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Updated on: 04 Jul 2019, 01:10
Originally posted by Asifpirlo on 20 Aug 2013, 12:37.
Last edited by Sajjad1994 on 04 Jul 2019, 01:10, edited 3 times in total.
Last edited by Sajjad1994 on 04 Jul 2019, 01:10, edited 3 times in total.
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
76% (02:03) correct
24%
(01:44)
wrong
based on 267
sessions
History
Date
Time
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Not Attempted Yet
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
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
Source: Nova GMAT
Difficulty Level: 600
Difficulty Level: 600
20 Aug 2013, 12:43
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Asifpirlo
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
20 Aug 2013, 12:45
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Asifpirlo
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.
Answer: A.
Similar question to practice: for-all-real-numbers-a-and-b-where-a-b-0-let-a-b-ab-102228.html
Hope this helps.
