SIMPLE INTEREST–
Simple interest is a very basics thing which is to be described for finding interest of a given principal within a stipulated time. Let us go over some terms and formula that you learnt in the previous class. Interest (I) is to be considered as the additional amount paid by the borrower to the lender for the use of a sum of money. Principal (P) is considered as the sum of money lent, borrowed or invested. Time (T) - the duration for which the sum of money is lent or invested, usually in years. Rate (R) is to be considered as the interest paid on $ 100 per unit of time, usually per annum (p.a). Amount (A) is to be considered as the sum of the principal and the interest.
There are some formula are given below –
Amount (A) = Principal (P) + Interest (I)
Principal (P) = Amount (A) – Interest (I)
Interest (I) = Amount (A) – Principal (P)
Simple Interest (I) – The interest paid on the original sum of money borrowed or invested.
Principal (P) X Rate (R) X Time (T)
Interest (I) = -------------------------------------
100
Rate (R) X Time (T)
Amount (A) = Principal (P) ( 1 + ---------------------- )
100
Example.-
1) Find the simple interest on $ 12,000 for 5 years at the rate of 1) 7.5% per annum, and 2) 1.5% per month. Find the amount in each case.
Answer) As per the given condition, if he put $ 100 then in 1 year he should get interest around $ 7.5
then in, if he put $ 1 then in 1 year he should get the interest around =
$ 7.5
------------ 100
then in, if he put $ 12000 then in 1 year he should get the interest around
7.5 X 12000
= ----------------- 100
in, if he put $ 12000 then in 5 year he should get the interest around
7.5 X 12000 X 5
= --------------------- = $ 4500……………….(1) 100
if he put $ 100 then in 1 month he should get interest around $ 1.5
then in, if he put $ 1 then in 1 month he should get the interest around
$ 1.5
= ------------- 100
then in, if he put $ 1 then in 12 month ( 1 year) he should get the interest around
$ 1.5 X 12
= --------------- 100
then in, if he put $ 12000 then in 1 year he should get the interest around
1.5 X 12 X 12000
= ----------------------
100
in, if he put $ 12000 then in 5 year he should get the interest around
1.5 X 12 X 12000 X 5
= -------------------------- = $ 10800……………….(2)
100
Another way –
As per the given condition, he is getting the interest around $ 7.5 per annum, so now we have to calculate the obtained interest over $ 12000 for 5 years.
Here, Principal (P) = $ 12000, Rate (R) = 7.5 %, Time (T) = 5 years
Principal (P) X Rate (R) X Time (T)
As per the formula, Interest (I) = --------------------------------
100
12000 X 7.5 X 5 450000
= ------------------- = ------------ = $ 4500 ……………(1)
100 100
As per the given condition, he is getting the interest around $ 1.5 per month, so now we have to calculate the obtained interest over $ 12000 for 5 years.
So, first we have to calculate the interest per annum, so we get = 1.5 X 12 = 18 % per annum
Here, Principal (P) = $ 12000, Rate (R) = 18 %, Time (T) = 5 years
Principal (P) X Rate (R) X Time (T)
As per the formula, Interest (I) = --------------------------------
100
12000 X 18 X 5 1080000
= ------------------ = ------------ = $ 10800 ……………(2) (Ans.)
100 100
2) Find the simple interest on, $ 1500 from 25 April 2018 to 5
2
December 2019 at the rate of 5 ------- % per annum.
3
Ans.) The number of days from 25 April 2018 to 5 December 2019 are given below –
25th Apr'18 – 30th Apr'18 = 6 days
1st May’18 – 31st May’18 = 31 days
1st Jun’18 – 30th Jun’18 = 30 days
1st Jul’18 – 31st Jul’18 = 31 days
1st Aug’18 – 31st Aug’18 = 31 days
1st Sep’18 – 30th Sep’18 = 30 days
1st Oct’18 – 31st Oct’18 = 31 days
1st Nov’18 – 30th Nov’18 = 30 days
1st Dec’18 – 31st Dec’18 = 31 days
1st Jan’19 – 31st Jan’19 = 31 days
1st Feb’19 – 28th Feb’19 = 28 days
1st Mar’19 – 31st Mar’19 = 31 days
1st Apr’19 – 30th Apr'19 = 30 days
1st May’19 – 31st May’19 = 31 days
1st Jun’19 – 30th Jun’19 = 30 days
1st Jul’19 – 31st Jul’19 = 31 days
1st Aug’19 – 31st Aug’19 = 31 days
1st Sep’19 – 30th Sep’19 = 30 days
1st Oct’19 – 31st Oct’19 = 31 days
1st Nov’19 – 30th Nov’19 = 30 days
1st Dec’19 – 5th Dec’19 = 5 days
Total days = 6 + 31 + 30 + 31 + 31 + 30 + 31 + 30 + 31 + 31 + 28 + 31 + 30 + 31 + 30 + 31 + 30 + 31 + 31 + 30 + 5 = 590 days
590 118
If we calculate total obtained days in year = -------- = -------- years
365 73
Now we have to find out the desired interest where, Principal (P) = $ 1500, Time (T) = 118/73 years,
2
Rate (R) = 5 ------- % = 17/3 %
3
Principal (P) X Rate (R) X Time (T)
As per the formula, Interest (I) = -------------------------------
100
17 118
1500 X ------- X --------
3 73
= -------------------------------------
100
1500 X 17 X 118
= -------------------- = $ 137.4 (Ans.)
3 X 73 X 100
3) Find the simple interest on, $ 2500 from 20th Oct'19 to 10 Apr'20
1
at the rate of 3 ------- % per month.
2
Ans.) The number of days from 20 October 2019 to 10 April 2020 are given below –
20th Oct’19 – 31st Oct’19 = 12 days
1st Nov’19 – 30th Nov’19 = 30 days
1st Dec’19 – 31st Dec’19 = 31 days
1st Jan’20 – 31st Jan’20 = 31 days
1st Feb’20 – 29th Feb’20 = 29 days ( Feb'2020 is considered as Lip year )
1st Mar’19 – 31st Mar’19 = 31 days
1st Apr’19 – 10th Apr = 10 days
Total days = 12 + 30 + 31 + 31 + 29 + 31 + 10 = 174 days
174
If we calculate total obtained days in year = --------- years
365
1
Given Rate = 3 ------- % = 7/2 % per month, if we convert this given
2
per month rate into per annum
7
Rate, then it will be = --------- X 12 months = 42 %
2
Now we have to find out the desired interest where, Principal (P) = $ 2500, Time (T) = 174/365 years,
Rate (R) = 42 % per annum
Principal (P) X Rate (R) X Time (T)
As per the formula, Interest (I) = ------------------------------
100 174
2500 X 42 X ---------
365
= ----------------------------------
100
2500 X 42 X 174
= -------------------- = $ 500.55 (Ans.)
365 X 100