In what time will the simple interest on Rs 400 at 10% per annum be the same as the simple interest on Rs 1000 for 4 year at 4 % per annum?

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

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.

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 .

According to question , we can say that
Total number of odd days = 1600 years have 0 odd day + 300 years have 1 odd day + 49 years (12 leap years + 37 ordinary years) have 5 odd days + 26 days of January have 5 odd days = 0 + 1 + 5 + 5 = 4 odd days.
So, the day was Thursday.

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.

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

Let the amount instalment be Rs ' x '
Then According to question,
(Amount of 'x' for 3 yrs) + (Amount of 'x' for 2 yrs) + (Amount of 'x' for 1 yrs) + x =3220
or, [x+(x × 10 × 3)/100] + [x+(x × 10 × 2)/100] + [x+(x × 10 × 1)/100] + x=3220
? 4x+ (30x/100)+(20x/100)+(10x/100)=3220
? 460x=322000
? x=Rs 700

? Each Instalment= Rs 700

Simple Interest in 1 year= Rs (1729 - 1586)
= Rs 143
now, SI in 2 year = Rs 286

Principal P= Rs(1586 - 286)
= 1300
And R= (100 × SI)/(P × T)
= (100 × 143)/(1300 × 1)
= 11 %
where, R = rate
SI= Simple Interest
P = Principal
T= Time

Let the sum of money be Rs y
So Amount = y +[( y x 5 x 4 )/100]
But Amount = Rs 900
? 900 = y +(20y)/100
? 900 = 6y / 5
? y = ( 900 x 5) /6
= Rs 750

Let the sum be Rs 'y' , so amount = 2y
Simple interest =Rs y
Let R be the rate of interest,
R= (100 x SI)/(P x T) = (100 x y) / (y x 8) = 12.5 %
where SI= Simple Interest
P = Principal
T = Time

now, the needed amount = Rs 4y
since SI = Rs (4x-x)= Rs 3y
since T= (100 x SI)/(P x R)
= (100 x 3y )/(y x 125)= 24 yr