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# For all real numbers a and b, a ⋅ b ≠ 0, let a◊b = ab − 1

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Senior Manager
Joined: 10 Jul 2013
Posts: 326

Kudos [?]: 430 [0], given: 102

For all real numbers a and b, a ⋅ b ≠ 0, let a◊b = ab − 1 [#permalink]

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20 Aug 2013, 11:37
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Difficulty:

25% (medium)

Question Stats:

74% (01:35) correct 26% (01:04) wrong based on 163 sessions

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For all real numbers a and b, a ⋅ b ≠ 0, let a◊b = ab − 1. Which of the following must be true?

I. a◊b = b◊a
II. (a◊a)/a = 1◊1
III. ( a◊b)◊c = a◊(b◊c)

(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I and III only
[Reveal] Spoiler: OA

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Asif vai.....

Last edited by Bunuel on 20 Aug 2013, 11:40, edited 1 time in total.
Edited the question.

Kudos [?]: 430 [0], given: 102

Director
Joined: 14 Dec 2012
Posts: 832

Kudos [?]: 1644 [0], given: 197

Location: India
Concentration: General Management, Operations
GMAT 1: 700 Q50 V34
GPA: 3.6
Re: For all real numbers a and b, a ⋅ b ≠ 0, let a◊b = ab − 1 [#permalink]

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20 Aug 2013, 11:43
Asifpirlo wrote:
For all real numbers a and b, a ⋅ b ≠ 0, let a◊b = ab − 1. Which of the following must be true?
I. a◊b = b◊a
II. (a◊a)/a = 1◊1
III. ( a◊b)◊c = a◊(b◊c)
(A) I only (B) II only (C) III only (D) I and II only (E) I and III only

I. a◊b = b◊a
$$ab-1 = ba-1$$==>correct.

II. (a◊a)/a = 1◊1
$$\frac{(a^2-1)}{a}= 1-1$$ ==>not correct

III. ( a◊b)◊c = a◊(b◊c)
$$(ab-1)c -1 =(bc-1)a - 1$$ not correct.

hence A
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Kudos [?]: 1644 [0], given: 197

Math Expert
Joined: 02 Sep 2009
Posts: 42585

Kudos [?]: 135508 [0], given: 12697

Re: For all real numbers a and b, a ⋅ b ≠ 0, let a◊b = ab − 1 [#permalink]

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20 Aug 2013, 11:45
Asifpirlo wrote:
For all real numbers a and b, a ⋅ b ≠ 0, let a◊b = ab − 1. Which of the following must be true?

I. a◊b = b◊a
II. (a◊a)/a = 1◊1
III. ( a◊b)◊c = a◊(b◊c)

(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I and III only

I. a◊b = b◊a --> a◊b=ab-1 and b◊a=ba-1 --> ab-1=ba-1. This option is always true.

II. (a◊a)/a = 1◊1 --> (a◊a)/a=(a^1-1)/a and 1◊1=1-1=0. These 2 expressions are not always equal. Discard.

III. (a◊b)◊c = a◊(b◊c) --> (a◊b)◊c=(ab-1)◊c=(ab-1)c-1 and a◊(b◊c)=a◊(bc-1)=a(bc-1)-1. These 2 expressions are not always equal. Discard.

Similar question to practice: for-all-real-numbers-a-and-b-where-a-b-0-let-a-b-ab-102228.html

Hope this helps.
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Kudos [?]: 135508 [0], given: 12697

Non-Human User
Joined: 09 Sep 2013
Posts: 14872

Kudos [?]: 287 [0], given: 0

Re: For all real numbers a and b, a ⋅ b ≠ 0, let a◊b = ab − 1 [#permalink]

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28 Oct 2017, 23:16
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).

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Kudos [?]: 287 [0], given: 0

Re: For all real numbers a and b, a ⋅ b ≠ 0, let a◊b = ab − 1   [#permalink] 28 Oct 2017, 23:16
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