P@ = \frac{P}{P-1}\). For example, \(2@ = \frac{2}{2-1}=2\).

I think similar questions should help. Check Operations/functions defining algebraic/arithmetic expressions.
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Re: M05-05  [#permalink]

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New post 16 Feb 2018, 02:49
Hi Bunuel,

I also plugged in 2 for P...and got pretty lost from there once I found out that P=2 as well.

I followed your explanation and understand P@ = 2 since when you plug in 2...the result is 2.

However, I don't follow what happens from there...how exactly is P@@=P?

As for solution 2: how did you solve this algebraically? In other words, how did you end up knowing that P@@ required you to dividing P/P-1 P/P-1 -1

Thanks
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Re: M05-05  [#permalink]

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New post 16 Feb 2018, 06:40
ttaiwo wrote:
Hi Bunuel,

I also plugged in 2 for P...and got pretty lost from there once I found out that P=2 as well.

I followed your explanation and understand P@ = 2 since when you plug in 2...the result is 2.

However, I don't follow what happens from there...how exactly is P@@=P?

As for solution 2: how did you solve this algebraically? In other words, how did you end up knowing that P@@ required you to dividing P/P-1 P/P-1 -1

Thanks


1. If \(P = 2\). Then \(P@ = \frac{2}{2 - 1} = 2\).

So, we got that for p = 2, \(P@ = P\), then \((P@)@ =P@ = P\).


2. \(P@ = \frac{P}{(P-1)}\).

\(P@@ = (\frac{P}{(P-1)})@=\frac{\frac{P}{(P-1)}}{\frac{P}{(P-1)} - 1}\)
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Re: M05-05  [#permalink]

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New post 17 Feb 2018, 03:56
Bunuel wrote:
ttaiwo wrote:
Hi Bunuel,

I also plugged in 2 for P...and got pretty lost from there once I found out that P=2 as well.

I followed your explanation and understand P@ = 2 since when you plug in 2...the result is 2.

However, I don't follow what happens from there...how exactly is P@@=P?

As for solution 2: how did you solve this algebraically? In other words, how did you end up knowing that P@@ required you to dividing P/P-1 P/P-1 -1

Thanks


1. If \(P = 2\). Then \(P@ = \frac{2}{2 - 1} = 2\).

So, we got that for p = 2, \(P@ = P\), then \((P@)@ =P@ = P\).


2. \(P@ = \frac{P}{(P-1)}\).

\(P@@ = (\frac{P}{(P-1)})@=\frac{\frac{P}{(P-1)}}{\frac{P}{(P-1)} - 1}\)


Thanks bro
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Re: M05-05  [#permalink]

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New post 20 Apr 2018, 02:38
Hi can you please explain the algebra of how you simplified the P@@ expression to:
(P/(P−1))/(1/(P−1))?

I was able to translate the P@@ expression but not simplify correctly. thanks!
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Re: M05-05  [#permalink]

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New post 20 Apr 2018, 03:17
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Re: M05-05   [#permalink] 20 Apr 2018, 03:17
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