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If x is an integer, what is the value of x? (1) [#permalink]
 25 Apr 2008, 19:53
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If √x is an integer, what is the value of √x?
(1) 11<x<17
(2) 2<√x<5
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getting C. plz confirm.
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Director
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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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Senior Manager
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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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Manager
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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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Manager
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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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Intern
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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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Director
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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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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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Director
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guys i am confused now.  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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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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Director
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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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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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Keep on asking, and it will be given you;
keep on seeking, and you will find;
keep on knocking, and it will be opened to you.
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Director
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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!
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