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

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TOUGH GUY
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The operation Θ is defined by x Θ y = 1/x + 1/y for all [#permalink] New post   Updated on: 23 Jul 2018, 22:47
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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. \(z \theta (-z) = 0\)

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
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E

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.
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Re: The operation Θ is defined by x Θ y = 1/x + 1/y for all [#permalink] New post   28 Aug 2015, 05:25
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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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Re: The operation Θ is defined by x Θ y = 1/x + 1/y for all [#permalink] New post   08 Sep 2013, 06:48
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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.
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Re: The operation Θ is defined by x Θ y = 1/x + 1/y for all [#permalink] New post   07 Mar 2015, 08:39
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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 \(\hspace{10} z \hspace{2} \theta \hspace{2} \frac{z}{(z-1)}=\frac{1}{3}+\frac{3}{3-1}=/=1\)
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Re: The operation is defined by x y = 1/x + 1/y for all [#permalink] New post   22 Mar 2023, 16:45
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A hint for anyone looking at this in 2023 and beyond. It's much easier to solve this question if you think about it this way...
The question is essentially asking you the sum of the reciprocal of each term on either side of the unique symbol. You can skip a few steps if you go straight to writing out the equations.
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Re: The operation Θ is defined by x Θ y = 1/x + 1/y for all [#permalink] New post   08 Sep 2013, 04:19
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 Θ is defined by x Θ y = 1/x + 1/y for all [#permalink] New post   07 Mar 2015, 08:48
Expert Reply
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
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Re: The operation Θ is defined by x Θ y = 1/x + 1/y for all [#permalink] 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)?
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Re: The operation Θ is defined by x Θ y = 1/x + 1/y for all [#permalink] New post   07 Mar 2015, 09:04
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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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Re: The operation Θ is defined by x Θ y = 1/x + 1/y for all [#permalink] 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?

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

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

https://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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Re: The operation Θ is defined by x Θ y = 1/x + 1/y for all [#permalink] New post   07 Mar 2015, 09:47
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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?

https://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] New post   07 Mar 2015, 09:54
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] 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
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Re: The operation is defined by x y = 1/x + 1/y for all [#permalink] New post   22 Apr 2022, 03:16
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 brunel, could you help expand II and III please ? I did not quite get the calculation? Thanks
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