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Re D0142 [#permalink]
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15 Sep 2014, 23:13
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Official Solution: If x is an integer and \(9 \lt x^2 \lt 99\), then what is the value of maximum possible value of \(x\) minus minimum possible value of \(x\)?A. \(5\) B. \(6\) C. \(7\) D. \(18\) E. \(20\) Notice that \(x\) can take positive, as well as negative values to satisfy \(9 \lt x^2 \lt 99\), hence \(x\) can be: 9, 8, 7, 6, 5, 4, 4, 5, 6, 7, 8, or 9. We are asked to find the value of x(max)x(min), and since x(max)=9 and x(min)=9 then x(max)x(min)=9(9)=18. Answer: D
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Re: D0142 [#permalink]
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26 Dec 2014, 03:09
Bunuel wrote: Official Solution:
If x is an integer and \(9 \lt x^2 \lt 99\), then what is the value of maximum possible value of \(x\) minus minimum possible value of \(x\)?
A. \(5\) B. \(6\) C. \(7\) D. \(18\) E. \(20\)
Notice that \(x\) can take positive, as well as negative values to satisfy \(9 \lt x^2 \lt 99\), hence \(x\) can be: 9, 8, 7, 6, 4, 4, 5, 6, 7, 8, or 9. We are asked to find the value of x(max)x(min), and since x(max)=9 and x(min)=9 then x(max)x(min)=9(9)=18.
Answer: D but the questions is asking about x not x^2 though
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amenon55 wrote: Bunuel wrote: Official Solution:
If x is an integer and \(9 \lt x^2 \lt 99\), then what is the value of maximum possible value of \(x\) minus minimum possible value of \(x\)?
A. \(5\) B. \(6\) C. \(7\) D. \(18\) E. \(20\)
Notice that \(x\) can take positive, as well as negative values to satisfy \(9 \lt x^2 \lt 99\), hence \(x\) can be: 9, 8, 7, 6, 4, 4, 5, 6, 7, 8, or 9. We are asked to find the value of x(max)x(min), and since x(max)=9 and x(min)=9 then x(max)x(min)=9(9)=18.
Answer: D but the questions is asking about x not x^2 though The question asks what is the value of maximum possible value of \(x\) minus minimum possible value of \(x\)? x (not x^2) can be: 9, 8, 7, 6, 5, 4, 4, 5, 6, 7, 8, or 9: x(max)  x(min) = 9  (9) = 18.
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Re: D0142 [#permalink]
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02 Dec 2015, 21:39
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amenon55 wrote: Bunuel wrote: Official Solution:
If x is an integer and \(9 \lt x^2 \lt 99\), then what is the value of maximum possible value of \(x\) minus minimum possible value of \(x\)?
A. \(5\) B. \(6\) C. \(7\) D. \(18\) E. \(20\)
Notice that \(x\) can take positive, as well as negative values to satisfy \(9 \lt x^2 \lt 99\), hence \(x\) can be: 9, 8, 7, 6, 4, 4, 5, 6, 7, 8, or 9. We are asked to find the value of x(max)x(min), and since x(max)=9 and x(min)=9 then x(max)x(min)=9(9)=18.
Answer: D but the questions is asking about x not x^2 though 9 < x^2 < 99 Take the square root of all sides. Now when you take the square root of a number you get a positive and a negative root: √9 = ±3 and √99 ≅ ±9.95. This produces two intervals for x: 9.95 < x < 3 and 3 < x < 9.95 Since x is an integer, its maximum value is x = 9 and its minimum value is x = 9 (Max Value of x)  (Min Value of x) = 9  (9) = 9 + 9 = 18



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Re: D0142 [#permalink]
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10 Dec 2015, 02:48
I think this is a highquality question and the explanation isn't clear enough, please elaborate. since 9< x^2 < 99 means +/ 3 <x< +/10 (square root) therefore max minus min value would be +10  (3) which is 10 + 3 = 13. Please tell me where am i going wrong with my approach. thank you



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Re: D0142 [#permalink]
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12 Dec 2015, 07:38



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Re: D0142 [#permalink]
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26 Apr 2016, 22:58
Good question. Need to consider the negative values as well...



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Re: D0142 [#permalink]
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16 Jul 2016, 11:36
x can take positive, as well as negative values
One should always consider all possible cases in mind.



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Re: D0142 [#permalink]
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22 Aug 2016, 11:48
I think this is a highquality question and I agree with explanation. GG. Well Played...



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Re: D0142 [#permalink]
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02 Jan 2017, 03:59
I think this is a poorquality question and the explanation isn't clear enough, please elaborate. Since 9<x2 <99 so values of x should range
3,2,1,0,1,2,3,4,5,6,7,8,9



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Re: D0142 [#permalink]
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02 Jan 2017, 04:18
C3BISHT wrote: I think this is a poorquality question and the explanation isn't clear enough, please elaborate. Since 9<x2 <99 so values of x should range
3,2,1,0,1,2,3,4,5,6,7,8,9 I have to disagree. The question is correct and of a good quality. As for your doubt. If 9 < x^2 < 99 how can x be 3, 2, 1, or 0? The do not satisfy 9 < x^2 < 99. x can be: 9, 8, 7, 6, 4, 4, 5, 6, 7, 8, or 9.
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Re: D0142 [#permalink]
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21 Feb 2017, 17:46
This is a highquality and "tricky" question.



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Re: D0142 [#permalink]
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27 May 2017, 11:28
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Bunuel, Is there a reason why you skipped 5 among the possible values of x?



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Re: D0142 [#permalink]
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27 May 2017, 11:57



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Re D0142 [#permalink]
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06 Feb 2018, 13:03
I think this is a poorquality question and I agree with explanation.



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