The variables B and C are functions of the value of A such that with

Hi..

So if A increases by 1, say A+1, the value of B increases by a factor of 3, that is it becomes 3B and the value of C increases by a factor of 5, that is 5C..

(1) When A = 5, B = 2C.
A=6..\(C_6=5C\) and \(B_6=3*(2C_6)=3*(2*5C)=30C\)
A=7..\(C_7=5C_6=5*5C=25C\) and \(B_7=3*(2C_7)=3*(2*25C)=150C\)
A=8..\(C_8=5C_7=5*25C=125C\) and \(B_8=3*(2C_8)=3*(2*125C)=750C\)
Hence \(\frac{B_8}{(C_8-B_8)}=\frac{750C}{125C-750C}=\frac{750C}{-625C}=\frac{-6}{5}\)
Suff

(2) When A = 6, B = C + 15.
Now , here you have a term 15 which is free of variable so the answer will be in terms of variables, and NOT sufficient..
A=7..\(C_7=5C\) and \(B_7=3*(C_7+15)=3*(5C+15)=15(C+3)\)
A=8..\(C_8=5C_7=25C\) and \(B_8=3*(C_8+15)=3*(25C+15)=15(5C+3)\)
Hence \(\frac{B_8}{(C_8-B_8)}=\frac{15(5C+3)}{25C-15(5C+3)}\)
Insuff

A