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Math Expert V
Joined: 02 Sep 2009
Posts: 59628

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15 00:00

Difficulty:   15% (low)

Question Stats: 71% (01:04) correct 29% (02:03) wrong based on 208 sessions

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If for all numbers $$P$$, where $$P \neq 1$$, the function $$@$$ is defined by $$P@ = \frac{P}{P-1}$$, 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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Math Expert V
Joined: 02 Sep 2009
Posts: 59628

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2
1
Official Solution:

If for all numbers $$P$$, where $$P \neq 1$$, the function $$@$$ is defined by $$P@ = \frac{P}{P-1}$$, 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 $$(P@)@$$..

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

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

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Intern  Joined: 03 Dec 2014
Posts: 14
GMAT 1: 620 Q43 V32 GPA: 2.9

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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 V
Joined: 02 Sep 2009
Posts: 59628

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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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Intern  B
Joined: 13 Oct 2017
Posts: 38

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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 V
Joined: 02 Sep 2009
Posts: 59628

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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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Intern  B
Joined: 13 Oct 2017
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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  B
Joined: 12 Mar 2018
Posts: 2

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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 V
Joined: 02 Sep 2009
Posts: 59628

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1
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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Manager  B
Joined: 20 Jul 2019
Posts: 55

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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

Posted from my mobile device. In question he gave p(x) =p/p-1 ,he is asking what is p(p(x)) just substitute p with p/p-1
Intern  B
Joined: 06 May 2019
Posts: 3
Location: India
GMAT 1: 710 Q49 V36 GPA: 4

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1
1
Good question. But it should be mentioned in question that @ basically stands for function. Otherwise most people will consider AS a variable.
Intern  B
Joined: 07 Oct 2018
Posts: 7

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1
Bunuel wrote:
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.

Bunuel

This question took me a minute to understand. I do not mean to come out questioning your explanation, but I am not sure it was complete.

The picking number explanation might not be complete. It is my understanding that 2 is a unique number and should be avoided when using the testing cases approach. For example, if we set 4 = P the explanation P@=P, so P@@ = P does not work.

For example if P = 4: 4/(4-1) -> P@ = 4/3 this does not yield the result P@=P so you cannot then use the logic suggested above to find that P@@ = P. It is not until we substitute again that we find P@@=P.

If P@ = 4/3 then P@@ = (4/3)/((4/3)-1) -> (4/3)/((4/3)-(3/3)) -> (4/3)/(1/3) -> 4 which = P so therefore P@@=P

I think the above offers a more complete explanation of the substitution above. Bunuel, please correct me if my explanation is inefficient or incorrect.
Math Expert V
Joined: 02 Sep 2009
Posts: 59628

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harshb1995 wrote:
Good question. But it should be mentioned in question that @ basically stands for function. Otherwise most people will consider AS a variable.

Edited. Thank you for the suggestion.
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Math Expert V
Joined: 02 Sep 2009
Posts: 59628

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bcal95 wrote:
Bunuel wrote:
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.

Bunuel

This question took me a minute to understand. I do not mean to come out questioning your explanation, but I am not sure it was complete.

The picking number explanation might not be complete. It is my understanding that 2 is a unique number and should be avoided when using the testing cases approach. For example, if we set 4 = P the explanation P@=P, so P@@ = P does not work.

For example if P = 4: 4/(4-1) -> P@ = 4/3 this does not yield the result P@=P so you cannot then use the logic suggested above to find that P@@ = P. It is not until we substitute again that we find P@@=P.

If P@ = 4/3 then P@@ = (4/3)/((4/3)-1) -> (4/3)/((4/3)-(3/3)) -> (4/3)/(1/3) -> 4 which = P so therefore P@@=P

I think the above offers a more complete explanation of the substitution above. Bunuel, please correct me if my explanation is inefficient or incorrect.

Edited algebraic approach and made it more detailed. Thank you for the suggestion.
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# M05-05

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