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A number is to be selected at random from the set above. What is the

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A number is to be selected at random from the set above. What is the  [#permalink]

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New post 25 Oct 2010, 07:43
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{-10, -6, -5, -4, -2.5, -1, 0, 2.5, 4, 6, 7, 10}

A number is to be selected at random from the set above. What is the probability that the number selected will be a solution of the equation \((x - 5)(x + 10)(2x - 5) = 0\)?

A. 1/12
B. 1/6
C. 1/4
D. 1/3
E. 1/2
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Re: A number is to be selected at random from the set above. What is the  [#permalink]

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New post Updated on: 01 May 2012, 02:41
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{-10, -6, -5, -4, -2.5, -1, 0, 2.5, 4, 6, 7, 10}

A number is to be selected at random from the set above. What is the probability that the number selected will be a solution of the equation (x-5)(x+10)(2x-5) = 0?

A. 1/12
B. 1/6
C. 1/4
D. 1/3
E. 1/2
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Originally posted by Stiv on 01 May 2012, 02:30.
Last edited by Bunuel on 01 May 2012, 02:41, edited 1 time in total.
Edited the question and added the OA
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Re: A number is to be selected at random from the set above. What is the  [#permalink]

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New post 25 Oct 2010, 07:50
vanidhar wrote:
{-10, -6, -5, -4, -2.5, -1, 0, 2.5, 4, 6, 7, 10}

A number is to be selected at random from the set above. What is the probability that the number selected will be a solution of the equation (x - 5)(x + 10)(2x - 5) = 0?

A. 1/12
B. 1/6
C. 1/4
D. 1/3
E. 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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New post 26 Oct 2010, 00:39
it contains 2 roots: -10 and 2.5 ??
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Re: A number is to be selected at random from the set above. What is the  [#permalink]

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New post 26 Oct 2010, 00:45
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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New post 01 May 2012, 02:44
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22. Probability



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New post 02 May 2012, 00:39
Aaaaaa OMG, I actually didn't notice that there isn't number 5(positive) in this bunch of numbers. I assumed it, and I got it wrong!

My advice to everyone: Don't assume anything!!! ;)
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Re: A number is to be selected at random from the set above. What is the  [#permalink]

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New post 20 Jan 2018, 16:07
Hi All,

We're asked for the probability of choosing a number (from a given set of 12 numbers) that will "fit" a given equation. That equation is set equal to 0, which makes solving the equation rather easy. Since we're dealing with the PRODUCT of three "terms", if ANY of those terms equals 0, then the product will equal 0.

IF... we select X = -10 from the set, then we have (-15)(0)(-25) = 0, so X = -10 IS a solution.
IF... we select X = 0 from the set, then we have (-5)(10)(-5) = 250, which is NOT a solution.

The given equation has 3 solutions: +5, -10 and +5/2. However, only two of those solutions appear in the set (-10 and 2.5). Thus, the probability of selecting a number that fits the equation is 2/12 = 1/6

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New post 14 Jun 2018, 05:05
Bunuel wrote:
vanidhar wrote:
{-10, -6, -5, -4, -2.5, -1, 0, 2.5, 4, 6, 7, 10}

A number is to be selected at random from the set above. What is the probability that the number selected will be a solution of the equation (x - 5)(x + 10)(2x - 5) = 0?

A. 1/12
B. 1/6
C. 1/4
D. 1/3
E. 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.




Hi pushpitkc

can you please explain what i did wrong

i did this \(x^2+10-5x-50(2x-5)\) now left with it

\(x^2+10x-5x-100x+250 = 0\)

\(x^2+10x-5x-100x+250 = 0\)

\(x^2 -95x+250 = 0\) now what :?

thank you :-)
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Re: A number is to be selected at random from the set above. What is the  [#permalink]

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New post 14 Jun 2018, 06:39
1
dave13 wrote:
Bunuel wrote:
vanidhar wrote:
{-10, -6, -5, -4, -2.5, -1, 0, 2.5, 4, 6, 7, 10}

A number is to be selected at random from the set above. What is the probability that the number selected will be a solution of the equation (x - 5)(x + 10)(2x - 5) = 0?

A. 1/12
B. 1/6
C. 1/4
D. 1/3
E. 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.


Hi pushpitkc

can you please explain what i did wrong

i did this \(x^2+10-5x-50(2x-5)\) now left with it

\(x^2+10x-5x-100x+250 = 0\)

\(x^2+10x-5x-100x+250 = 0\)

\(x^2 -95x+250 = 0\) now what :?

thank you :-)


Hi dave13

You have 3 values of x possible and we need to check how many of these values are part
of the set. I don't seem to understand why are you trying to form a quadratic equation :(

Btw, as for your question, the highlighted portion is wrong

\((x-5)(x+10)(2x-5) = 0\) -> \((x^2 - 5x +10x - 50)(2x - 5) = 0\) -> \((x^2 + 5x - 50)(2x - 5) = 0\)
The entire equation will need to be multiplied and we will have a cubic equation

Hope this clears your confusion!
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New post 14 Jun 2018, 09:10
pushpitkc thanks can you please show how do we find 2 roots: -10 and 2,5, :? i am kind of confused what to do with highlighted part :-)
\((x^2 + 5x - 50)(2x - 5) = 0\)
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Re: A number is to be selected at random from the set above. What is the  [#permalink]

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New post 14 Jun 2018, 09:45
Count there are 12 items
x = 5 or x = -10 or x = 2.5
Two are in the set
B.
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Re: A number is to be selected at random from the set above. What is the  [#permalink]

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New post 14 Jun 2018, 10:54
1
dave13 wrote:
pushpitkc thanks can you please show how do we find 2 roots: -10 and 2,5, :? i am kind of confused what to do with highlighted part :-)
\((x^2 + 5x - 50)(2x - 5) = 0\)


Hey dave13

If 2x - 5 = 0 -> 2x = 5 -> x = 5/2 = 2.5
If x - 5 = 0 -> x = 5
If x + 10 = 0 -> x = -10

Therefore, the roots of the quadratic equation are -10, 5, and 2.5
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Re: A number is to be selected at random from the set above. What is the  [#permalink]

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New post 17 Jun 2018, 19:36
vanidhar wrote:
{-10, -6, -5, -4, -2.5, -1, 0, 2.5, 4, 6, 7, 10}

A number is to be selected at random from the set above. What is the probability that the number selected will be a solution of the equation \((x - 5)(x + 10)(2x - 5) = 0\)?

A. 1/12
B. 1/6
C. 1/4
D. 1/3
E. 1/2


According to the given equation, x can equal the following:

x - 5 = 0

x = 5

or

x + 10 = 0

x = -10

or

2x - 5 = 0

2x = 5

x = 2.5

Since -10 and 2.5 are in the set, the probability of selecting one of those solutions is 2/12 = 1/6.

Answer: B
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Re: A number is to be selected at random from the set above. What is the &nbs [#permalink] 17 Jun 2018, 19:36
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