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GMAT Diagnostic Test Question 42 [#permalink]
29 Sep 2013, 21:00

Expert's post

1

This post was BOOKMARKED

GMAT Diagnostic Test Question 42

Field: Algebra Difficulty: 700

If x is an integer and 9<x^2<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 \(9<x^2<99\), hence \(x\) can be: -9, -8, -7, -6, -5, -4, 4, 5, 6, 7, 8, or 9. We asked to find the value of \(x_{max}-x_{min}\), ans since \(x_{max}=9\) and \(x_{min}=-9\) then \(x_{max}-x_{min}=9-(-9)=18\).

Re: GMAT Diagnostic Test Question 42 [#permalink]
18 Aug 2014, 20:38

bb wrote:

Explanation Official Answer: D

Notice that \(x\) can take positive, as well as negative values to satisfy \(9<x^2<99\), hence \(x\) can be: -9, -8, -7, -6, -5, -4, 4, 5, 6, 7, 8, or 9. We asked to find the value of \(x_{max}-x_{min}\), ans since \(x_{max}=9\) and \(x_{min}=-9\) then \(x_{max}-x_{min}=9-(-9)=18\).

Answer: D.

Can you explain how Xmin = (-9). IMO Xmin = (-2). Where did i go wrong ???

Re: GMAT Diagnostic Test Question 42 [#permalink]
18 Aug 2014, 20:45

Ashishmathew01081987 wrote:

bb wrote:

Explanation Official Answer: D

Notice that \(x\) can take positive, as well as negative values to satisfy \(9<x^2<99\), hence \(x\) can be: -9, -8, -7, -6, -5, -4, 4, 5, 6, 7, 8, or 9. We asked to find the value of \(x_{max}-x_{min}\), ans since \(x_{max}=9\) and \(x_{min}=-9\) then \(x_{max}-x_{min}=9-(-9)=18\).

Answer: D.

Can you explain how Xmin = (-9). IMO Xmin = (-2). Where did i go wrong ???

My understanding is that since 9 < x^2, it implies that (+/-) 3 < x and since x has to be an integer x = (+/-) 2. Also, since X^2 < 99, it implies that x < (+/-) 9 but x cannot be less than (-9) since (-3) < x

So Xmax = 9 and Xmin = -2 therefore Xmax - Xmin = 9 -(-2) = 11

Re: GMAT Diagnostic Test Question 42 [#permalink]
19 Aug 2014, 01:49

Expert's post

Ashishmathew01081987 wrote:

Ashishmathew01081987 wrote:

bb wrote:

Explanation Official Answer: D

Notice that \(x\) can take positive, as well as negative values to satisfy \(9<x^2<99\), hence \(x\) can be: -9, -8, -7, -6, -5, -4, 4, 5, 6, 7, 8, or 9. We asked to find the value of \(x_{max}-x_{min}\), ans since \(x_{max}=9\) and \(x_{min}=-9\) then \(x_{max}-x_{min}=9-(-9)=18\).

Answer: D.

Can you explain how Xmin = (-9). IMO Xmin = (-2). Where did i go wrong ???

My understanding is that since 9 < x^2, it implies that (+/-) 3 < x and since x has to be an integer x = (+/-) 2. Also, since X^2 < 99, it implies that x < (+/-) 9 but x cannot be less than (-9) since (-3) < x

So Xmax = 9 and Xmin = -2 therefore Xmax - Xmin = 9 -(-2) = 11

The easiest way to check your reasoning is to plug -9 there and see whether the inequality holds: (-9)^2=81 < 99, (x cannot be -10 because 10^2=100>99). So, the least value is -9 not -2 (notice that -9 is less than -2).

Also: \(9 < x^2\) means that \(x < -3\) or \(x > 3.\) \(x^2 < 99\) means that \(-\sqrt{99}<x<\sqrt{99}\)

It seems that you need to brush up fundamentals on inequalities: