The sum of all two digit numbers which when divided by $4$, yied unity as remainder, is

The correct option is D $1210$
According to given conditions , such numbers are

$13,17,21,........................97$

which is an AP with $a=13,d=4$

so, that $a_0=13,a_1=\mathrm{17......}{a}_n=a+(n-1)d$

$97=13+(n-1)4\Rightarrow n=22$

Now, Sum of AP

$=\frac{n}{2}(2a+(n-1\left)d\right)=\frac{22}{2}(2\ast 13+(22-1\left)4\right)=11\ast 110=1210$