In a two digit number, the units digit is 7 more than the ten’s digit. Sum of the digits is $\frac{1}{2}$ of the whole number, then the number is

Let the two digit number $=10x+y$

Hence unit digit is $y$, ten’s digit is $x$

Given the unit digit is 7 more than the ten’s digit

i.e. $y=x+7$

$y-x=7\to \left(1\right)$

And sum of the digits is $\frac{1}{2}$ of the whole number

i.e. $x+y=\frac{1}{2}(10x+y)$

$2x+2y=10x+y$

$y=8x$

From (1), $y-x=7$

$8x-x=7$

$7x=7$

$x=1$

From $y-1=7$

$y=8$

$\therefore$ The 2-digit number is $=10x+y$

$=10\left(1\right)+8$

$=18$