Question

A two digit number is 36 more than the number obtained by reversing the digit. If difference between tens digit and unit digit is 4. Find the number.

Answer 1

let the two digit number be 10a+b
thus, the reversed number = 10b+a

according to problem,

10a+b = 10b+a+36
9a = 9b + 36
9a = 9(b + 4)

a = b+4
thus, 
a  = 5 or 6 or  7 or 8 or 9
b  = 1 or 2 or 3 or 4 or 5 
the numbers can be 51, 62, 37,84,95

Answer 2

$\huge\underline\pink{Required\:Answer}$

There is no singularly unique two-digit number that will satisfy this question. Instead, there are multiple two-digit numbers that conform to the conditions set in the question. I will break it down step-by-step.

A two digit number

10a+b

is 36 more than the number obtained by reversing the digits.

10a+b=10b+a+36

⟹9a=9b+36

⟹a=b+4

If the difference between the tens digit and units digit is 4,

This part of the question doesn't add any new information because that's already been demonstrated as the only possibility. After all:

a=b+4⟹a−b=4

Since this part is a foregone conclusion, adding it adds nothing to the question.

then what is the number?

The value for a can be 4−9 (inclusively), while b can be as little as 0 , and cannot exceed 5 . So the numbers that satisfy this are 40,51,62,73,84, and 95.

Addendum: Others have answered that 40 is not a valid response because the reverse of it isn't 4 , but rather 04 , and that they are not the same thing. My take on it is that a preceding zero doesn't distinguish the number from just plain 4 because either way it represents zero tens and four units.