We need to find the total number of two-digit numbers whose remainder when divided by 10 is 1, and whose remainder when divided by 6 is 5

Let's solve the problem using two methods

Let n be the two digit number which gives 1 remainder when divided by 10 and 5 remainder when divided by 6

Dividend = Divisor * Quotient + Remainder
n when divided by 10 gives 1 remainder
Dividend = n
Divisor = 10
Quotient = a (Assume)
Remainder = 1
=> n = 10a + 1 ...(1)

n when divided by 6 gives 5 remainder
Dividend = n
Divisor = 6
Quotient = b (Assume)
Remainder = 5
=> n = 6b + 5 ...(2)

Method 1: Writing Values

n = 10a + 1 [ From (1) ]
Putting a = 1, 2,... we get values of n as
n = 11, 21, 31, 41, ...., 91 (we need to consider only two digit values)

n = 6b + 5 [ From (2) ]
Putting b = 1, 2,... we get values of n as
n = 11, 17, 23, 29, ...., 95 (we need to consider only two digit values)
For common values we need unit's digit to be 1
=> units' digit of 6b + 5 = 1
=> units' digit of 6b = 11-5 = 6
This will happen when b = 1, 6, 11, 16,
=> n = 6b + 5 = 11 (for b=1)
n = 41 (for b=6)
n = 71 (for b=11)
n = 101 (for b= 16.. but we need only two digit values of n)

=> Common values of n in both the cases is 11, 41 and 71
=> 3 values

Method 2: Algebra

n = 10a + 1 [ From (1) ] and n = 6b + 5 [ From (2) ]
=> 10a + 1 = 6b + 5
=> 10a = 6b + 4
=> a = \(\frac{6b+4}{10}\)
Now, only those values of "b" which also make "a" an integer will give us common values of n in both the cases

=> 6b + 4 has to be a multiple of 10
=> unit's digit of 6b = 10-4 = 6
=> b = 1, 6, 11 (As shown above)

So, Answer will be A
Hope it helps!

Watch the following video to learn the Basics of Remainders