Simple Interest (S.I.) :
If the interest on a sum borrowed for certain period is reckoned uniformly, then it is called simple interest.
Let the principal = P, Rate = R% per annum (p.a) and Time = T years. Then ,
1. S.I. = (P x R x T / 100)
2. P = (100 x S.I. / R x T)
3. R = (100 x S.I / P x T)
4. T = (100 x S.I. / P x R)
1. Principal :
The money borrowed or lent out for a certain period is called the principal or the sum
Extra money paid for using other’s money is called interest.
What is the present worth of Rs. 132 due in 2 years at 5% simple interest per annum?
Let the present worth be Rs.x
Then,S.I.= Rs.(132 - x)
=> (x× 5 × 2/100)
=> 132 - x
=> 10x = 13200 - 100x
=> 110x = 13200
Mr. Thomas invested an amount of Rs. 13,900 divided in two different schemes A and B at the simple interest rate of 14% p.a. and 11% p.a. respectively. If the total amount of simple interest earned in 2 years be Rs. 3508, what was the amount invested in Scheme B?
Let the sum invested in Scheme A be Rs. x and that in Scheme B be Rs. (13900 - x)
Then, [(x * 14* 2)/100]+[((13900-x)*11*2)/100] = 3508
28x - 22x = 350800 - (13900 x 22)
6x = 45000
x = 7500.
So, sum invested in Scheme B = Rs. (13900 - 7500) = Rs. 6400
A sum of Rs. 725 is lent in the beginning of a year at a certain rate of interest. After 8 months, a sum of Rs. 362.50 more is lent but at the rate twice the former. At the end of the year, Rs. 33.50 is earned as interest from both the loans. What was the original rate of interest?
Let the original rate be R%. Then, new rate = (2R)%
Note: Here, original rate is for 1 year(s); the new rate is for only 4 months i.e. 1/3 year(s)
=> [(725*R*1)/100]+[(362.50*2R*1)/100*3] = 33.50
=> (2175 + 725) R = 33.50 x 100 x 3
=> (2175 + 725) R = 10050
=> (2900)R = 10050
=> R = 10050/2900 =3.46
Original rate = 3.46%
Monet paid in cash = Rs. 1000
Balance amount = Rs. (20000 - 1000)
=> Rs. 19000