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The mean of the set of the positive integers {4, 4, 5, 5, 6, x} is x^2

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The mean of the set of the positive integers {4, 4, 5, 5, 6, x} is x^2  [#permalink]

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New post 03 Jul 2017, 02:43
00:00
A
B
C
D
E

Difficulty:

  35% (medium)

Question Stats:

66% (01:55) correct 34% (01:51) wrong based on 133 sessions

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The mean of the set of the positive integers {4, 4, 5, 5, 6, x} is x^2  [#permalink]

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New post 03 Jul 2017, 02:57
2
1
Assume that the missing value is x

We know that mean = \(\frac{x^2}{2}\)
\(\frac{24 + x}{6} = \frac{x^2}{2}\)
\(3x^2 - x - 24 = 0\)

Solving for x,
x=3 and -8/3(not possible since as x is an integer)

Range : Maximum value - Minimum value = 6-3 = 3(Option C)
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Re: The mean of the set of the positive integers {4, 4, 5, 5, 6, x} is x^2  [#permalink]

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New post 06 Oct 2018, 12:38
Bunuel, can you please explain this one? I can't understand from the already given explanation. Thanks.
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The mean of the set of the positive integers {4, 4, 5, 5, 6, x} is x^2  [#permalink]

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New post 06 Oct 2018, 13:51
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Hi MsInvBanker

Mean formula is \(\frac{sum of elements}{number of elements}\)

from this you will get \(\frac{24+x}{6}\) . Equate this to given mean \(\frac{x^2}{2}\) to get below quadratic equation :

\(3x^2−x−24=0\)
Factorize this equation : \(3x^2 -9x+8x-24=0\) ==> \(3x(x-3)+8(x-3)=0\) ==> \((x-3)(3x+8)=0\)==> \(x=3\) as in question it is given x is +ve

Range is Maximum value in set- Minimum value in set ==> \(6 - 3 = 3\)

Please +1 kudos if you liked my post :)
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Re: The mean of the set of the positive integers {4, 4, 5, 5, 6, x} is x^2  [#permalink]

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New post 06 Oct 2018, 23:25
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Thank you very much pranjal123

I was messing up on the factorization step. I was mistakenly putting them in for \(+x\) instead of \(-x\). Thanks again.
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Re: The mean of the set of the positive integers {4, 4, 5, 5, 6, x} is x^2  [#permalink]

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New post 07 Oct 2018, 05:29
Bunuel wrote:
The mean of the set of the positive integers {4, 4, 5, 5, 6, x} is x^2/2. What is the range of the above set of integers?

(A) 1
(B) 2
(C) 3
(D) 4
(E) 5



\(\frac{24+x}{6} =\frac{x^2}{2}\)

\((24+x)2 = 6x^2\) (cross multiply)

\(48+2x=6x^2\)

\(48+2x-6x^2=0\) (rearrange and multiply by -1)

\(6x^2-2x-48=0\) (divide by 2)

\(3x^2 - x - 24 = 0\) (factorize ) -24 *3 = -72 ---> -9*8 = -72

\((3x^2 - 9x)+(8x - 24) = 0\) factor out

\(3x(x-3)+8(x-3)=0\)

\((3x+8)(x-3)=0\)
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Re: The mean of the set of the positive integers {4, 4, 5, 5, 6, x} is x^2 &nbs [#permalink] 07 Oct 2018, 05:29
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The mean of the set of the positive integers {4, 4, 5, 5, 6, x} is x^2

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