We have learnt about compound interest in previous topics of this chapter. Under this topic, we’ll be dealing from formulae that are useful in calculating compound interest in different cases. Following are the cases and formulae used in them to calculate the amount payable at the principal sum.
If ‘P’ is the principal sum, i.e., amount taken as loan.
‘R’ is the rate percent which the bank/ lender is charging at the principal amount.
‘T’ is the time duration in which you have to repay the amount,
And ‘A’ will be the amount to be paid in following cases using following formulae:
Case 1: When the interest is compounded yearly:
A = \(P(1+\frac{R}{100})^{T}\)
Case 2: When the interest is compounded half yearly:
A = \(P(1+\frac{\frac{R}{2}}{100})^{2T}\)
Case 3: When the interest is compounded quarterly:
A = \(P(1+\frac{\frac{R}{4}}{100})^{4T}\)
Case 4: When the time is in fraction of a year, say \{2^{\frac{1}{5}}\), then:
A = \(P(1+\frac{R}{100})^{2}(1+\frac{\frac{R}{5}}{100})\)
Case 5: If the rate of interest in 1st year, 2nd year, 3rd year,…, nth year are R1%, R2%, R3%,…, Rn% respectively. Then,
A = \(P(1+\frac{R_{1}}{100})(1+\frac{R_{2}}{100})(1+\frac{R_{3}}{100})...(1+\frac{R_{n}}{100})\)
Case 6: Present worth of Rs x due ‘n’ years hence is given by:
Present worth = \(\frac{1}{1+\frac{R}{100}}\)
A fact that we all know very well is that interest is the difference between amount and principal sum, i.e.,
Interest = Amount – Principal
Now let us solve some problems based upon these formulae:
1. A man borrowed $20,000 from a bank on an interest of 10% p.a. compounded annually for 3 years. Calculate the compound amount and interest.
Solution:
R = 10%
P = $20,000
T = 3 years
We know that, A = \(P(1+\frac{R}{100})^{T}\)
A = \(20,000(1+\frac{10}{100})^{3}\)
A = \(20,000(\frac{110}{100})^{3}\)
A = \(20,000(\frac{11}{10})^{3}\)
A = \(20,000(\frac{1331}{1000})\)
A = 26,620
So, amount = $26,620
Interest = amount – principal amount
= $26,620 – $20,000
= $6,620
2. Find the compound amount on $10,000 if the interest rate is 7% per annum compounded annually for 5 years. Also calculate the compound interest.
Solution:
principal, P = $10,000
R = 7%
T = 5 years
We know that, A = \(P(1+\frac{R}{100})^{T}\)
A = \(10,000(1+\frac{7}{100})^{5}\)
A = \(10,000(\frac{107}{100})^{5}\)
A = $14,025.51
Also, interest = amount - principal
= $14,025.51 - $10,000
= $4,025.51
3. Find compound interest on amount $2,00,000 invested at 6% per annum, compound semi-annually for 10 years.
Solution:
we know that:
A = \(P(1+\frac{R}{100})^{T}\)
A = \(2,00,000(1+\frac{6}{100})^{20}\)
A = \(2,00,000(\frac{106}{100})^{20}\)
A = $6,41,427.09
Also, interest = amount – principal
= $6,41,427.09 - $2,00,000
= $4,41,427.09
4. If the interest rates for 1st, 2nd and 3rd are 5%, 10% and 15% respectively on a sum of $5,000. Then calculate the amount after 3 years.
Solution:
Principal = $5,000
R\(_{1}\) = 5%
R\(_{2}\) = 10%
R\(_{3}\) = 15%
We know that,
A = \(P(1+\frac{R_{1}}{100})(1+\frac{R_{2}}{100})(1+\frac{R_{3}}{100})...(1+\frac{R_{n}}{100})\)
A = \(5000(1+\frac{5}{100})(1+\frac{10}{100})(1+\frac{15}{100})\)
So, A = \(5000(\frac{105}{100})(\frac{110}{100})(\frac{115}{100})\)
A = $6,641.25
Also, interest = amount – principal
= $6,641.25 - $5,000
= $1.641.25
Compound Interest
Introduction to Compound Interest
Formulae for Compound Interest
Worksheet on Use of Formula for Compound Interest
9th Grade Math
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