I have another suggestion, building on what sukrut was saying it might be clearer.
We could proceed by elimination.
The quickest thing we can say is that D isn't possible because if it were true, c and d (or a and b) would be of different signs, the ratio cannot be >1. --> eliminate D.
Then we look at answer A.
if $\frac{1}{a}$ and $\frac{1}{b}$ are positive this means that a,b are positive. But if a,b are positive and $\frac{1}{a} > \frac{1}{b}$ this means $\frac{b}{a}>1$ which is a contradiction --> eliminate A.
Answer B isn't ruled out so far.
Answer C is similar to B, so to see which one applies, we can check whether it's possible to have that $0> \frac{1}{d} > \frac{1}{c}$.
If it were the case we would have $d,c <0$ and thus $0>c>d$ which would mean that $\frac{c}{d} <1$ (take 0 > c=-2 > d=-4 to convince yourself). --> eliminate C.
And similarly we can also eliminate E because b is the analogous of d and a the analogous of c in this last assertion so the argument above still works. --> eliminate E.
So B could work and we could plug in the numbers by sukrut to complete the answer.