GMAT Question of the Day - Daily to your Mailbox; hard ones only

It is currently 10 Dec 2019, 05:40

Close

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel

M05-05

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Find Similar Topics 
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 59628
M05-05  [#permalink]

Show Tags

New post 16 Sep 2014, 00:24
15
00:00
A
B
C
D
E

Difficulty:

  15% (low)

Question Stats:

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

HideShow timer Statistics

Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 59628
Re M05-05  [#permalink]

Show Tags

New post 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 \((P@)@\)..

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

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


Answer: C
_________________
Intern
Intern
avatar
Joined: 03 Dec 2014
Posts: 14
GMAT 1: 620 Q43 V32
GPA: 2.9
GMAT ToolKit User Reviews Badge
Re: M05-05  [#permalink]

Show Tags

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
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 59628
Re: M05-05  [#permalink]

Show Tags

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.
_________________
Intern
Intern
avatar
B
Joined: 13 Oct 2017
Posts: 38
Re: M05-05  [#permalink]

Show Tags

New post 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
User avatar
V
Joined: 02 Sep 2009
Posts: 59628
Re: M05-05  [#permalink]

Show Tags

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

\(P@@ = (\frac{P}{(P-1)})@=\frac{\frac{P}{(P-1)}}{\frac{P}{(P-1)} - 1}\)
_________________
Intern
Intern
avatar
B
Joined: 13 Oct 2017
Posts: 38
Re: M05-05  [#permalink]

Show Tags

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
Intern
Intern
avatar
B
Joined: 12 Mar 2018
Posts: 2
Re: M05-05  [#permalink]

Show Tags

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!
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 59628
Re: M05-05  [#permalink]

Show Tags

New post 20 Apr 2018, 03:17
1
Manager
Manager
avatar
B
Joined: 20 Jul 2019
Posts: 55
Re: M05-05  [#permalink]

Show Tags

New post 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
Intern
Intern
avatar
B
Joined: 06 May 2019
Posts: 3
Location: India
GMAT 1: 710 Q49 V36
GPA: 4
Re: M05-05  [#permalink]

Show Tags

New post 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.
Intern
Intern
avatar
B
Joined: 07 Oct 2018
Posts: 7
GMAT ToolKit User
Re: M05-05  [#permalink]

Show Tags

New post 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.


Answer: C


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
User avatar
V
Joined: 02 Sep 2009
Posts: 59628
Re: M05-05  [#permalink]

Show Tags

New post 26 Aug 2019, 02:02
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 59628
Re: M05-05  [#permalink]

Show Tags

New post 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.


Answer: C


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.
_________________
GMAT Club Bot
Re: M05-05   [#permalink] 26 Aug 2019, 02:03
Display posts from previous: Sort by

M05-05

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  

Moderators: chetan2u, Bunuel






Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne