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Compound Interest is the interest that we pay on a loan or get from bank on our saved deposits. It is the addition of interest to the principal sum of a loan or deposit. In simpler words, Compound Interest is the interest on interest. Compound Interest depends on the principal amount and the interest gained over the years. It is denoted by C.I.
Compared by simple interest, the value of compound interest on the same amount of principal is more. That means, for the same amount of loan one will pay lesser interest if based on simple interest but will have to pay a higher amount if interest is calculated as compound interest.
Also Read: Types of Matrices
| Table of Content |
Key Terms: Compound Interest, Time Period, Rate, Principal, Simple Interest
What is Compound Interest?
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Compound interest is defined as the interest calculated on a principal amount and on the interests that have been accumulated over the previous periods. Compound interest finds application in many important sectors such as banking and finance sector.
Denoted by C.I. compond interest is basically the difference between the principle and the amount. Some of the applications of compound interest are following:
- Calculating rise or depriciation in value of an item.
- Evaluating growth of bacteria.
- Calculating increase or decrease in the population.
Compound Interest Formula
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As mentioned above, compound interest is based on the initial principal amount and the interest gained over the years. Compound Interest is given by:
| C.I. = Amount - Principal |
Where,
- A = Amount
- P = Principal
- r = Rate of Interest
- n = Number of times interest is compounded per year
- t = Time (in years)
Therefore, C.I. becomes:
⇒ C.I. = P { 1 + \( {{r} \over n}\)}nt – P
The above-mentioned formula is called the Periodic Compounding formula.
Also, if the interest is compounded annually, the formula for amount becomes:
⇒ A = P{1 + \( {{r} \over 100}\)}t
Where,
- A = Amount
- P = Principal
- r = Rate of Interest
- t = Time (in years)
| Example: Find the rate of interest if the CI on a sum of Rs 1000 in 2 years is Rs 440. Solutions: P = 1000, t = 2 years, CI = 440 CI = A - P 440 = A - 1000 A = 1440 Now, A = P{1 + r/100}t 1440 = 1000{1 + r/100}2 1440/1000 = {1 + r/100}2 √1440/1000 = 1 + r/100 12/10 - 1 = r/100 2/10 = r/100 r = 20% |
Also Read: Differential Equation
Derivation of Compound Interest Formula
The derivation of Compound Interest requires the use of Simple Interest. We all know that the S.I. for one year is equal to the C.I. for one year.
Let P =Principal Amount, r = Rate of Interest, t = Time (in years)
Simple Interest for the first year:
S.I1 = P x r x t / 100
Amount after first year:
P + S.I1
P + P x r x t / 100
P {1 + r/100} = P2
SI for the second year:
S.I2 = P2 x r x t / 100
Amount after second year:
P2 + S.I2
P2 + P2 x r x t / 100
P2 {1 + r/100}
P {1 + r/100} {1 + r/100}
P{1 + r/100} ^ 2
In the same way, if we continue for n years, the amount becomes:
A = P{1 + r/100} ^ n
Therefore, Compound Interest comes out to be:
⇒CI = A - P = P{1 + \( {{r} \over 100}\)}n - P = P[{1 + \( {{r} \over 100}\)}n - 1]
Compound Interest Formula for Different Years
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Here are the formulas of Compound Interest for different years:
| Years | Amount | Compound Interest |
|---|---|---|
| 1 | P{1 + \( {{r} \over 100}\)} | P\( {{r} \over 100}\) |
| 2 | P{1 + \( {{r} \over 100}\)}2 | P{1 + \( {{r} \over 100}\)}2 - P |
| 3 | P{1 +\( {{r} \over 100}\)}3 | P{1 +\( {{r} \over 100}\)}3 - P |
| 4 | P{1 + \( {{r} \over 100}\)}4 | P{1 + \( {{r} \over 100}\)}4 - P |
| n | P{1 + \( {{r} \over 100}\)}n | P{1 + \( {{r} \over 100}\)}n - P |
| Half-yearly | P{1 + \( {{r} \over 200}\)}2 | P{1 +\( {{r} \over 200}\)}2 - P |
| Quarterly | P{1 +\( {{r} \over 400}\)}4 | P{1 + \( {{r} \over 400}\)}4 - P |
It is to be noted that the amounts in the above table are calculated at the end of 1 year. Therefore, t = 1.
Periodic Compounding Formula
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We already learned above that the formula for Periodic Compounding is given by:
| C.I. = P’ - P P’ = P { 1 + \( {{r} \over n}\)}nt |
where
P = Principal
P’ = New Principal
r = Annual Interest Rate
n = Number of times the interest is compounding
t = Time (in years)
Also Read:
| Topic Related Concepts | ||
|---|---|---|
| Ratio to Percentage | ||
Things to Remember
- Compound Interest is the addition of interest to the principal sum of a loan or deposit. In simpler words, it is the interest on interest.
- Compound Interest is given by: C.I. = Amount - Principal
- Periodic Compounding formula: P { 1 + \( {{r} \over n}\)}nt - P
- Compound Interest if the interest is compounded annually: P{1 +\( {{r} \over 100}\)}t - P
- Compound Interest if the interest is compounded half-yearly: P{1 + \( {{r} \over 200}\)}2 - P
- Compound Interest if the interest is compounded quarterly: P{1 + \( {{r} \over 400}\)}4 - P
Also Read:
| Chapter Related Concepts | ||
|---|---|---|
| Multiplication and Division of Integers | ||
Sample Questions
Ques. Find the rate of interest if the CI on a sum of Rs 1000 in 2 years is Rs 440. (3 marks)
Ans. P = 1000, t = 2 years, CI = 440
CI = A - P
440 = A - 1000
A = 1440
Now,
A = P{1 + r/100}t
1440 = 1000{1 + r/100}2
1440/1000 = {1 + r/100}2
√1440/1000 = 1 + r/100
12/10 - 1 = r/100
2/10 = r/100
r = 20%
Ques. What will be the principal amount if a sum amount of Rs 7200 is compounded annually for 2 years at 6% per annum? (2 marks)
Ans. A = 7200, t = 2 years, r = 6%
A = P{1 + r/100}t
7200 = P{1 + 6/100}2
7200 = P{106/100}2
7200 = P{1.1236}
P = 7200/1.1236
P = Rs 6408
Ques. Derive the formula for Compound Interest. (4 marks)
Ans. Let P =Principal Amount, r = Rate of Interest, t = Time (in years)
Simple Interest for the first year:
S.I1 = P x r x t / 100
Amount after first year:
P + S.I1
P + P x r x t / 100
P {1 + r/100} = P2
SI for the second year:
S.I2 = P2 x r x t / 100
Amount after second year:
P2 + S.I2
P2 + P2 x r x t / 100
P2 {1 + r/100}
P {1 + r/100} {1 + r/100}
P{1 + r/100}2
In the same way, if we continue for n years, the amount becomes:
A = P{1 + r/100}n
Therefore, Compound Interest comes out to be:
CI = A - P = P{1 + r/100}n - P = P[{1 + r/100}n - 1]
Ques. Two friends Radha and Shyam borrow Rs 10,000 from the bank at simple interest and compound interest annually, respectively. If the rate of interest is 10%, who pays more interest and by how much at the end of two years? (3 marks)
Ans. Here, P = Rs 10,000, r = 10%, t = 2 years
Amount Radha will have:
SI = P x r x t / 100 = 10000 x 10 x 2 / 100 = Rs 2000
Amount Shyam will have:
CI = P{1 + r/100}t - P = 10000{1 + 10/100}2 - 10000 = 10000 x 1.1 x 1.1 - 10000 = Rs 2,100
Ques. If the principal amount is Rs 12,000, the rate of interest is 8%, find the compound interest quarterly and half-yearly. (2 marks)
Ans. Here, P = Rs 12,000, r = 8%
Half-yearly CI:
CI = P{1 + r/200}2 - P = 12000{1 + 8/200}2 - 12000 = 12,979.2 - 12000 = Rs 979.2
Quarterly CI:
CI = P{1 + r/400}4 - P = 12000{1 + 8/400}4 - 12000 = 12989.19 - 12000 = Rs 989.19
Ques. The city's population at the end of 2010 was 5,00,000. The population rises at the rate of 6% per year. If the trend continues, what will be the population of the city at the beginning of 2019? (2 marks)
Ans. Here, P will be the population in 2010 i.e. 5,00,000, r is 6% and t is 8 years.
The population at the beginning of 2019:
P{1 + r/100}8 = 500000{1 + 6/100}8 = 7,96,924
Therefore, the population of the city at the beginning of 2019 is 7,96,924 people.
Ques. If the rate of growth of bacteria is 3% per hour, find the bacteria count after 4 hours if the initial count was 8,00,000. (2 marks)
Ans. P = 800000, r = 3%, t = 2 hours
Bacteria Count after 4 hours:
P{1+r/100}t = 800000{1+3/100}t = 9,00,407
Therefore, the bacteria count after 4 hours is 9,00,407.
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