## 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

## 2 thoughts on “Calculating Compound Interest By Using Formulae”

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