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

- Amount = \(P(1 + \frac{R}{100})^{n}\)

**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.

- Amount = \(P(1 + \frac{(R/2)}{100})^{2n}\)

**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.

- Amount = \(P(1 + \frac{(R/4)}{100})^{4n}\)

**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.

- 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 = \(\frac{x}{(1 + \frac{R}{100})^n}\)

- Given that,

Principal = Rs. 1600

Rate = 7\(\frac{1}{4}\)% = \( \frac{29}{4} \)

Time = 2 years

Now, consider

Amount = P\((1 + \frac{R}{100})^n\)

â‡’ Amount = 1600\((1 + \frac{29}{4 * 100})^2\)

â‡’ Amount = 1600 x \( \frac{429}{400} \) x \( \frac{429}{400} \)

â‡’ 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\((1 + \frac{R}{100})^n\)

â‡’ 882 = 800\((1 + \frac{5}{100})^n\)

â‡’ \( \frac{882}{800} \) = \( (\frac{21}{20})^n \)

â‡’ \( (\frac{21}{20})^2 \) = \( (\frac{21}{20})^n \)

â‡’ 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\((1 + \frac{R}{100})^n\)

â‡’ 4732 = 4375\((1 + \frac{R}{100})^2\)

â‡’ \((1 + \frac{R}{100})^2\) = \(\frac{4732}{4375}\)

â‡’ \((1 + \frac{R}{100})^2\) = \(\frac{676}{625}\)

â‡’ \((1 + \frac{R}{100})\) = \(frac{26}{25}\)

â‡’ \(\frac{R}{100}\) = \(\frac{26}{25}\) – 1

â‡’ \(\frac{R}{100}\) = \(\frac{1}{25}\)

â‡’ R = \(\frac{100}{25}\)

â‡’ 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

â‡’ Compound interest = 100\( (1 + \frac{10}{100})^2 \) – 100

â‡’ Compound interest = [100 x \( \frac{11}{10} \) x \( \frac{11}{10} \) – 100]

â‡’ Compound interest = 121 -100

â‡’ 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

â‡’ \((1 + \frac{R}{100})^5\) = 2

Let P\((1 + \frac{R}{100})^n\) = 8P then,

â‡’ \((1 + \frac{R}{100})^n\) = 8

â‡’ \((1 + \frac{R}{100})^n\) = \(2^3\)

â‡’ \((1 + \frac{R}{100})^n\) = \([(1 + \frac{R}{100})^5]^3\)

â‡’ \((1 + \frac{R}{100})^n\) = \((1 + \frac{R}{100})^15\)

As the bases are equal, equate the powers i.e. n = 15

Therefore, n = 15 years