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Math Expert
Joined: 02 Sep 2009
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16 Sep 2014, 00:13
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9
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Difficulty:

45% (medium)

Question Stats:

54% (01:12) correct 46% (01:11) wrong based on 256 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. $$20$$

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Posts: 58434

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16 Sep 2014, 00:13
5
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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26 Dec 2014, 04:09
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
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Posts: 58434

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26 Dec 2014, 08:01
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, 22:39
5
3
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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10 Dec 2015, 03:48
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
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12 Dec 2015, 08:38
1
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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26 Apr 2016, 23:58
1
Good question.
Need to consider the negative values as well...
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16 Jul 2016, 12:36
x can take positive, as well as negative values

One should always consider all possible cases in mind.
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22 Aug 2016, 12:48
I think this is a high-quality question and I agree with explanation. GG. Well Played...
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02 Jan 2017, 04:59
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
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02 Jan 2017, 05:18
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, 18:46
This is a high-quality and "tricky" question.
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27 May 2017, 12:28
1
Bunuel,

Is there a reason why you skipped -5 among the possible values of x?
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27 May 2017, 12:57
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, 14:03
I think this is a poor-quality question and I agree with explanation.
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06 Feb 2018, 20:44
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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25 Jun 2018, 08:49
This is a definitely a high quality question and I agree with the explanation.
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19 Sep 2018, 15:44
1
Very good, tricky and gmat kind of a question. Sad to see people are judging it as a poor quality question without properly understanding/thinking through the solution
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26 Jan 2019, 11:43
I think this is a high-quality question and I agree with explanation.
Re D01-42   [#permalink] 26 Jan 2019, 11:43

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