The first Republic Day of India was celebrated on January 26, 1950. What was the day of the week on that date?

As per the given above question , we know that
Year 2000 was a leap year.
Number of days remaining in 1999 = 365 - [31 days of January + 28 days of February + 5 days March] = 301 days = 43 weeks, i.e., 0 odd day.
Number of days passed in 2000 = January 31 days have 3 odd days.
February 29 days (being leap year) have 1 odd day
March 5 days have 5 odd days.
? Total number of odd days = 0 + 3 + 1 + 5 = 9 days = 1 week + 2 odd days
Therefore, March 5, 2000 would be two days beyond Friday, i.e., on Sunday.

As we can say that ,
There are 3 intervals when the clock strikes 4 Time taken in 3 intervals = 9 seconds
? Time taken for 1 interval = 3 seconds
In order to strike 12, there are 11 intervals, for which the time taken is 11 × 3 seconds = 33 seconds.
Therefore , Required time taken is 33 seconds .

Given that , The day after tomorrow is Sunday.
Therefore today is Friday.
Hence, the day on tomorrow's day before yesterday is given by:
= Friday ? 1 day = Thursday
Therefore , required day will be Thursday .

We know that ,
In every 30 minutes the time of watch increased by 3 minutes = 12 × 3 = 36 minutes
So the time after 6 hours = 5 am + 6 hours + 30 minutes = 11 : 36 am.
Therefore , required time will be 11 : 36 am.

Total interest needed in a year = Rs 400 × 12
= Rs 4800
Principal = (100 × SI)/R × T
where, R = Rate
T = Time
SI= Simple Interest

Here , P= Rs 1000
T= 4 yrs
R= 4 %
where, P= Principal
T= Time
R= Rate
Since , Simple Interest on Rs 1000=(1000 × 4 × 4)/100
= Rs 160

now, simple interest=Rs 160
P = Rs 400
R = 10 %
then, T=(100 × SI)/P × R
= (100 × 160)/(400 × 10)
= 4 yr

Let Sum = P, Then SI=P
As Amount A = 2 × P
where , P = Principal

Rate R = (100 × SI)/(P × T)
= (100 × P)/(P × 8) %
= 12.5 %
where , SI= Simple Interest
T= Time