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

It is currently 14 Oct 2019, 16:51

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

The operation Θ is defined by x Θ y = 1/x + 1/y for all

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

Hide Tags

Find Similar Topics 
Manager
Manager
avatar
Joined: 10 Oct 2005
Posts: 103
Location: Hollywood
The operation Θ is defined by x Θ y = 1/x + 1/y for all  [#permalink]

Show Tags

New post Updated on: 23 Jul 2018, 22:47
1
12
00:00
A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

63% (01:39) correct 37% (02:05) wrong based on 583 sessions

HideShow timer Statistics

The operation Θ is defined by x Θ y = 1/x + 1/y for all nonzero numbers x and y. If z is a number greater than 1, which of the following must be true?

I. \(z \theta (-z) = 0\)

II. \(z \theta \frac{z}{(z-1)}= 1\)

III. \(\frac{2}{z} \theta \frac{2}{z}= z\)

A. I only
B. I and II only
C. I and III only
D. II and III only
E. I, II, and III

_________________
The GMAT, too tough to be denied.
Beat the tough questions...

Originally posted by TOUGH GUY on 28 Sep 2006, 03:17.
Last edited by Bunuel on 23 Jul 2018, 22:47, edited 4 times in total.
Edited the question.
Intern
Intern
avatar
B
Joined: 01 Feb 2013
Posts: 36
Location: India
Concentration: Technology, Leadership
GMAT 1: 750 Q50 V41
GPA: 3.49
WE: Engineering (Computer Software)
Reviews Badge
Re: The operation Θ is defined by x Θ y = 1/x + 1/y for all  [#permalink]

Show Tags

New post 08 Sep 2013, 04:19
1
Can anyone verify the question? I think that first option should be z Θ (-z) =0 for this to be true!..
So either D or E! cheers~!!~
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 58320
Re: The operation Θ is defined by x Θ y = 1/x + 1/y for all  [#permalink]

Show Tags

New post 08 Sep 2013, 06:48
1
rskanumuri wrote:
Can anyone verify the question? I think that first option should be z Θ (-z) =0 for this to be true!..
So either D or E! cheers~!!~


Edited the question.

The operation Θ is defined by x Θ y = 1/x + 1/y for all nonzero numbers x and y. If z is a number greater than 1, which of the following must be true?

I. \(\hspace{10} z \hspace{2} \theta \hspace{2} (-z) = 0\)

II. \(\hspace{10} z \hspace{2} \theta \hspace{2} \frac{z}{(z-1)}= 1\)

III. \(\hspace{10} \frac{2}{z} \hspace{2} \theta \hspace{2} \frac{2}{z}= z\)

A. I only
B. I and II only
C. I and III only
D. II and III only
E. I, II, and III

I. \(\hspace{10} z \hspace{2} \theta \hspace{2} (-z) =\frac{1}{z}-\frac{1}{z}=0\) --> true.

II. \(\hspace{10} z \hspace{2} \theta \hspace{2} \frac{z}{(z-1)}=\frac{1}{z}+\frac{z-1}{z}=1\) --> true.

III. \(\hspace{10} \frac{2}{z} \hspace{2} \theta \hspace{2} \frac{2}{z}= \frac{z}{2}+\frac{z}{2}=z\) --> true.

Answer: E.
_________________
Manager
Manager
avatar
Joined: 26 Feb 2015
Posts: 110
Re: The operation Θ is defined by x Θ y = 1/x + 1/y for all  [#permalink]

Show Tags

New post 07 Mar 2015, 08:39
1
Bunuel wrote:
rskanumuri wrote:
Can anyone verify the question? I think that first option should be z Θ (-z) =0 for this to be true!..
So either D or E! cheers~!!~


Edited the question.

The operation Θ is defined by x Θ y = 1/x + 1/y for all nonzero numbers x and y. If z is a number greater than 1, which of the following must be true?

I. \(\hspace{10} z \hspace{2} \theta \hspace{2} (-z) = 0\)

II. \(\hspace{10} z \hspace{2} \theta \hspace{2} \frac{z}{(z-1)}= 1\)

III. \(\hspace{10} \frac{2}{z} \hspace{2} \theta \hspace{2} \frac{2}{z}= z\)

A. I only
B. I and II only
C. I and III only
D. II and III only
E. I, II, and III

I. \(\hspace{10} z \hspace{2} \theta \hspace{2} (-z) =\frac{1}{z}-\frac{1}{z}=0\) --> true.

II. \(\hspace{10} z \hspace{2} \theta \hspace{2} \frac{z}{(z-1)}=\frac{1}{z}+\frac{z-1}{z}=1\) --> true.

III. \(\hspace{10} \frac{2}{z} \hspace{2} \theta \hspace{2} \frac{2}{z}= \frac{z}{2}+\frac{z}{2}=z\) --> true.

Answer: E.


I don't understand the algebraic form for option "II". I mean, lets say z = 3. Then I'd get \(\hspace{10} z \hspace{2} \theta \hspace{2} \frac{z}{(z-1)}=\frac{1}{3}+\frac{3}{3-1}=/=1\)
Math Expert
avatar
V
Joined: 02 Aug 2009
Posts: 7954
Re: The operation Θ is defined by x Θ y = 1/x + 1/y for all  [#permalink]

Show Tags

New post 07 Mar 2015, 08:48
erikvm wrote:
Bunuel wrote:
rskanumuri wrote:
Can anyone verify the question? I think that first option should be z Θ (-z) =0 for this to be true!..
So either D or E! cheers~!!~


Edited the question.

The operation Θ is defined by x Θ y = 1/x + 1/y for all nonzero numbers x and y. If z is a number greater than 1, which of the following must be true?

I. \(\hspace{10} z \hspace{2} \theta \hspace{2} (-z) = 0\)

II. \(\hspace{10} z \hspace{2} \theta \hspace{2} \frac{z}{(z-1)}= 1\)

III. \(\hspace{10} \frac{2}{z} \hspace{2} \theta \hspace{2} \frac{2}{z}= z\)

A. I only
B. I and II only
C. I and III only
D. II and III only
E. I, II, and III

I. \(\hspace{10} z \hspace{2} \theta \hspace{2} (-z) =\frac{1}{z}-\frac{1}{z}=0\) --> true.

II. \(\hspace{10} z \hspace{2} \theta \hspace{2} \frac{z}{(z-1)}=\frac{1}{z}+\frac{z-1}{z}=1\) --> true.

III. \(\hspace{10} \frac{2}{z} \hspace{2} \theta \hspace{2} \frac{2}{z}= \frac{z}{2}+\frac{z}{2}=z\) --> true.

Answer: E.


I don't understand the algebraic form for option "II". I mean, lets say z = 3. Then I'd get \(\hspace{10} z \hspace{2} \theta \hspace{2} \frac{z}{(z-1)}=\frac{1}{3}+\frac{3}{3-1}=/=1\)



hi erikvm,

hi option ll says we are adding reciprocals of x and y.....
it will not be \(\frac{1}{z}+\frac{z}{(z-1)}\) but \(\frac{1}{z}+\frac{(z-1)}{z}\) =1/3+2/3=1
hope it helped
_________________
Manager
Manager
avatar
Joined: 26 Feb 2015
Posts: 110
Re: The operation Θ is defined by x Θ y = 1/x + 1/y for all  [#permalink]

Show Tags

New post 07 Mar 2015, 08:57
Okay well why isnt it then (1/3) + (1/3) / (-2/3)?

I mean, because 1/3 - 1 = -2/3. So it should be 1/3 + (1/3)*(3/-2)?
Math Expert
avatar
V
Joined: 02 Aug 2009
Posts: 7954
Re: The operation Θ is defined by x Θ y = 1/x + 1/y for all  [#permalink]

Show Tags

New post 07 Mar 2015, 09:04
erikvm wrote:
Okay well why isnt it then (1/3) + (1/3) / (-2/3)?

I mean, because 1/3 - 1 = -2/3. So it should be 1/3 + (1/3)*(3/-2)?


hi,
it is so because u have taken z as 3 and not 1/3.... replace 3 for z.. and u will get the answer
_________________
Manager
Manager
avatar
Joined: 26 Feb 2015
Posts: 110
Re: The operation Θ is defined by x Θ y = 1/x + 1/y for all  [#permalink]

Show Tags

New post 07 Mar 2015, 09:10
I never get the fractions to work so I just uploaded a picture of it instead, where do I go wrong?

http://i.imgur.com/ISdtWL7.png
Math Expert
avatar
V
Joined: 02 Aug 2009
Posts: 7954
Re: The operation Θ is defined by x Θ y = 1/x + 1/y for all  [#permalink]

Show Tags

New post 07 Mar 2015, 09:17
erikvm wrote:
I never get the fractions to work so I just uploaded a picture of it instead, where do I go wrong?

http://i.imgur.com/ISdtWL7.png


in the attached fig...x Θ y = 1/x + 1/y
here x is z and y is\(\frac{z}{(z-1)}\)...
you have taken z as 3...
so x=z=3..and y=\(\frac{z}{(z-1)}\)=3/(3-1)=3/2...
1/x + 1/y=1/3+1/(3/2)=1/3+2/3=1
you are taking z as 3 for x and z as 1/3 for y.. that is why u are going wrong
_________________
Manager
Manager
avatar
Joined: 26 Feb 2015
Posts: 110
Re: The operation Θ is defined by x Θ y = 1/x + 1/y for all  [#permalink]

Show Tags

New post 07 Mar 2015, 09:34
chetan2u wrote:
erikvm wrote:
I never get the fractions to work so I just uploaded a picture of it instead, where do I go wrong?

http://i.imgur.com/ISdtWL7.png


in the attached fig...x Θ y = 1/x + 1/y
here x is z and y is\(\frac{z}{(z-1)}\)...
you have taken z as 3...
so x=z=3..and y=\(\frac{z}{(z-1)}\)=3/(3-1)=3/2...
1/x + 1/y=1/3+1/(3/2)=1/3+2/3=1
you are taking z as 3 for x and z as 1/3 for y.. that is why u are going wrong


Im sorry man, I'm sure you are really annoyed, but I really dont understand..
What values am I suppose to take?
Math Expert
avatar
V
Joined: 02 Aug 2009
Posts: 7954
Re: The operation Θ is defined by x Θ y = 1/x + 1/y for all  [#permalink]

Show Tags

New post 07 Mar 2015, 09:47
erikvm wrote:
chetan2u wrote:
erikvm wrote:
I never get the fractions to work so I just uploaded a picture of it instead, where do I go wrong?

http://i.imgur.com/ISdtWL7.png


in the attached fig...x Θ y = 1/x + 1/y
here x is z and y is\(\frac{z}{(z-1)}\)...
you have taken z as 3...
so x=z=3..and y=\(\frac{z}{(z-1)}\)=3/(3-1)=3/2...
1/x + 1/y=1/3+1/(3/2)=1/3+2/3=1
you are taking z as 3 for x and z as 1/3 for y.. that is why u are going wrong


Im sorry man, I'm sure you are really annoyed, but I really dont understand..
What values am I suppose to take?


does not matter...
look dont take any values and work with z itself..
z Θ \(\frac{z}{(z-1)}\)....
since x Θ y = 1/x + 1/y...
z Θ \(\frac{z}{(z-1)}\)=\(\frac{1}{z}\)+1/\(\frac{z}{(z-1)}\)= \(\frac{1}{z}\)+\(\frac{(z-1)}{z)}\)....
\(\frac{(1+z-1)}{z}\)=z/z=1
_________________
Manager
Manager
avatar
Joined: 26 Feb 2015
Posts: 110
Re: The operation Θ is defined by x Θ y = 1/x + 1/y for all  [#permalink]

Show Tags

New post 07 Mar 2015, 09:54
Thanks man. I get it now, appreciate your patience
Intern
Intern
avatar
Joined: 26 Jun 2015
Posts: 2
Concentration: Finance, Strategy
GMAT 1: 740 Q48 V44
GPA: 3.67
Re: The operation Θ is defined by x Θ y = 1/x + 1/y for all  [#permalink]

Show Tags

New post 04 Jul 2015, 18:18
Bunuel wrote:
rskanumuri wrote:
Can anyone verify the question? I think that first option should be z Θ (-z) =0 for this to be true!..
So either D or E! cheers~!!~


Edited the question.

The operation Θ is defined by x Θ y = 1/x + 1/y for all nonzero numbers x and y. If z is a number greater than 1, which of the following must be true?

I. \(\hspace{10} z \hspace{2} \theta \hspace{2} (-z) = 0\)

II. \(\hspace{10} z \hspace{2} \theta \hspace{2} \frac{z}{(z-1)}= 1\)

III. \(\hspace{10} \frac{2}{z} \hspace{2} \theta \hspace{2} \frac{2}{z}= z\)

A. I only
B. I and II only
C. I and III only
D. II and III only
E. I, II, and III

I. \(\hspace{10} z \hspace{2} \theta \hspace{2} (-z) =\frac{1}{z}-\frac{1}{z}=0\) --> true.

II. \(\hspace{10} z \hspace{2} \theta \hspace{2} \frac{z}{(z-1)}=\frac{1}{z}+\frac{z-1}{z}=1\) --> true.

III. \(\hspace{10} \frac{2}{z} \hspace{2} \theta \hspace{2} \frac{2}{z}= \frac{z}{2}+\frac{z}{2}=z\) --> true.

Answer: E.


Hi Bunuel,

I don't know if that is the ideal place to ask my question, but I haven't found any post gathering the broader topic, i.e Functions / Expressions.

In such exercise, can we consider the assumption right in any case if it did work using 1 number (respecting the constraint given by the wording, obviously). I used 2 for that one, but lost time double checking with 8 and 9, probably a bad reflex acquired after thoroughly testing numbers in DS questions!

Thanks in advance,

RCM
Intern
Intern
avatar
Joined: 02 Jan 2015
Posts: 22
GMAT ToolKit User
Re: The operation Θ is defined by x Θ y = 1/x + 1/y for all  [#permalink]

Show Tags

New post 28 Aug 2015, 05:25
3
The easiest way to answer this is to use substitution... I used 5 to check
so
I) Z Θ (-Z) = \(\frac{1}{5}\) - \(\frac{1}{5}\) = 0

II) Z Θ \(\frac{Z}{Z-1}\) = \(\frac{1}{5}\) + 1/\(\frac{5}{4}\) = \(\frac{1}{5}\) + \(\frac{4}{5}\) = 1

III) \(\frac{2}{Z}\) Θ \(\frac{2}{Z}\) = \(\frac{5}{2}\)+\(\frac{5}{2}\) = \(\frac{10}{2}\) = 5 =Z

Hence I,II,III are all possible


------------------------------------------------------
Kindly press"+1 Kudos" to appreciate
Non-Human User
User avatar
Joined: 09 Sep 2013
Posts: 13143
Re: The operation Θ is defined by x Θ y = 1/x + 1/y for all  [#permalink]

Show Tags

New post 01 Nov 2018, 07:56
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
GMAT Club Bot
Re: The operation Θ is defined by x Θ y = 1/x + 1/y for all   [#permalink] 01 Nov 2018, 07:56
Display posts from previous: Sort by

The operation Θ is defined by x Θ y = 1/x + 1/y for all

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





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