Is 1/x greater than 1/y?

The answer should be C as follows.

(1) x is greater than 1.
Since only value of x is give, y can be anything
INSUFFICIENT

(2) x is less than y
Here, there will be 3 cases

Case 1: When both x and y are positive
E.G. let x=2 and y=3
\(\frac{1}{x}=\frac{1}{2}=0.5\)
\(\frac{1}{y}=\frac{1}{3}=0.333\)
in this case \(\frac{1}{x}>\frac{1}{y}\)

Case 2: When x is negative and y is positive
E.G. x = -2 and y=3
\(\frac{1}{x}=\frac{1}{(-2)}= -0.5\)
\(\frac{1}{y}=\frac{1}{3}=0.333\)
in this case \(\frac{1}{x}<\frac{1}{y}\)

Case 2: When both x and y are negative
E.G. x = -2 and y=-1 (Since x<y)
\(\frac{1}{x}=\frac{1}{(-2)}= -0.5\)
\(\frac{1}{y}=\frac{1}{(-1)}= -1\)
in this case \(\frac{1}{x}<\frac{1}{y}\)

ICUFFICIENT

Note: Case 3 was really not required to be considered as case 1 and case 2 themselves produced the conflicting results.

Combining both, the condition will become, x<y and x is greater than 1. This will be an spacial scenario of case 1 of statement 2 in the question.
Since x>1, both x and y will be positive
E.G. x=2 and y=3
\(\frac{1}{x}=\frac{1}{2}=0.5\)
\(\frac{1}{y}=\frac{1}{3}=0.333\)
in this case \(\frac{1}{x}>\frac{1}{y}\)

SUFFICIENTS

Hence, answer is C