SSC CHSL 11 Aug 2021 Shift 2 Paper

In $2000\mathit{y}\mathit{r}$, number of odd days $=0$
Now in $8\mathit{y}\mathit{r}$, two are leap years.
$\therefore$ Number of odd days in $8\mathit{y}\mathit{r}$
$=6\times 1+2\times 2=10$

Now, from 1 Jan 2009 to 31 Jan 2009 ,
Number of odd days $=3$
$\therefore$ Total odd days $=3+3=6$ and required day $=$ Saturday