If x is a positive integer, is √x an integer?
(1) √(4x) is an integer.
(2) √(3x) is not an integer
Statement 1: √(4x) is an integer
Since x must be a positive integer, the smallest possible value for x is 1, with the result that √(4x) = √(4*1) = 2.
If x=1, then √x = √1 = 1.
The next greatest possible integer value for x is 4, with the result that √(4x) = √(4*4) = 4.
If x=4, then √x = √4 = 2.
The next greatest possible integer value for x is 9, with the result that √(4x) = √(4*9) = 6.
If x=9, then √x = √9 = 3.
In every case, √x is an INTEGER.
SUFFICIENT.
Statement 2: √(3x) is not an integer
Case 1: x=1, with the result that √(3x) = √3 = non-integer.
In this case, √x = √1 = 1, which is an integer.
Case 2: x=2, with the result that √(3x) = √6 = non-integer.
In this case, √x = √2, which is NOT an integer.
Since √x is an integer in Case 1 but a non-integer in Case 2, INSUFFICIENT.
The correct answer is
A.
pims wrote:Why is the OA A?
2*sqrt(x)=INT
2*sqrt(x)=1 --> sqrt(x)=0.5
2*sqrt(x)=2 -- sqrt(x)=1 ??
The case in red violates that constraint that x must be a POSITIVE INTEGER.
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