Compound Interest Concepts

1. Basic Compound Interest

• Formula: $A=P\left(1+\frac{R}{100}\right)^T$

  $CI=A-P$

• Example: Question: Find the compound interest on ₹10,000 at 10% per annum for 2 years.

  Solution:

  $A=10000\left(1+\frac{10}{100}\right)^2$

  $A=10000(1.1)^2=12100$

  Therefore,

  $CI=12100-10000=\boxed{₹2100}$

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2. Finding Principal, Rate or Time

• Formula: $A=P\left(1+\frac{R}{100}\right)^T$

• Example: Question: A sum becomes ₹12,100 in 2 years at 10% compound interest. Find the principal.

  Solution:

  $12100=P(1.1)^2$

  $12100=1.21P$

  $P=\frac{12100}{1.21}=\boxed{₹10,000}$

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3. Difference Between CI and SI

• Formula for 2 years:

  $CI-SI=P\left(\frac{R}{100}\right)^2$

• Example: Question: Find the difference between compound interest and simple interest on ₹10,000 at 10% per annum for 2 years.

  Solution:

  $CI-SI=10000\left(\frac{10}{100}\right)^2$

  $=10000(0.01)$

  $=\boxed{₹100}$

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4. Compound Interest for 3 Years

• Formula:

  $A=P\left(1+\frac{R}{100}\right)^3$

• Example: Question: Find the amount on ₹20,000 at 5% compound interest for 3 years.

  Solution:

  $A=20000(1.05)^3$

  $=20000(1.157625)$

  $=\boxed{₹23,152.50}$

  Therefore,

  $CI=23152.50-20000=\boxed{₹3,152.50}$

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5. Principal Becomes a Multiple of Itself

• Formula:

  $A=P\left(1+\frac{R}{100}\right)^T$

  If the amount becomes (k) times the principal:

  $k=\left(1+\frac{R}{100}\right)^T$

• Example: Question: A sum becomes 8 times itself in 3 years at compound interest. In how many years will it become 64 times?

  Solution:

  $8P=P(1+r)^3$

  $8=(1+r)^3$

  $1+r=2$

  Therefore,

  $64=2^T$

  $T=6$

  Answer: (\boxed{6\text{ years}})

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6. Half-Yearly Compounding

• Formula:

  $A=P\left(1+\frac{R}{200}\right)^{2T}$

• Example: Question: Find the amount on ₹10,000 at 12% per annum for 1 year, compounded half-yearly.

  Solution:

  Rate for each half-year:

  $\frac{12}{2}=6$

  Number of periods:

  $2\times1=2$

  Therefore,

  $A=10000(1.06)^2$

  $A=\boxed{₹11,236}$

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7. Quarterly Compounding

• Formula:

  $A=P\left(1+\frac{R}{400}\right)^{4T}$

• Example: Question: Find the amount on ₹8,000 at 8% per annum for 1 year, compounded quarterly.

  Solution:

  Quarterly rate:

  $\frac{8}{4}=2$

  Number of quarters:

  $4$

  $A=8000(1.02)^4$

  $A\approx\boxed{₹8,659.46}$

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8. Changing Compounding Frequency

• Formula:

  $A=P\left(1+\frac{R}{100n}\right)^{nT}$

  where (n) is the number of compounding periods per year.

• Example: Question: ₹20,000 is invested at 10% per annum for 2 years. Find the amount when interest is compounded quarterly.

  Solution:

  $A=20000\left(1+\frac{10}{400}\right)^8$

  $A=20000(1.025)^8$

  $A\approx\boxed{₹24,367.18}$

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9. Loan Repayment in Equal Installments

• Formula:

  $P=\frac{X}{1+r}+\frac{X}{(1+r)^2}+\cdots+\frac{X}{(1+r)^n}$

  Therefore,

  $X=P\frac{r(1+r)^n}{(1+r)^n-1}$

• Example: Question: A loan of ₹50,000 is taken at 10% compound interest and is repaid in 3 equal annual installments. Find each installment.

  Solution:

  $X=50000\frac{0.1(1.1)^3}{(1.1)^3-1}$

  $X\approx\boxed{₹20,105}$

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10. Installment Paid at the End of Each Year

• Idea: In installment questions, the first payment earns interest for fewer years than the later payments. Therefore, convert every installment to its present value.

• Formula:

  $P=\frac{X}{1+r}+\frac{X}{(1+r)^2}+\cdots+\frac{X}{(1+r)^n}$

• Example: Question: A loan of ₹33,100 is repaid in 2 equal annual installments at 10% compound interest. Find each installment.

  Solution:

  Let each installment be (X).

  $33100=\frac{X}{1.1}+\frac{X}{1.1^2}$

  $33100=\frac{X}{1.1}+\frac{X}{1.21}$

  Solving,

  $X=\boxed{₹20,000}$

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11. Different Rates in Different Years

• Formula:

  \left(1+\frac{R_2}{100}\right)\cdots$$

• Example: Question: ₹10,000 is invested at 8% for the first year and 10% for the second year. Find the amount.

  Solution:

  $A=10000(1.08)(1.10)$

  $A=\boxed{₹11,880}$

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12. Different Rates for Different Time Periods

• Formula:

  $A=P\prod\left(1+\frac{R_i}{100}\right)^{T_i}$

• Example: Question: ₹10,000 is invested at 8% for 1 year and 10% for the next 2 years. Find the final amount.

  Solution:

  $A=10000(1.08)(1.10)^2$

  $A=10000(1.08)(1.21)$

  $A=\boxed{₹13,068}$

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13. Depreciation

• Formula:

  $V=P\left(1-\frac{R}{100}\right)^T$

• Example: Question: A machine costs ₹50,000 and depreciates by 10% every year. Find its value after 2 years.

  Solution:

  $V=50000(1-0.10)^2$

  $V=50000(0.9)^2$

  $V=\boxed{₹40,500}$

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14. Population Growth

• Formula:

  $P_T=P\left(1+\frac{R}{100}\right)^T$

• Example: Question: A city’s population is 1,00,000 and grows by 5% every year. Find its population after 2 years.

  Solution:

  $P_T=100000(1.05)^2$

  $=\boxed{1,10,250}$

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15. Population Decrease

• Formula:

  $P_T=P\left(1-\frac{R}{100}\right)^T$

• Example: Question: A population of 80,000 decreases by 10% every year. Find the population after 2 years.

  Solution:

  $P_T=80000(0.9)^2$

  $=\boxed{64,800}$

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Advanced Variants

16. Finding the Rate When Amount Becomes a Multiple

• Formula:

  $\left(1+\frac{R}{100}\right)^T=\frac AP$

• Example: Question: A sum doubles in 5 years at compound interest. Find the approximate annual rate.

  Solution:

  $2=\left(1+\frac R{100}\right)^5$

  $1+\frac R{100}=2^{1/5}$

  R\approx\boxed{14.87%}

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17. Difference Between Amounts at Two Different Times

• Formula:

  $A_n=P(1+r)^n$

  $A_{n+1}-A_n=Pr(1+r)^n$

• Example: Question: The amount after 3 years is ₹13,310 at 10% compound interest. Find the interest earned during the 4th year.

  Solution:

  Amount after 3 years:

  $A_3=13310$

  Interest during the 4th year:

  $13310\times10$

  $=\boxed{₹1,331}$

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18. Compound Interest for Fractional Years

• Formula: When the compounding period is specified, convert the time into the corresponding number of periods.

• Example: Question: Find the amount on ₹10,000 at 12% per annum compounded half-yearly for 1.5 years.

  Solution:

  Half-yearly rate:

  $12/2=6$

  Number of half-years:

  $1.5\times2=3$

  $A=10000(1.06)^3$

  $A=\boxed{₹11,910.16}$

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19. CI When the Rate Changes After a Certain Period

• Formula:

  $A=P(1+r_1)^{t_1}(1+r_2)^{t_2}$

• Example: Question: ₹20,000 is invested for 3 years. The rate is 10% for the first year and 20% for the next 2 years. Find the final amount.

  Solution:

  $A=20000(1.10)(1.20)^2$

  $A=20000(1.10)(1.44)$

  $=\boxed{₹31,680}$

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20. Present Value of a Future Amount

• Formula:

  $P=\frac{A}{(1+r)^T}$

• Example: Question: What amount should be invested today at 10% compound interest to obtain ₹13,310 after 3 years?

  Solution:

  $P=\frac{13310}{(1.1)^3}$

  $P=\frac{13310}{1.331}$

  $=\boxed{₹10,000}$

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21. Comparing Simple Interest and Compound Interest

• Formula:

  For 2 years:

  $CI-SI=P\left(\frac R{100}\right)^2$

  For 3 years:

  $CI-SI=P\left[\left(1+\frac R{100}\right)^3-1-\frac{3R}{100}\right]$

• Example: Question: The difference between CI and SI on ₹20,000 for 2 years is ₹200. Find the rate.

  Solution:

  $200=20000\left(\frac R{100}\right)^2$

  $0.01=\left(\frac R{100}\right)^2$

  $R=10$

  Answer: (\boxed{10%})

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