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M05-05

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Math Expert
Joined: 02 Sep 2009
Posts: 59628

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16 Sep 2014, 00:24
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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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Posts: 59628

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16 Sep 2014, 00:24
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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Joined: 03 Dec 2014
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GMAT 1: 620 Q43 V32
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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
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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
1
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: 59628

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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
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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!
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20 Apr 2018, 03:17
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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10 Aug 2019, 07:41
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
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10 Aug 2019, 22:40
1
1
Good question. But it should be mentioned in question that @ basically stands for function. Otherwise most people will consider AS a variable.
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23 Aug 2019, 11:05
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.
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Posts: 59628

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26 Aug 2019, 02:02
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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26 Aug 2019, 02:03
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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