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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, 05:00

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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, 08:39

1

This post was BOOKMARKED

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 \(\hspace{10} z \hspace{2} \theta \hspace{2} \frac{z}{(z-1)}=\frac{1}{3}+\frac{3}{3-1}=/=1\)

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 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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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
_________________

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?

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
_________________

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

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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!

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

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07 Jul 2017, 10:26

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.
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