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probablity for an eq

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probablity for an eq [#permalink] New post 25 Oct 2010, 06:43
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B
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E

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  35% (medium)

Question Stats:

78% (01:52) correct 22% (00:27) wrong based on 9 sessions
{-10,-6,-5,-4,-2.5,-1,0,2.5,4,6,7,10}

A number is selectedat random from the set above. what is the probablitiy that the number selected will be a solution of the equation (x-5)(x+10)(2x-5)=0?
1/12
1/6
1/4
1/3
1/2
[Reveal] Spoiler: OA
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Re: probablity for an eq [#permalink] New post 25 Oct 2010, 06:50
Expert's post
vanidhar wrote:
{-10, -6, -5, -4, -2.5, -1, 0, 2.5, 4, 6, 7, 10}

A number is selectedat random from the set above. what is the probablitiy that the number selected will be a solution of the equation (x-5)(x+10)(2x-5)=0?
1/12
1/6
1/4
1/3
1/2


Solutions of the equation (x-5)(x+10)(2x-5)=0 are: x=5, x=-10 and x=2.5.

There are total of 12 numbers in the set {-10, -6, -5, -4, -2.5, -1, 0, 2.5, 4, 6, 7, 10} and it contains 2 roots: -10 and 2,5, so the probablitiy that the number selected will be a solution of the given equation is P=\frac{2}{12}=\frac{1}{6}.

Answer: B.
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Re: probablity for an eq [#permalink] New post 25 Oct 2010, 23:39
it contains 2 roots: -10 and 2.5 ??
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Re: probablity for an eq [#permalink] New post 25 Oct 2010, 23:45
vanidhar wrote:
it contains 2 roots: -10 and 2.5 ??


Roots = Solution to the equation

We know the only solutions are -10,2.5,5
And of these, two are present in the set at hand
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Re: probablity for an eq   [#permalink] 25 Oct 2010, 23:45
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