enigma123 wrote:
If k is a positive integer, is \(\sqrt{k}\) an integer?
(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?
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.
what a brilliant explanation...I answered E, but now I see why I made the mistake...
i thought that k^2 = 4, therefore sqrt(k) can be either a non-integer or an integer...