Find the sum of all prime numbers between $1$ and $100$ that are simultaneously $1$ greater than a multiple of $4$ and $1$ less than a multiple of $5$. 
$\mathrm{(A) \ } 118 \qquad \mathrm{(B) \ }137 \qquad \mathrm{(C) \ } 158 \qquad \mathrm{(D) \ } 187 \qquad \mathrm{(E) \ } 245$
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 $\boxed{A}$.