What is the smallest possible number which when divided by 7, 9 and 11 leaves a remainder 2 in each case?

Correct option is B

Given:
We are asked to find the smallest possible number which when divided by 7, 9, and 11 leaves a remainder of 2 in each case.
Formula Used:
LCM (Least Common Multiple) is used to find the smallest number divisible by multiple numbers.
If a number leaves the same remainder 'r' when divided by a, b, c, then:
Required number = LCM(a, b, c) + r
Solution:
Since 7, 9, and 11 are co-prime (no common factors),
LCM = 7 × 9 × 11 = 693
Required number = 693 + 2 = 695