Theory: Dividend = Divisor*Quotient + Remainder

When positive integer n is divided by 5, the remainder is 2
=> n = 5x + 2 [ x being the quotient] ...(1)

When positive integer n is divided by 6, the remainder is 3
=> n = 6y + 3 [ y being the quotient] ...(2)

There are multiple ways to solve the problem. Two of them are listed below

Method 1: Compare the two equations for n (Faster Method)
n = 5x + 2 = 6y + 3
=> x = \(\frac{6y+1}{5}\)
So, only those values of y which will give x as integer will give common values for n in (1) and (2)
We need to get units digit of 6y as 4 or 9 to make 6y+1 a multiple of 5
So, y can be 4 giving x as \(\frac{6*4+1}{5}\) = 5 => n = 6*4 + 3 = 27
y can also be 9 giving 6y as 54 and x as \(\frac{6*9+1}{5}\) = 11. => n = 6*9 + 3 = 57
y can be 14 giving x as an integer. n = 6*14 + 3 = 87
y = 19 will become too big for n < 100
So, 3 values of n such that 0 < n < 100 are possible.

Method 2: Write all values of n in both the cases and compare them
n = 5x + 2
Put x=0,1,2,3,4... and find the corresponding values of n
x=0, n = 5*0 + 2 = 2 and so on..
So values n will be
2, 7, 12, 17, 22,27.... So numbers ending with 2 and 7

n = 6y + 3
Put y=0,1,2,3,4... and find the corresponding values of n
x=0, n = 6*0 + 3 = 3 and so on..
So values n will be
3, 9, 15, 21, 27,....
If we write all the numbers then we can find the common numbers as 27,57,87

So, Answer will be C
Hope it helps!

Watch the following video to MASTER Remainders