smc

Numbers that are $1$ less than a multiple of $5$ all end in $4$ or $9$. No prime number ends in $4$, since all numbers that end in $4$ are divisible by $2$. Thus, we are only looking for numbers that end in $9$. Writing down the ten numbers that so far qualify, we get $9,19,29,39,49,59,69,79,89,99$. Crossing off multiples of $3$ gives $19,29,49,59,79,89$. Crossing off numbers that are not $1$ more than a multiple of $4$ (in other words, numbers that are $1$ less than a multiple of $4$, since all numbers are odd), we get: $29,49,89.$ Noting that $49$ is not prime, we have only $29$ and $89$, which give a sum of $118$, so the answer is $\boxed{118}$.