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
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