Question

In U.P. on 17th October, 1996 President's rule was declared. Find the day of week on that date.

• A. Tuesday
• B. Friday
• C. Wednesday
• D. Thursday

Answer 1

The correct option is D Thursday

We use the concept of odd days, for such type of questions.

The number of days more than the complete weeks are called odd days.

$1$ ordinary year $=365$ days $=(52$ weeks $+1$ odd day$)$

$1$ leap year $=366$ days $=(52$ weeks $+2$ odd days$)$

$100$ years $=76$ ordinary years $+24$ leap years $=76\left(1\right)+24\left(2\right)=124$ odd days $=17$ weeks $+5$ odd days $=5$ odd days

Number of odd days in $200$ years $=5\times 2=3$ odd days

Number of odd days in $300$ years $=5\times 3=1$ odd day

Number of odd days in $400$ years $=5\times 4+1=0$ odd days

If number of odd days come as $0$, that means given date is on a Sunday, $1$ for Monday , $2$ for Tuesday, $3$ for Wednesday, $4$ for Thursday, $5$ for Friday and $6$ for Saturday.

$17$ oct. $1996$ $=$ $1995$ years $+9$ months $+17$ days
No. of odd days in $1600$ years $=0$
No. of odd days in $300$ years $=1$
No. of odd days in $95$ years $=6$
$\therefore$ Total no. of odd days upto $31$st Dec${}^{\prime }95$ $=7=0$ odd days

Number of odd days in $1996$ till $17$ Oct $=31+29+31+30+31+30+31+31+30+17=291$ days $=41$ weeks $+4$ odd days $=4$ odd days

$\Rightarrow 17$ Oct $1996$ is a Thursday.