The number of $4$ digit numbers that can be formed with $0,1,2,3,5$ which are divisible by $2$ or $5$ is

Considering $0,1,2,3,5$ and non repetitions of digits.
Number of four digit numbers divisible by $2$ or $5$ should contain $0,2$ or $5$ in the end of the number.
$\to$  _ _ _ 5
Implies $4(3!)$ numbers.
Now eliminating the numbers with $0$ at the starting, we get
$4(3!)-3!$
$=3(3!)$
$=18$. ...(i)