Let the digit at unit’s place be a and ten’s place be b. Thus, the number to be found is $10b+a$.

From given question, the sum of two digit number is equal to $5$.

Thus we can write equation according to question,

$a+b=5$ ………….. (i)

On interchanging the place of digits, the new number so formed will be $10a+b$.

It is also given in question that, the new number obtained after interchanging the digits is greater by $9$ from the original number.

Therefore, this can be written as;

$10a+b=10a+b+9$

⇒ $10a+b–10b–a=9$

⇒ $9a–9b=9$

⇒ $9(a–b)=9$

⇒ $a–b=1$………………. (ii)

On solving (i) and (ii), we get value of a and b

Adding the eq. (i) and (ii), we get;

$(a+b)+(a–b)=5+1$

⇒ $a+b+a–b=5+1$

⇒ $2a=6$

⇒ $a=6/2$

⇒ $a=3$

Putting the value of a in the equation (i), we get;

$3+b=5$

⇒ $b=5-3$

⇒ $b=2$

Thus, the required number is $10\times 2+3=23$