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What is the value of integer x ? [#permalink ]
20 Apr 2010, 06:04
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What is the value of integer x ? (1) 4 < (x-1)*(x-1) < 16 (2) 4 < (x+1)*(x-1) < 16

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Re: Tough Inequation DS: What is the value of integer x ? [#permalink ]
20 Apr 2010, 07:00
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MBAUncle wrote:

What is the value of integer x ? (1) 4 < (x-1)*(x-1) < 16 (2) 4 < (x+1)*(x-1) < 16 OA is C If it is asking for the value of X, should an interval be enough to answer the question ?

Note:

x is an integer.

(1)

4<(x-1)*(x-1)<16 -->

4<(x-1)^2<16 -->

(x-1)^2 is a perfect square between 4 and 16 --> there is only one perfect square: 9 -->

(x-1)^2=9 -->

x-1=3 or

x-1=-3 -->

x=4 or

x=-2 . Two answers, not sufficient.

(2)

4<(x+1)*(x-1)<16 -->

4<x^2-1<16 -->

5<x^2<17 -->

x^2 is a perfect square between 5 and 17 --> there are two perfect squares : 9 and 16 -->

x^2=9 or

x^2=16 -->

x=3 or

x=-3 or

x=4 or

x=-4 . Four answers, not sufficient.

(1)+(2) Intersection of values from (1) and (2) is

x=4 . Sufficient.

Answer: C.

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Re: Tough Inequation DS: What is the value of integer x ? [#permalink ]
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Re: Tough Inequation DS: What is the value of integer x ? [#permalink ]
24 Apr 2010, 09:23

Good question with a great explanation... Bunnel you are the champ

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Re: Tough Inequation DS: What is the value of integer x ? [#permalink ]
24 Apr 2010, 16:06

Bunuel, Just excellent!!!.

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What is the value of integer x ? (1) 4 < (x-1)*(x-1) < 16 [#permalink ]
07 Nov 2011, 00:36

I have a doubt with that of choosing option 1. 4 < (x-1)*(x-1) < 16 4 < (x-1)^2 < 16 Taking square root. 2 < (x-1) < 4 This gives 2 scenarios. x>3 and x< 5 Since x is an integer, the option must be 4. So the answer must be A rite . Any inputs ??

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Re: DS question from 700 level DS qstns by Bunnel [#permalink ]
07 Nov 2011, 02:54
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ksp wrote:

Question source :

tough-inequation-ds-what-is-the-value-of-integer-x-93008.html The question is

What is the value of integer x ?

(1) 4 < (x-1)*(x-1) < 16

(2) 4 < (x+1)*(x-1) < 16

I have a doubt with that of choosing option 1.

4 < (x-1)*(x-1) < 16

4 < (x-1)^2 < 16

Taking square root. 2 < (x-1) < 4 When you take the square root (which you can because all terms are positive), you get

2 < |x - 1| < 4

If x is to be an integer, |x - 1| = 3

x can be 4 or -2

Remember,

\sqrt{x^2}\neq x Instead,

\sqrt{x^2} = |x|

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Re: What is the value of integer x ? [#permalink ]
17 Jun 2013, 04:05

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Re: What is the value of integer x ? [#permalink ]
28 Sep 2013, 05:41

I tried solving the first inequality and got this: Statement 1: 4 < (x-1)*(x-1) < 16 --> x^2 -2x -3 < 12 --> x^2 - 2x - 15<0 --> (x+3) (x-5)<0 If x is an integer it means the function will be negative for the values x= -2, -1, 0, 1, 2, 3, 4. What is the mistake here?

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Re: What is the value of integer x ? [#permalink ]
28 Sep 2013, 05:52
emailmkarthik wrote:

I tried solving the first inequality and got this: Statement 1: 4 < (x-1)*(x-1) < 16 --> x^2 -2x -3 < 12 --> x^2 - 2x - 15<0 --> (x+3) (x-5)<0 If x is an integer it means the function will be negative for the values x= -2, -1, 0, 1, 2, 3, 4. What is the mistake here?

Why would you expand and manipulate 4 < (x-1)^2 < 16, when its already in its simplest form?

Anyway:

4 < (x-1)^2 < 16 ;

0 < x^2-2x-3 < 12 or

-12 < x^2-2x-15 < 0 ;

-3<x<-1 or

3<x<5 -->

x=-2 or

x=4 .

Hope this helps.

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Re: What is the value of integer x ?
[#permalink ]
28 Sep 2013, 05:52