Correct Answer: 30

Solution : The first number greater than 201, divisible by 5, is 205, and the last number less than 401, divisible by 5, is 400.
The common difference in this arithmetic sequence is 5.
We can use the formula for the nth term of an arithmetic sequence, which is $a_n = a_1 + (n - 1)d$, where $a_n$ is the $n$th term, $a_1$ is the first term, $d$ is the common difference, and $n$ is the number of terms.
$⇒400 = 205 + (n - 1) 5$
$⇒n = 40$
So, 40 numbers between 201 and 401 are divisible by 5.
However, we only want to exclude those divisible by 4 and 5, i.e., divisible by 20. The first number greater than 201, divisible by 20, is 220, and the last number less than 401, divisible by 20, is 400.
The common difference in this arithmetic sequence is 20.
$⇒400 = 220 + (p - 1)20$
$⇒p = 10$
So, 10 numbers between 201 and 401 are divisible by 20.
So, the count of numbers between 201 and 401 that are divisible by 5 but not by 4 = 40 – 10 = 30
Hence, the correct answer is 30.