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07 Mar 2012, 15:55
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
49% (01:46) correct
51%
(01:35)
wrong
based on 180
sessions
History
Date
Time
Result
Not Attempted Yet
If k is a positive integer, is \(\sqrt{k}\) an integer?
(1) \(1 < \sqrt{k} < 3\)
(2) \(k^2 < 16\)
(1) \(1 < \sqrt{k} < 3\)
(2) \(k^2 < 16\)
How come the answer is C?
Statement 1 - The only value K can take is 2. So this is sufficient.
Statement 2 - is not sufficient as -4 < k < 4.
Where I am incorrect guys?
Statement 1 - The only value K can take is 2. So this is sufficient.
Statement 2 - is not sufficient as -4 < k < 4.
Where I am incorrect guys?
07 Mar 2012, 16:06
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enigma123
If k is a positive integer, is \(\sqrt{k}\) an integer?
(1) \(1 < \sqrt{k} < 3\). Square the given inequality: \(1<k<9\) --> \(k\) can be 2, 3, 4, 5, 6, 7, or 8. If \(k=4\) then \(\sqrt{k}\) is an integer, for other values it's not an integer. Not sufficient.
(2) \(k^2 < 16\) --> \(-4<k<4\), since it's also given that \(k\) is a positive integer, then \(k\) can be 1, 2 or 3. If \(k=1\) then the answer is YES but if \(k\) is 2 or 3 then the answer is NO. Not sufficient.
(1)+(2) Intersection of the values of \(k\) from (1) and (2) is 2 and 3, \(\sqrt{k}\) is NOT an integer for either of them. Sufficient.
Answer: C.
General Discussion
stonecold
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13 Mar 2016, 08:50
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The other way to look at this question is to find the values of k =>
statement 1 => k=> 2,3,4,5,6,7,8
statement 2 => k=> 1,2,3
combining them k=> 2,3
so √k will never be an integer
statement 1 => k=> 2,3,4,5,6,7,8
statement 2 => k=> 1,2,3
combining them k=> 2,3
so √k will never be an integer
