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# prob X-by

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Senior Manager
Joined: 27 May 2009
Posts: 265

Kudos [?]: 571 [0], given: 18

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30 Aug 2009, 11:11
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OA = 1/6 how???
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Kudos [?]: 571 [0], given: 18

Manager
Joined: 10 Aug 2009
Posts: 129

Kudos [?]: 77 [0], given: 10

Re: prob X-by [#permalink]

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30 Aug 2009, 12:26
You have 4C2=6 different products of 2 numbers.
If you write down this six options and simplify, you will see that only one option fits the give expression.
this option is $$(x-y)\times (x+y)=x^2-y^2$$...All other products will be of incorrect form.
Hence the prob = 1/6

Kudos [?]: 77 [0], given: 10

Manager
Joined: 25 Aug 2009
Posts: 174

Kudos [?]: 113 [0], given: 12

Re: prob X-by [#permalink]

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30 Aug 2009, 16:42
total possibilities = $$4C2$$ = 6

Only one selection yields the favorable outcome i.e. (x + y) and (x - y) = 1

Prob. = 1/6

Kudos [?]: 113 [0], given: 12

Manager
Joined: 05 Jul 2009
Posts: 181

Kudos [?]: 55 [0], given: 5

Re: prob X-by [#permalink]

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30 Aug 2009, 20:13
This is not a data sufficiency problem. Posted in the wrong section.

Kudos [?]: 55 [0], given: 5

Math Expert
Joined: 02 Sep 2009
Posts: 43292

Kudos [?]: 139143 [0], given: 12776

Re: prob X-by [#permalink]

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04 Sep 2009, 18:53
LenaA wrote:
You have 4C2=6 different products of 2 numbers.
If you write down this six options and simplify, you will see that only one option fits the give expression.
this option is $$(x-y)\times (x+y)=x^2-y^2$$...All other products will be of incorrect form.
Hence the prob = 1/6

Think it's not necessary to write down six options, x^2-(bx)^2=(x+bx)(x-bx)
We can see that there is only one pair like this one: (x+y)(x-y).

_________________

Kudos [?]: 139143 [0], given: 12776

Re: prob X-by   [#permalink] 04 Sep 2009, 18:53
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# prob X-by

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