Sum of the digits

Let $t$ and $u$ be the ten's digit and unit's digit respectively. Then

$10t+u$ $\implies$ the number

$10u+t$ $\implies$ the reversed number

From the first statement, we have

$t=u+5$ $\color{#D61F06}(1)$

From the second statement, we have

$10u+t=10t+u-5(t+u)=10t+u-5t-5u=5t-4u$

or

$14u-4t=0$ $\color{#D61F06}(2)$

Substitute $\color{#D61F06}(1)$ in $\color{#D61F06}(2)$ .

$14u-4(u+5)=0$

$14u-4u-20=0$

$10u=20$

$u=2$

It follows that

$t=u+5=2+5=7$

The sum of the digits is $2+7=$ $\boxed{11}$

let the no. in the units place be x. let the no. in the tens place be x+5. so, the number can be written as [10 (x+5) ]+ x . now, subtracting 5 times the sum of the digits from the no. we get. [10 (x+5)]+x - [5 (x+x+5)] --- A. after this, the digits of the no. are reversed. ie: the new no. we get is. [(10 x)+x+5] ---B. equating equations A & B. we get. x=2. x+5=7. therefore the sum of the digits of the no. is 2+7=9 .