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# D01-42

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Math Expert
Joined: 02 Sep 2009
Posts: 43894

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15 Sep 2014, 23:13
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Difficulty:

55% (hard)

Question Stats:

53% (00:48) correct 47% (00:44) wrong based on 204 sessions

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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. $$20$$
[Reveal] Spoiler: OA

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Math Expert
Joined: 02 Sep 2009
Posts: 43894

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15 Sep 2014, 23:13
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2
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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.

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Joined: 27 Apr 2014
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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.

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Math Expert
Joined: 02 Sep 2009
Posts: 43894

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26 Dec 2014, 07:01
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.

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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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.

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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Joined: 09 Nov 2015
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10 Dec 2015, 02:48
I think this is a high-quality 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
Math Expert
Joined: 02 Sep 2009
Posts: 43894

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12 Dec 2015, 07:38
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Expert's post
Bhavanasg wrote:
I think this is a high-quality 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

Hope it helps.
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Joined: 16 Feb 2016
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26 Apr 2016, 22:58
Good question.
Need to consider the negative values as well...
Intern
Joined: 09 Sep 2014
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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.
Manager
Joined: 03 Dec 2013
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Location: United States (HI)
GMAT 1: 660 Q49 V30
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22 Aug 2016, 11:48
I think this is a high-quality question and I agree with explanation. GG. Well Played...
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Joined: 12 May 2016
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02 Jan 2017, 03:59
I think this is a poor-quality 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
Math Expert
Joined: 02 Sep 2009
Posts: 43894

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02 Jan 2017, 04:18
C3BISHT wrote:
I think this is a poor-quality 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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21 Feb 2017, 17:46
This is a high-quality and "tricky" question.
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Joined: 28 Dec 2016
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Location: United States
GMAT 1: 640 Q42 V36
GMAT 2: 700 Q47 V38
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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?
Math Expert
Joined: 02 Sep 2009
Posts: 43894

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27 May 2017, 11:57
conneryeon001 wrote:
Bunuel,

Is there a reason why you skipped -5 among the possible values of x?

It was a typo. Yes, c can be -5 too. Edited. Thank you for noticing.
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06 Feb 2018, 13:03
I think this is a poor-quality question and I agree with explanation.
Math Expert
Joined: 02 Sep 2009
Posts: 43894

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06 Feb 2018, 19:44
shashant wrote:
I think this is a poor-quality question and I agree with explanation.

Care to elaborate a bit? The question s 100% mathematically correct and GMAT-like.
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Re: D01-42   [#permalink] 06 Feb 2018, 19:44
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# D01-42

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