Contents

## When the Interest is Compounded Annually

**Formula**

Let principal = P, rate = R% per annum and time = n years.

Then, the amount A is given by the formula

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

**Illustrative Examples**

**Example 1:** Find the amount of Rs 8000 for 3 years, compounded annually at 10% per annum. Also,find the compound interest.

**Solution.** Here, P = Rs 8000, R =10% per annum and n =3 years.

Using the formula A = P \( { (1+\frac { R }{ 100 } ) }^{ n } \) , we get

Amount after 3Years = {\( { 8000 \times (1+\frac { 10 }{ 100 } ) }^{ 3 } \) }

= Rs (8000 x \(\frac{11}{10}\) x \(\frac{11}{10}\) x \(\frac{11}{10}\))

= Rs 10648.

Thus, Amount after 3 years = Rs 10648.

And, compound interest = Rs (10648 – 8000) = Rs 2648

**Example 2:** Rakesh lent Rs 8000 to his friend for 3 years at the rate of 5% per annum compound interest. What amount does Rakesh get after 3 years?

**Solution.** Here, P = Rs 8000, R = 5% per annum and n =3.

Amount after 3 year = P\( { (1+\frac { R }{ 100 } ) }^{ n } \)

= Rs 8000 x \( { (1+\frac { 5 }{ 100 } ) }^{ 3 } \)

= Rs 8000 x \( { (1+\frac { 1 }{ 20 } ) }^{ 3 } \)

= Rs 8000 x \( { (\frac { 21 }{ 20 } ) }^{ 3 } \)

= Rs 8000 x \(\frac{21}{20}\) x \(\frac{21}{20}\) x \(\frac{21}{20}\)

= Rs 9261

**Example 3:** Find the amount and compound interest on Rs 5000 for 2 years at 10%, interest being pay yearly.

**Solution.** Here, P= Rs 5000, R = 10%, n = 2 years

Using the formula, A (Amount) = P \( { (1+\frac { R }{ 100 } ) }^{ n } \), we have

Therefore, A = Rs 5000 \( { (1+\frac { 10 }{ 100 } ) }^{ 2 } \)

= Rs 5000 x \(\frac{110}{100}\) x \(\frac{110}{100}\)

= Rs 6050

Therefore, Compound Interest = A – P = Rs 6050 – Rs 5000 = Rs 1050.

## When the Interest is Compounded Half-Yearly

**Formula**

If the interest is paid half-yearly, then in the formula A = P \( { (1+\frac { R }{ 100 } ) }^{ n } \), for R we take \(\frac{R}{2}\) , because R% p.a. means \(\frac{R}{2}\) % half-yearly and for n we take 2n, because n years is equal to 2n half-years.

Therefore, A = P \( { (1+\frac { R }{ 200 } ) }^{ 2n } \)

**Illustrative Examples**

**Example 1:** compute the compound interst on Rs 10000 for 2 years at 10% per annum when compounded half-yearly.

**Solution.**

Here, Principal P = Rs 10000, R = 10% per annum, and n = 2 years

Amount after 2 years

= P \( { (1+\frac { R }{ 200 } ) }^{ 2n } \)

= Rs 10000 x P \( { (1+\frac { 10 }{ 200 } ) }^{ 2 \times 2 } \)

= Rs 10000 x P \( { (1+\frac { 1 }{ 20 } ) }^{ 4 } \)

= Rs 10000 x P \( { (\frac { 21 }{ 20 } ) }^{ 4 } \)

= Rs 10000 x \(\frac{21}{20 }\) x \(\frac{21}{20}\) x \(\frac{21}{20}\) x \(\frac{21}{20}\)

= Rs 10000 x \(\frac{194481}{160000}\)

= Rs 12155.06

Therefore, Compound Interest = Ra(12155.06 – 10000) = Rs 2155.065

**Example 2:** How much will Rs 256 amount to in one year at \(12\frac{1}{2}\)% per annum, when the interest is compounded half-yearly.

**Solution. **P= Rs 256, 1 year= 2 half years, n = 2, Annual rate = 12 %

= \(12\frac{1}{2}\)%

Therefore, Half-yearly rate = \(\frac{1}{2}\)(\(\frac{25}{2}\) %) = \(\frac{25}{4}\)%

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

= Rs 256 \( (1+{ \frac { \frac { 25 }{ 4 } }{ 100 } ) }^{ 2 } \)

= Rs 256 \( { (1+\frac { 1 }{ 16 } ) }^{ 2 } \)

= Rs 256 x \(\frac{17}{16}\) x \(\frac{17}{16}\)

= Rs 289

**Example 3:** How much would a sum of Rs 16000 amount to in 2 years time at 10% per annum compounded interest, interest being payable half-yearly?

**Solution.** Here, P = Rs 16000, R = 10% per annum and n = 2 years.

Amount after 2 years

= P\( { (1+\frac { R }{ 200 } ) }^{ 2n } \)

= Rs 16000 x \( { (1+\frac { 10 }{ 200 } ) }^{ 2 \times 2 } \)

= Rs 16000 x \( { (1+\frac { 1 }{ 20 } ) }^{ 4 } \)

= Rs 16000 x \( { (\frac { 21 }{ 20 } ) }^{ 4 } \)

= Rs 16000 x \(\frac{21}{20}\) x \(\frac{21}{20}\) x \(\frac{21}{20}\) x \(\frac{21}{20}\)

= Rs 19448.10

Hence, a sum of Rs 16000 amounts to Rs 19448. 10 in 2 years.

## When the Interest is Compounded Quarterly

**Formula**

If P = Principal, R = Interest rate percent per annum and n = number of years, then

A = \( { (1+\frac { R }{ 400 } ) }^{ 4n } \)

C.I. = A – P

**Illustrative Examples**

**Example 1:** Find the compound interest on Rs 360000 for one year at the rate of 10% per annum, if the interest is compounded quarterly.

**Solution.** Here, P = Rs 360000, R = 10% per annum and n= 1 year

Amount after 1 year

= P\( { (1+\frac { R }{ 400 } ) }^{ 4n } \)

= Rs 360000 x \( { (1+\frac { 10 }{ 400 } ) }^{ 4 \times 1 } \)

= Rs 360000 x \( { (1+\frac { 1 }{ 40 } ) }^{ 4 } \)

= Rs 360000 x \( { (\frac { 41 }{ 40 } ) }^{ 4 } \)

= Rs 360000 x \(\frac{41}{40}\) x \(\frac{41}{40}\) x \(\frac{41}{40}\) x \(\frac{41}{40}\)

= Rs 397372.64

Therefore, Compound Interest = Rs 397372.64 – Rs 360000

= Rs 37372.64

**Example 2:** Sharukh deposited in a bank Rs 8000 for 6 months at the rate of 10% interest compounded quarterly. Find the amount he received after 6 months.

**Solution.** Here, P = Rs 8000, R = 10% per annum and n= 6 months

= \(\frac{6}{12}\)

= \(\frac{1}{2}\) year

Amount after 6 months

= P\( { (1+\frac { R }{ 400 } ) }^{ 4n } \)

= Rs 8000 x \((1+{ \frac { 10 }{ 400 } ) }^{ 4\times \frac { 1 }{ 2 } } \)

= Rs 8000 x \( { (1+\frac { 1 }{ 40 } ) }^{ 2 } \)

= Rs 8000 x \( { (\frac { 41 }{ 40 } ) }^{ 2 } \)

**Example 3:** Ramesh deposited Rs 7500 in a bank which pays him 12% interest per annum compounded quarterly. What is the amount which he receives after 9 months?

**Solution.** Here, P = Rs 7500, R = 10% per annum and n = 9 months

= \(\frac{9}{12}\)

= \(\frac{3}{4}\) year

Amount after 9 months

= P\( { (1+\frac { R }{ 400 } ) }^{ 4n } \)

= Rs 7500 x \( (1+{\frac { 12 }{ 400 } ) }^{ 4 \times \frac{3}{4} } \)

= Rs 8000 x \( { (1+\frac { 3 }{ 100 } ) }^{ 3 } \)

= Rs 8000 x \( { (\frac { 103 }{ 100 } ) }^{ 2 } \)

= Rs 8000 x \(\frac{103}{100}\) x \(\frac{103}{100}\) x \(\frac{103}{100}\)

= Rs 8195.45

= Rs 8000 x \(\frac{41}{40}\) x \(\frac{41}{40}\)

= Rs 8405

## When the Rate of Interest for Successive years are Different

**Formula**

If the rate of interest is different for every year say, \( { R }_{ 1 } \), \( { R }_{ 2 } \), \( { R }_{ 3 } \)…\( { R }_{ n } \) for the first, second, third year… nth year then the amount is given by

A = P \( (1+\frac { { R }_{ 1 } }{ 100 } ) \)\( (1+\frac { { R }_{ 2 } }{ 100 } ) \)\( (1+\frac { { R }_{ 3 } }{ 100 } ) \)….\( (1+\frac { { R }_{ n } }{ 100 } ) \)

**Illustrative Examples**

**Example 1**: Find the amount of Rs 50000 after 2 years, compounded annually; the rate interest being 8% p.a. during the first year and 9% p.a. during the second ear. Also,find the compound Interest.

**Solution.** Here, P = Rs 50000, \( { R }_{ 1 } \)= 8% p.a. and \( { R }_{ 2 } \) = 9% p.a.

Using the formula A = P \( (1+\frac { { R }_{ 1 } }{ 100 } ) \)\( (1+\frac { { R }_{ 2 } }{ 100 } ) \) we have:

Amount after 2 years

= Rs 50000 \( (1+\frac { 8 }{ 100 } ) \) \( (1+\frac { 9 }{ 100 } ) \)

= Rs 50000 \(\frac{27}{25}\) x \(\frac{109}{100}\)

= Rs 58860

Thus, amount after 2 years = Rs 58860.

And, compound interest = Rs (58860 – 50000) = Rs 8860.

**Example 2:** Find the compound interest on Rs 80,000 for 3 years if the rates 4%, 5% and 10% respectively.

**Solution.** Here, P = Rs 80000, \( { R }_{ 1 } \) = 4%, \( { R }_{ 2 } \) = 5% and \( { R }_{ 3 } \) = 10%

amount after 3 years

= Rs 80000 \( (1+\frac { 4 }{ 100 } ) \) \( (1+\frac { 5 }{ 100 } ) \) \( (1+\frac { 10 }{ 100 } ) \)

= Rs 80000 x \(\frac{104}{100}\) x \(\frac{105}{100}\) x \(\frac{110}{100}\)

= Rs 96096

Therefore, Compound interest = Rs (96096 — 80000) = Rs 16096.

## When Interest is Compounded Annually but Time is a Fraction

**Formula**

If P = Principal, R = Rate % per annum and Time = \(3\frac{3}{4}\) years (say), then

A = P\((1+{ \frac { R }{ 100 } ) }^{ 3} \) x \( (1+\frac { \frac { 3 }{ 4 } \times R }{ 100 } ) \)

**Illustrative Examples**

**Example 1:** Find the compound interest on Rs 31250 at 8% per annum for \(2\frac{3}{4}\) years.

**Solution.** Amount after \(2\frac{3}{4}\) years

= Rs 31250 x \((1+{ \frac { 8 }{ 100 } ) }^{ 2} \) x \( (1+\frac { \frac { 3 }{ 4 } \times 8 }{ 100 } ) \)

= Rs 31250 x \((1+{ \frac { 27 }{ 25 } ) }^{ 2} \) x \( (\frac { 53 }{ 50 } ) \)

= 31250 x \( (1+\frac { 27 }{ 25 } ) \) x \( (1+\frac { 27 }{ 25 } ) \) x \( (1+\frac { 53 }{ 50 } ) \)

= Rs 38637

Therefore, Amount = Rs 38637.

Hence, compound interest = Rs (38637 — 31250) = Rs 7387.

**Example 2:** Find the compound interest on Ra 24000 at 15% per annum for \(2\frac{1}{3}\) years.

**Solution.**

Here, P = Rs 24000, R =15% per annum and Time = \(2\frac{1}{3}\) years.

Amount after \(2\frac{1}{3}\)years

= P\((1+{ \frac { R }{ 100 } ) }^{ 2} \) x \( (1+\frac { \frac { 1 }{ 3 } \times R }{ 100 } ) \)

= Rs 24000 x \((1+{ \frac { 15 }{ 100 } ) }^{ 2} \) x \( (1+\frac { \frac { 1 }{ 3 } \times 15 }{ 100 } ) \)

= Rs 24000 x \((1+{ \frac { 3 }{ 20 } ) }^{ 2} \) x \( (1+\frac { 1 }{ 20 } ) \)

= Rs 24000 x \(({ \frac { 23 }{ 20 } ) }^{ 2} \) x \( (\frac { 21 }{ 20 } ) \)

= Rs 33327

Therefore, Compound interest = Rs (33327 — 24000) = Rs 9327

Vishal landge says

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