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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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03 Jul 2017, 03:43
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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
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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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03 Jul 2017, 03:57
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 = 63 = 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
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04 Oct 2018, 00:35
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 Mean = Sum \({4, 4, 5, 5, 6, x} / 6 = (24+x)/6 = x^2/2\) i.e. \(24+x = 3x^2\) i.e. \(x(3x1) = 24\) i.e. \(x = 3\) I.e. range = Highest value  Lowest value = 63 = 3 Answer: 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
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06 Oct 2018, 13:38
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
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06 Oct 2018, 14:51
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 : \(3x^2 9x+8x24=0\) ==> \(3x(x3)+8(x3)=0\) ==> \((x3)(3x+8)=0\)==> \(x=3\) 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
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07 Oct 2018, 00:25
Thank you very much pranjal123I 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
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07 Oct 2018, 06:29
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) \(48+2x=6x^2\) \(48+2x6x^2=0\) (rearrange and multiply by 1) \(6x^22x48=0\) (divide by 2) \(3x^2  x  24 = 0\) (factorize ) 24 *3 = 72 > 9*8 = 72 \((3x^2  9x)+(8x  24) = 0\) factor out \(3x(x3)+8(x3)=0\) \((3x+8)(x3)=0\)



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Re: The mean of the set of the positive integers {4, 4, 5, 5, 6, x} is x^2
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13 Jan 2019, 23:47
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 Dear Moderator, This same question has been duplicated in the link below, you may wish to merge the same, also let me know if I am missing anything.Thank you. https://gmatclub.com/forum/themeanof ... 43880.html
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Re: The mean of the set of the positive integers {4, 4, 5, 5, 6, x} is x^2
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13 Jan 2019, 23:50
stne wrote: 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 Dear Moderator, This same question has been duplicated in the link below, you may wish to merge the same, also let me know if I am missing anything.Thank you. https://gmatclub.com/forum/themeanof ... 43880.html____________________ Merged. Thank you.
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Re: The mean of the set of the positive integers {4, 4, 5, 5, 6, x} is x^2
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14 Jan 2019, 04:01
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 solve for x we get 3x^29x+8x24=0 x= 3 and 8/3 range would be 63 = 3 IMO C
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Re: The mean of the set of the positive integers {4, 4, 5, 5, 6, x} is x^2
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