# Compound Interest

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# Compound Interest

### Introduction

Compound Interest deals with the amount, rate of interest per annum, and principal until the given period.

### Methods

Compound Interest: Money is said to be at interest when the interest is to be paid to the lender at the end of a year or other fixed period. In compound interest calculation, the interest is added to the sum lent and the amount thus obtained becomes the principal for the next unit of time or the period fixed. This process is continued until the last period. After a certain period, the difference between the amount and the principal is called the compound interest.
Examples 1 Find compund intrest on Rs. 7500 at 4% per annum for 2 years, compounded annually.
Solution:
Amount = Rs. [7500 x $(1 + \frac{4}{100})^{2}$] = Rs. (7500 x $\frac{26}{25}$ x $\frac{26}{25}$) = Rs. 8112.
∴ C.I. = Rs. (8112 - 7500) = Rs. 612.

Examples 2 Find compund intrest on Rs. 8000 at 15% per annum for 2 years 4 months, compounded annually.
Solution:
Time = 2 years 4 months = 2$\frac{4}{12}$ years = 2$\frac{1}{3}$ years.
Amount = Rs. [8000 x $(1 + \frac{15}{100})^{2}$ x $(1 + \frac{\frac{1}{3} \times 15}{100})$] = Rs. (8000 x $\frac{23}{20}$ x $\frac{23}{20}$ x $\frac{21}{20}$)
= Rs. 11109.
∴ C.I. = Rs. (11109 - 8000) = Rs. 3109.

Examples 1 Find the compund intrest on Rs. 10,000 in 2 years at 4% per annum, the intrest being compunded half-yearly.
Solution:
Principal = Rs. 10000; Rate = 2% per half-year; Time = 2 years = 4 half-years.
∴ Amount = Rs. [10000 x ($1 + \frac{2}{100})^{4}$] = Rs. (10000 x $\frac{51}{50}$ x $\frac{51}{50}$ x $\frac{51}{50}$ x $\frac{51}{50}$)
= Rs. 10824.32.
∴ C.I. = Rs. (10824.32 - 10000) = Rs. 824.32.

Examples 2 What is the diffrence between the compound intrests on Rs. 5000 for 1$\frac{1}{2}$ years at 4% per annum compounded yearly and half-yearly?
Solution:
C.I. when intrest is comppounded half-yearly
= Rs. [5000 x (1 + $\frac{4}{100}$) x (1 + $\frac{\frac{1}{2} \times 4}{100}$)] = Rs. (5000 x $\frac{26}{25}$ x $\frac{51}{50}$) = Rs. 5304.
C.I. when intrest is compunded half-yearly
= Rs. [5000 x (1 + $\frac{2}{100})^{3}$] = Rs. (5000 x $\frac{51}{50}$ x $\frac{51}{50}$ x $\frac{51}{50}$) = Rs. 5306.04.
∴ Diffrence = Rs. (5306.04 - 5304) = Rs. 2.04.

Examples 1 Find the compunf intrest on Rs. 16,000 at 20% per annum for 9 months compunded quarterly.
Solution:
Principal = Rs. 16000; Time = 9 months = 3 quarters;
Rate = 20% per annum = 5% per quarter.
∴ Amount = Rs. [16000 x $(1 + \frac{5}{100})^{3}$] = Rs. (16000 x $\frac{21}{20}$ x $\frac{21}{20}$ x $\frac{21}{20}$) = Rs. 18522.
∴ C.I. = Rs. (18522 - 16000) = Rs. 2522.

Examples 2 Find the compound intrest on Rs. 15,625 for 9 months at 16% per annum compounded quaterly.
Solution:
P = Rs.. 15625, n = 9 months = 3 quarters, R = 16% p.a. = 4% per quarter.
Amount = Rs.[15625 x $(1 + \frac{4}{100})^{3}$] = Rs. (15625 x $\frac{26}{25}$ x $\frac{26}{25}$ x $\frac{26}{25}$) = Rs. 17576.
∴ C.I. = Rs. (17576 -15625) = Rs. 1951.

### Formulae

Here principal = Rs. P time = n and rate = R% p.a. i.e. per annum.
1. If interest is compounded annually :
Amount = P($(1 + \frac{R}{100})^n$)
Compound interest = P[$(1 + \frac{R}{100})^n$- 1 ]

2. If interest is compounded half - yearly :
Amount = P$(1 + \frac{\frac{1}{2}R}{100})^2n$

3. If interest is compounded quarterly :
Amount = P$(1 + \frac{\frac{1}{4}R}{100})^4n$

4. When the rate are different for different years, say $$R_{1}$$%, $$R_{2}$$%, $$R_{3}$$% for first, second and third year respectively. Then,
Amount = P$(1 + \frac{R_{1}}{100}) x (1 + \frac{R_{2}}{100}) + (1 + \frac{R_{3}}{100})$

5. When interest is compounded annually but time is in fraction, say 3$$\frac{2}{5}$$ years.
Amount = P$(1 + \frac{R}{100})^3$ x $(1 + \frac{\frac{1}{4}R}{100})$

6. Present worth of Rs. $$x$$ due to $$n$$ years is given by:
Present worth = $\frac{x}{(1 + \frac{R}{100})^n}$

### Samples

1. Find the compound interest on Rs. 1600 at 7$$\frac{1}{4}$$% per annum for 2 years.
Solution:
Given that,
Principal = Rs. 1600
Rate = 7$$\frac{1}{4}$$% = $$\frac{29}{4}$$
Time = 2 years
Now, consider
Amount = P$(1 + \frac{R}{100})^n$
⇒ Amount = 1600$(1 + \frac{29}{4 * 100})^2$
⇒ Amount = 1600 x $$\frac{429}{400}$$ x $$\frac{429}{400}$$
⇒ Amount = Rs. 1840 .41
Therefore, Compound interest = 1840 .41 - 1600 = Rs. 240.41

2. Rs. 800 at 5% per annum compound interest would Rs. 4375 yield Rs. 357 in 2 years?
Solution:
Given that,
Rate = 5%
Amount = Rs. 882
Principal = Rs. 800
$n_{1}$ = 1 year
Now, consider
Amount = P$(1 + \frac{R}{100})^n$
⇒ 882 = 800$(1 + \frac{5}{100})^n$
⇒ $$\frac{882}{800}$$ = $$(\frac{21}{20})^n$$
⇒ $$(\frac{21}{20})^2$$ = $$(\frac{21}{20})^n$$
⇒ t = 2
Therefore, Time = 2 years

3. At what rate percent compound interest would Rs. 4375 yield Rs. 357 in 2 years?
Solution:
Given that,
Principal = Rs. 4375
Time = 2 years
Compound interest = Rs. 357
So, Amount = Compound interest + principal = Rs. 4375 + Rs. 357 = Rs. 4732
Now, consider
Amount = P$(1 + \frac{R}{100})^n$
⇒ 4732 = 4375$(1 + \frac{R}{100})^2$
⇒ $(1 + \frac{R}{100})^2$ = $\frac{4732}{4375}$
⇒ $(1 + \frac{R}{100})^2$ = $\frac{676}{625}$
⇒ $(1 + \frac{R}{100})$ = $frac{26}{25}$
⇒ $\frac{R}{100}$ = $\frac{26}{25}$ - 1
⇒ $\frac{R}{100}$ = $\frac{1}{25}$
⇒ R = $\frac{100}{25}$
⇒ R = 4%

4. The difference between compound interest and simple interest on a certain sum of money at 10% per annum for 2 years is Rs. 46. Find the sum?
Solution:
Given that,
Rate = 10 %
time = 2 years
Let the principal sum be Rs. 100. Then,
Simple interest(S.I) = $\frac{principal(P) * rate(R) * time(T)}{100}$
Simple interest = $\frac{100 * 10 * 2}{100}$ = Rs. 20
Then, Compound interest = P$$(1 + \frac{R}{100})^n$$ - P
⇒ Compound interest = 100$$(1 + \frac{10}{100})^2$$ - 100
⇒ Compound interest = [100 x $$\frac{11}{10}$$ x $$\frac{11}{10}$$ - 100]
⇒ Compound interest = 121 -100
⇒ Compound interest = 21
Difference in compound interest and simple interest = Rs. 21 - Rs. 20 = Rs. 1
If the difference in compound interest and simple interest is Rs. 1 then sum = Rs. 100
If the difference in compound interest and simple interest is Rs. 40 then sum = Rs. 100 x 40 = Rs. 4000

5. A sum of money placed at compound interest double itself in 5 years. It will amount to eight times itself in?
Solution:
Given that,
P$(1 + \frac{R}{100})^5$ = 2P
⇒ $(1 + \frac{R}{100})^5$ = 2
Let P$(1 + \frac{R}{100})^n$ = 8P then,
⇒ $(1 + \frac{R}{100})^n$ = 8
⇒ $(1 + \frac{R}{100})^n$ = $2^3$
⇒ $(1 + \frac{R}{100})^n$ = $[(1 + \frac{R}{100})^5]^3$
⇒ $(1 + \frac{R}{100})^n$ = $(1 + \frac{R}{100})^15$
As the bases are equal, equate the powers i.e. n = 15
Therefore, n = 15 years