The sum of a two-digit number and the number obtained by reversing the digits is 66.The digits of the number differ by 2. How many two-digit numbers meet these conditions?

Let the two-digit number be of the form ab.

Number = 10a+b     ----- (1)

On reversing, the number will be ba, i.e. 10b+a    ----- (2)

It is given that sum of the two-digit number and the number obtained by reversing the digits is 66.

=> 10a+b+10b+a = 66

=> 11a + 11b = 66

=> a + b = 6

Also, |a - b| = 2 

There are two cases possible

Case 1: a > b

=> a - b = 2 and a + b = 6

=> Solving, we get a = 4 and b = 2

Case 2: b > a

=> b - a = 2 and a + b = 6

=> Solving, we get a = 2 and b = 4

Therefore, a total of 2 numbers are possible 24 and 42.

The answer is option D.