Let the digits at unit’s place be a and ten’s place be v, respectively. Thus, the number we need to find is $10b+a$.

As per the given question, the sum of the two digit number is $15$. Thus, we have;

$a+b=15$ ……………(i)

Upon interchanging the digit’s place, the new number will  be $10a+b$.

It is also given that, the new number obtained exceeds from the original number by $9$. Therefore, we can write this as;

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

⇒ $10a+b–10b–a=9$

⇒ $9a–9b=9$

⇒ $9(a–b)=9$

⇒ $a–b=9/9$

⇒ $a–b=1$………………….. (ii)

On solving equation (i) and (ii), we can find value of a and b.

Now, adding the equations (i) and (ii), we get;

$(a+b)+(a–b)=15+1$

⇒ $a+b+a–b=16$

⇒ $2a=16$

⇒ $a=16/2$

⇒ $a=8$

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

$8+b=5$

⇒ $b=15–8$

⇒ $b=7$

Hence, the required number is, $10\times 7+8=78$