sanjoo wrote:
Is the square root of x is an integer?
(1) The unit digit of x is 2
(2) x is divisible by 3.
Sol: We need to find whether \(\sqrt{x}\) or x=I^2 where I is an Integer
St 1: Unit digit of x is 2...
Consider x=2^(Some positive integer ) so we get x=2,4,8,16 and
32 (2^5)\(\sqrt{32}\) is not an integer
Now x=3^ (some positive number, it will not end in 2 in any case)....
Consider x=4 ^(Some positive number) so we get x=4,16,64,256 (Ends in 4 or 6 )
Consider x=6 ^(Some positive number)...Always ends in 6
Consider x=8^(Some power) we get x=8,64,
512,---6,----8
Now x=512 then \(\sqrt{512}\) is not an integer
Based on above we can say that\(\sqrt{x}\) is not an integer.We did not consider value of x=5,7,9 (^ some positive number) because it can not end in zero.
St 1 is sufficient
St2: Possible value X=3,6,9,12....SO if X=9 then\(\sqrt{X}\) is an Integer but if x=6 then not.
Ans is A
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67% (01:24) correct 
