C+12=2R.
Is C<R?

1) R>5

say R=7
C+12=2*7=14, or C=2
=>C<R

say R=50
C+12=100, or C=88
=>C>R
not sufficient

2) C<12
say C=11
11+12=2R or R=11.5
=>C<R

say C=0 (min possible)
0+12=2R or R=6
=>C<R for all possible values of C
sufficient

B

I have a little different way of solving this.
C + 12 =2R

1) This condition puts a restriction on the value of R i.e it says its value has to be more than 5.

So we have to start checking the corresponding value of C by putting R=6, Then R=7 etc etc....

R=6
C+12=12
Thus C=0. so C<R

BUT ONCE R crosses the value of 12, let say R=13

C+12 = 26
C=14

So, now C>R.

Hence not sufficient.

2) This condition puts a restriction on the Max value of C.

So, let say C=12

12+12= 2R
R=12

But Since Value of C will always be less than 12 so the value of R will also be less than 12
as it has been shown in the previous POST

So answer is B

Please note in my process you are actually using the critcal values,this becomes important specially for the first part.That's the only difference

I guess there's no formula I can use. I had to use "plug in" and find the range of numbers.