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

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The operation Θ is defined by x Θ y = 1/x + 1/y for all [#permalink]

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28 Sep 2006, 02:17
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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
[Reveal] Spoiler: OA

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Last edited by Bunuel on 08 Sep 2013, 05:45, edited 3 times in total.
Edited the question.
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Re: The operation theta is defined by [#permalink]

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08 Sep 2013, 03:19
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~!!~
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Re: The operation theta is defined by [#permalink]

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08 Sep 2013, 05:48
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Expert's post
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.

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Re: The operation Θ is defined by x Θ y = 1/x + 1/y for all [#permalink]

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01 Nov 2014, 04:00
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The operation Θ is defined by x Θ y = 1/x + 1/y for all [#permalink]

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07 Mar 2015, 07:39
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.

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

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07 Mar 2015, 07:48
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.

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
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Re: The operation Θ is defined by x Θ y = 1/x + 1/y for all [#permalink]

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07 Mar 2015, 07:57
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)?
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Re: The operation Θ is defined by x Θ y = 1/x + 1/y for all [#permalink]

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07 Mar 2015, 08:04
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
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Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
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Re: The operation Θ is defined by x Θ y = 1/x + 1/y for all [#permalink]

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07 Mar 2015, 08:10
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
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Re: The operation Θ is defined by x Θ y = 1/x + 1/y for all [#permalink]

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07 Mar 2015, 08:17
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
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The operation Θ is defined by x Θ y = 1/x + 1/y for all [#permalink]

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07 Mar 2015, 08:34
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?
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The operation Θ is defined by x Θ y = 1/x + 1/y for all [#permalink]

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07 Mar 2015, 08:47
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
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Re: The operation Θ is defined by x Θ y = 1/x + 1/y for all [#permalink]

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07 Mar 2015, 08:54
Thanks man. I get it now, appreciate your patience
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Re: The operation Θ is defined by x Θ y = 1/x + 1/y for all [#permalink]

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04 Jul 2015, 17:18
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.

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!

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The operation Θ is defined by x Θ y = 1/x + 1/y for all [#permalink]

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28 Aug 2015, 04:25
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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

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The operation Θ is defined by x Θ y = 1/x + 1/y for all   [#permalink] 28 Aug 2015, 04:25
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