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# M05-05

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Math Expert
Joined: 02 Sep 2009
Posts: 49968

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16 Sep 2014, 00:24
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Difficulty:

25% (medium)

Question Stats:

73% (00:38) correct 27% (01:26) wrong based on 135 sessions

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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. $$P - 1$$

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Joined: 02 Sep 2009
Posts: 49968

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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.

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Joined: 03 Dec 2014
Posts: 15
GMAT 1: 620 Q43 V32
GPA: 2.9

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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
Math Expert
Joined: 02 Sep 2009
Posts: 49968

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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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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
Math Expert
Joined: 02 Sep 2009
Posts: 49968

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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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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
Intern
Joined: 12 Mar 2018
Posts: 1

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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!
Math Expert
Joined: 02 Sep 2009
Posts: 49968

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20 Apr 2018, 03:17
kramerjay13 wrote:
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!

Check here: https://gmatclub.com/forum/m05-183664.html#p2016692

Hope it helps.
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Re: M05-05 &nbs [#permalink] 20 Apr 2018, 03:17
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# M05-05

Moderators: chetan2u, Bunuel

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