Concept: Any number is divisible by 6 if
1. the unit digit is even
2. The summation of the digits is divisible by 3
We only need to check if b is even and 2+b is divisible by 3
Statement 1: 8b is divisible by 3
81 (b=1) is divisible by 3
84 (b=4) is divisible by 3
87 (b=7) is divisible by 3
B can hold the value 1,4 and 7 but remember that we need to check if all of these are even or not. Here it has both even and odd, statement 1 is not sufficient
Statement 2: 9b is divisible by 12
means it is divisible by 2*2*3
So the number have to nullify the entire denominator if 2*2*3
3 cuts 9, so 3 goes off from denominator
Now only if b has 4 or 8 then it can nullify 2*2 in the denominator
That means b can be 4 or 8
Now test for whether 2+b is divisible by 3
2+4=6 divisible by 3
2+8=10 not divisible by 3
So, statement 2 is not sufficient
Now if you check for statement A and B together, then we know from statement 1 that
b=1,4,7
Statement 2 that b=4,8
Means b=4 is only the common value
So the number 2b is 24
I can now determine whether it is divisible by 6
Answer C