If \(y\) is a positive integer is \(\sqrt{y}\) an integer?Note that as \(y\) is a positive integer then \(\sqrt{y}\) is either a positive integer or an irrational number. Also note that the question basically asks whether \(y\) is a perfect square.
(1) \(\sqrt{4*y}\) is not an integer --> \(\sqrt{4*y}=2*\sqrt{y}\neq{integer}\) --> \(\sqrt{y}\neq{integer}\). Sufficient.
(2) \(\sqrt{5*y}\) is an integer --> \(y\) cannot be a prefect square because if it is, for example if \(y=x^2\) for some positive integer \(x\) then \(\sqrt{5*y}=\sqrt{5*x^2}=x\sqrt{5}\neq{integer}\). Sufficient.
Answer: D.
Bunuel answer should be A, for the second statement is insufficient
if y = 20 then root( 5* y) root(100) is still an integer BUT root(20) is not an integer so i chose A
Your logic is not clear. If y = 20, then (2) is satisfied \(\sqrt{5*y}=\sqrt{100}=10=integer\) and \(\sqrt{y}\) is NOT an integer (so you have a NO answer to the question). But what value of y gives an YES answer to the question?