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# For all integers a nd b, where a is different from b

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Moderator
Joined: 18 Apr 2015
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Kudos [?]: 1113 [0], given: 5129

For all integers a nd b, where a is different from b [#permalink]  11 Apr 2018, 08:22
Expert's post
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Question Stats:

88% (00:16) correct 11% (00:32) wrong based on 9 sessions
For all integers and b, where a is $$a \neq b$$, $$a \ast b = |\frac{(a^2 - b^2)}{(a - b)}|$$. What is the value of $$4 \ast 2$$?

A. 2

B. 4

C. 6

D. 8

E. 10
[Reveal] Spoiler: OA

_________________
Manager
Joined: 26 Jan 2018
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Kudos [?]: 105 [0], given: 3

Re: For all integers a nd b, where a is different from b [#permalink]  11 Apr 2018, 09:09
There is a typo in the question.

(a+b)(a-b)/(a-) = a+b

(a+b)(a-b)=a^2+b^2

4*2 = 4+2

Hence C
Moderator
Joined: 18 Apr 2015
Posts: 5516
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Kudos [?]: 1113 [0], given: 5129

Re: For all integers a nd b, where a is different from b [#permalink]  11 Apr 2018, 13:15
Expert's post
Where is the typo ??

Please point out so I can fix it. If any
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Intern
Joined: 03 Apr 2018
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Kudos [?]: 7 [0], given: 13

Re: For all integers a nd b, where a is different from b [#permalink]  11 Apr 2018, 13:25
I don’t think there is one...
[(a-b)*(a+b)]/(a-b)= a+b=4+2=6
Re: For all integers a nd b, where a is different from b   [#permalink] 11 Apr 2018, 13:25
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