Let's look at stmt 1, say n = 45 remainder with 21 = 3, remainder with 7 = 3; say n = 51, remainder with 21 is 9, with 7 it is 2. hence 1, not sufficient.

Stmt 2 n = 28*x + 3, where x is an integer. This can also be written as n = 4*(7*p) + 3, we know that 28*x is always divisible by 7, hence whatever remainder remains from 28 will also remain from 7.
This can be verified by picking numbers like 31, 59, 87 etc..

Hence answer is B.

OA please?

n=7q+R

What is R?

n=21q+r

n can be any number like 22,24,26.insufficient

n=28q+3 can be rephrased as,

n=7(4q)+3

from this when n is divisible by 7,the remainder will be 3.

n=31,59,67,95

Hence B

Thanks very much guys! B is indeed the correct answer.....