Question

On what date of October, 1975 did 3rd Tuesday fall :

• A. $17/10/75$
• B. $18/10/75$
• C. $21/10/75$
• D. $23/10/75$

Answer 1

Answer: (C)

$21/10/75$

The answer is option $C$

Concept:

$[\text{Definition: Odd days}=Reminder\text{[no.of days when divided by 7]}]$

$\left[\text{If odd days =0, then day is Sunday, If odd days =1, then day is Monday}\right]$

Let's first find the day for $1^{st}Oct1975$

1 Oct 1975 = ($1600years+300years+74years+$period from $1^{st}Jan1975$ till $1^{st}Oct1975)$

Odd days in $1600years=0$ $\to \left(I\right)$

Odd days in $300years=1$ $\to \left(I\right)$

$74years$ has $18$ leap years and $561$ ordinary years

Each leap year has $2$ odd days and an ordinary year has $1$ odd day

$\therefore 74years$ has $(18\times 2+56\times 1),i.e.92$ odd days,$i.e.1$ odd days $\to \left(III\right)$

$\text{Jan Feb Mar Apr May Jun Jul Aug Sep Oct}$

($31$ + $28$ + $31$ + $30$ + $31$ + $30$ + $31$ + $31$ + $30$ + $1$) $=274$ days

Odd days from $274$ days $=1$ odd days $\to \left(IV\right)$

From $\left(I\right)\left(II\right)\left(III\right),\left(IV\right)$, we get number of odd days for $1^{st}Oct1975=0+1+1+1=3$

$\therefore$ $1^{st}Oct1975$ falls on a $Wednesday$

The $1^{st}$ Tuesday would fall on $7^{th}$

$\Rightarrow$ The $3^{rd}$ Tuesday would fall on ${21}^{st}$ october 1975.