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If x is an integer, what is the value of x? (1)

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If x is an integer, what is the value of x? (1) [#permalink] New post 25 Apr 2008, 18:53
00:00
A
B
C
D
E

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If √x is an integer, what is the value of √x?
(1) 11<x<17
(2) 2<√x<5
________________________________________________________________
getting C. plz confirm.
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Re: DS - root X [#permalink] New post 25 Apr 2008, 21:56
C for me too

If √x is an integer, what is the value of √x?
(1) 11<x<17
(2) 2<√x<5

Stmt -1 -> x could be 12,13,14,15,16 but Sqrt x is integer thus x has to be 16 but Sqrt of x could be +/- 4 thus insufficient

Stmt-B-> sqrt of x is 3,4

thus combining both we know x is 16 and Sqrt of x is +4
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Re: DS - root X [#permalink] New post 26 Apr 2008, 01:59
Getting A here. Essentially the questions needs to identify.

The reasoning is as under...
i) Means that the possible values of x are- 12, 13, 14, 15, 16 & 17. Out of these values only sqrt x(=16) would yield an integer. Hence sufficient.

ii) yields sqrtx= 3,4. Therefore insufficient. Hence A
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Re: DS - root X [#permalink] New post 26 Apr 2008, 10:16
sondenso wrote:
prasannar wrote:
Sqrt of x could be +/- 4


Prasannar,√16 = -4 ?


Square roots have only one answer and that must be a positive value. GMAT follows the convention that a root sign denotes only a non-negative root of a number :)
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Re: DS - root X [#permalink] New post 26 Apr 2008, 10:23
saravalli wrote:
If √x is an integer, what is the value of √x?
(1) 11<x<17
(2) 2<√x<5
________________________________________________________________
getting C. plz confirm.


Rephrasing \sqrt{x} must be an integer.

Statement (1) lists x as {12,13,14,15,16}. Only 16 will make \sqrt{x} must be an integer, which is 4. Sufficient. Eliminate B,C,E.
Statement (2) lists values of \sqrt{x} as {3,4}. Could be 3 or 4. No definite value. Not Sufficient. Elimimate D.

Ans: A
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Re: DS - root X [#permalink] New post 26 Apr 2008, 10:37
I will go with C
if you consider just 1st statementt you can narrow it down to 16...but sqrt(16) can be + or -4
if you consider 2nd statement it is 4
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Re: DS - root X [#permalink] New post 26 Apr 2008, 22:04
As per OG, the Sqrt(x) is the non negative square root of x.
i.e. sqrt(x) = |x|

So answer should be A.
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Re: DS - root X [#permalink] New post 27 Apr 2008, 05:00
i said A as well, because I recall reading that on the GMAT, you only need worry about the positive roots.
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Re: DS - root X [#permalink] New post 27 Apr 2008, 09:07
guys i am confused now. :shock:
root(16) = ONLY + 4? and NOT -4?

I really did not know this. Really appreciate if someone can provide an example from OG.
All these days I was thinking that root(X) is +/- ..of something.
Thanks.
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Re: DS - root X [#permalink] New post 27 Apr 2008, 11:20
a^2=x always have two roots, +sqrt(x) and -sqrt(x). However, when we refer to sqrt(x) itself, it is always positive.
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Re: DS - root X [#permalink] New post 27 Apr 2008, 11:40
HongHu wrote:
a^2=x always have two roots, +sqrt(x) and -sqrt(x). However, when we refer to sqrt(x) itself, it is always positive.


Sorry.I did not get it.
Are you saying C is wrong.
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Re: DS - root X [#permalink] New post 27 Apr 2008, 11:46
Yes, A would be sufficient. Before we are certain that sqrt(x) is 4.

(If x is 16, it has roots +sqrt(x) and -sqrt(x), which is +4 and -4, but sqrt(x) denotes 4.)
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Re: DS - root X [#permalink] New post 29 Apr 2008, 08:27
HongHu wrote:
Yes, A would be sufficient. Before we are certain that sqrt(x) is 4.

(If x is 16, it has roots +sqrt(x) and -sqrt(x), which is +4 and -4, but sqrt(x) denotes 4.)


Thankyou! :)
Re: DS - root X   [#permalink] 29 Apr 2008, 08:27
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