Menu Top
Complete Course of Mathematics
Topic 1: Numbers & Numerical Applications Topic 2: Algebra Topic 3: Quantitative Aptitude
Topic 4: Geometry Topic 5: Construction Topic 6: Coordinate Geometry
Topic 7: Mensuration Topic 8: Trigonometry Topic 9: Sets, Relations & Functions
Topic 10: Calculus Topic 11: Mathematical Reasoning Topic 12: Vectors & Three-Dimensional Geometry
Topic 13: Linear Programming Topic 14: Index Numbers & Time-Based Data Topic 15: Financial Mathematics
Topic 16: Statistics & Probability


Content On This Page
Simple Interest: Definition, Formula, and Calculations Compound Interest: Definition and Comparison with Simple Interest Compound Interest Formula and Calculation (Annual Compounding)
Compound Interest for Different Compounding Periods (Half-yearly, Quarterly, etc.) Applications of Compound Interest (Growth and Depreciation) Solving Problems on Simple and Compound Interest


Simple and Compound Interest



Simple Interest: Definition, Formula, and Calculations

In financial transactions involving borrowing or lending money, an additional sum is usually paid by the borrower to the lender as a charge for using the money for a specific period. This additional sum is called Interest.

The initial amount of money borrowed or lent is called the Principal (P).

The interest is typically calculated at a certain percentage of the principal amount per unit of time. This percentage is called the Rate of Interest (R). The rate is commonly quoted as 'per annum' (p.a.), meaning per year.

The duration for which the money is borrowed or lent is called the Time Period (T).

At the end of the time period, the borrower pays back the original principal amount along with the accumulated interest. The total sum paid back is called the Amount (A).

$\boldsymbol{\text{Amount (A) = Principal (P) + Interest}}$

... (i)


Simple Interest (SI)

Simple Interest (SI) is a type of interest calculation where the interest for each period is calculated solely on the original principal amount. The interest earned in previous periods is not added to the principal for calculating the interest in the current period. This results in the interest amount being the same for each unit of time (e.g., same interest every year if the rate is per annum).

Formula for Simple Interest

The Simple Interest is directly proportional to the Principal amount, the Rate of Interest, and the Time Period. The formula for calculating Simple Interest is:

$\boldsymbol{\text{SI} = \frac{\text{P} \times \text{R} \times \text{T}}{100}}$

... (ii)

Where:

Important Note on Units: For the formula $\text{SI} = \frac{\text{P} \times \text{R} \times \text{T}}{100}$ to work correctly, the units of Rate (R) and Time (T) must be compatible. If the Rate is given 'per annum', the Time must be in 'years'. If the Time is given in months or days, convert it to years:

Formula for Amount (A) with Simple Interest

The total Amount (A) to be repaid is the sum of the Principal (P) and the Simple Interest (SI) earned. Using formula (i) and substituting the formula for SI (ii):

$\text{A} = \text{P} + \text{SI}$

$\text{A} = \text{P} + \frac{\text{P} \times \text{R} \times \text{T}}{100}$

Factor out the Principal (P) from the right side:

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

... (iii)


Finding Principal, Rate, or Time from the SI Formula

The basic Simple Interest formula (ii) $\text{SI} = \frac{\text{P} \times \text{R} \times \text{T}}{100}$ can be rearranged to find any of the variables (P, R, or T) if the other three (SI, and the remaining two from P, R, T) are known.


Example 1. Calculate the Simple Interest on $\textsf{₹ } 8000$ at 12% per annum for 4 years.

Answer:

Given: Principal (P) $= \textsf{₹ } 8000$, Rate (R) $= 12\%$ p.a., Time (T) $= 4$ years.

The rate is per annum and time is in years, so units are compatible.

Using the formula for Simple Interest (ii):

$\text{SI} = \frac{\text{P} \times \text{R} \times \text{T}}{100}$

Substitute the values:

$\text{SI} = \frac{8000 \times 12 \times 4}{100}$

Simplify the expression:

$\text{SI} = \frac{\cancel{8000}^{\normalsize 80} \times 12 \times 4}{\cancel{100}^{\normalsize 1}}$

$\text{SI} = 80 \times 12 \times 4 = 80 \times 48$

Perform the multiplication:

$\begin{array}{cccc}& & & 4 & 8 \\ \times & & & 8 & 0 \\ \hline &&& 0 & 0 \\ & 3 & 8 & 4 & \times \\ \hline & 3 & 8 & 4 & 0 \\ \hline \end{array}$

$\boldsymbol{\text{SI} = \textsf{₹ } 3840}$

The Simple Interest is $\boldsymbol{\textsf{₹ } 3840}$.

The Total Amount to be repaid after 4 years would be $A = P + SI = \textsf{₹ } 8000 + \textsf{₹ } 3840 = \textsf{₹ } 11840$.


Example 2. At what rate per annum will $\textsf{₹ } 6000$ yield a Simple Interest of $\textsf{₹ } 1440$ in 3 years?

Answer:

Given: Principal (P) $= \textsf{₹ } 6000$, Simple Interest (SI) $= \textsf{₹ } 1440$, Time (T) $= 3$ years.

We need to find the Rate of Interest (R) per annum.

Using the formula for Rate (v):

$\text{R} = \frac{\text{SI} \times 100}{\text{P} \times \text{T}}$

Substitute the values:

$\text{R} = \frac{1440 \times 100}{6000 \times 3}$

Simplify the expression:

$\text{R} = \frac{1440 \times \cancel{100}^{\normalsize 1}}{\cancel{6000}^{\normalsize 60} \times 3}$

$\text{R} = \frac{1440}{60 \times 3} = \frac{1440}{180}$

Simplify the fraction:

$\frac{1440}{180} = \frac{144}{18} = 8$

$\boldsymbol{\text{R} = 8}$

Since the formula gives the rate as a percentage, the rate is $\boldsymbol{8\%}$ per annum.


Competitive Exam Notes:

Simple Interest is a fundamental concept. Problems usually involve straightforward application of the formula or finding one missing variable.

  • Key Formula: $\text{SI} = \frac{\text{P} \times \text{R} \times \text{T}}{100}$. Memorize this.
  • Unit Consistency: ALWAYS ensure R is per annum and T is in years before using the formula. Convert units if necessary.
  • Amount: Amount = Principal + Simple Interest.
  • Finding P, R, or T: Use the rearranged formulas or solve the basic SI formula algebraically.
  • Percentage Increase/Decrease relation: Simple interest can be seen as a fixed percentage increase of the principal per year. If R% is the rate per annum, the principal increases by R% every year.
  • Variations: Sometimes problems might involve different rates for different time periods. In such cases, calculate SI for each period separately and sum them up.

Compound Interest: Definition and Comparison with Simple Interest

Compound Interest (CI) is a type of interest calculation where the interest earned in each period is added to the principal amount for the next period. This accumulated interest from previous periods then also starts earning interest in subsequent periods. This process of adding the interest back to the principal is called compounding, and it leads to interest earning 'interest on interest'.

Compound interest is the standard method used in most financial transactions today, including savings accounts, loans, and investments, because it reflects the true growth of money over time.


Comparison between Simple Interest and Compound Interest

The fundamental difference between Simple Interest (SI) and Compound Interest (CI) lies in the base on which the interest is calculated in subsequent periods:

Feature Simple Interest (SI) Compound Interest (CI)
Principal for Interest Calculation Remains constant throughout the entire time period. It is always the original principal amount. Changes after each interest period. The interest from the previous period is added to the principal to form the new principal for the next period.
Interest Earned Over Time Increases linearly with time. The interest amount earned per unit of time (e.g., per year) is constant. Increases exponentially with time. The interest amount earned is higher in each successive period.
Total Interest Over Time > 1 Year Lower than Compound Interest for the same Principal, Rate, and Time (if compounded more than once). Higher than Simple Interest for the same Principal, Rate, and Time (if compounded more than once).
Applicable Formulas $\text{SI} = \frac{\text{P} \times \text{R} \times \text{T}}{100}$
$\text{Amount (A)} = \text{P} + \text{SI} = \text{P} \left(1 + \frac{\text{R} \times \text{T}}{100}\right)$
$\text{Amount (A)} = \text{P} \left(1 + \frac{\text{R}}{100}\right)^n$
$\text{Compound Interest (CI)} = \text{A} - \text{P}$
Compounding Effect No compounding. Interest is not added to the principal. Interest is added to the principal (compounded), leading to exponential growth.

Crucially, for a time period of exactly one year, the Simple Interest and Compound Interest calculated annually on the same principal and rate are equal. The difference between SI and CI becomes significant and increases as the time period extends beyond one year and the number of compounding periods increases.


Example 1. Compare the Simple Interest and Compound Interest on $\textsf{₹ } 5000$ at 10% per annum for 3 years.

Answer:

Given: Principal (P) $= \textsf{₹ } 5000$, Rate (R) $= 10\%$ p.a., Time (T or n) $= 3$ years.

Simple Interest Calculation:

Using the formula $\text{SI} = \frac{\text{P} \times \text{R} \times \text{T}}{100}$:

$\text{SI} = \frac{5000 \times 10 \times 3}{100}$

Calculate SI:

$\text{SI} = \frac{\cancel{5000}^{\normalsize 50} \times 10 \times 3}{\cancel{100}^{\normalsize 1}} = 50 \times 10 \times 3 = 500 \times 3 = 1500$

Simple Interest for 3 years $= \boldsymbol{\textsf{₹ } 1500}$.

Compound Interest Calculation (Step-by-step for understanding the concept):

Assume interest is compounded annually.

Year 1: Interest is calculated on the initial principal $\textsf{₹ } 5000$.

Interest (Year 1) $= \frac{5000 \times 10 \times 1}{100} = \textsf{₹ } 500$

Amount at the end of Year 1 $= \text{Principal} + \text{Interest} = \textsf{₹ } 5000 + \textsf{₹ } 500 = \textsf{₹ } 5500$

Year 2: The principal for the second year is the amount at the end of Year 1, which is $\textsf{₹ } 5500$.

Interest (Year 2) $= \frac{5500 \times 10 \times 1}{100} = \textsf{₹ } 550$

Amount at the end of Year 2 $= \text{Amount (end of Year 1)} + \text{Interest (Year 2)} = \textsf{₹ } 5500 + \textsf{₹ } 550 = \textsf{₹ } 6050$

Year 3: The principal for the third year is the amount at the end of Year 2, which is $\textsf{₹ } 6050$.

Interest (Year 3) $= \frac{6050 \times 10 \times 1}{100} = \textsf{₹ } 605$

Amount at the end of Year 3 $= \text{Amount (end of Year 2)} + \text{Interest (Year 3)} = \textsf{₹ } 6050 + \textsf{₹ } 605 = \textsf{₹ } 6655$

Total Compound Interest (CI) = Total Amount at the end of 3 years - Original Principal

CI $= \textsf{₹ } 6655 - \textsf{₹ } 5000 = \boldsymbol{\textsf{₹ } 1655}$.

Compound Interest for 3 years $= \boldsymbol{\textsf{₹ } 1655}$.

Comparison:

Simple Interest (SI) $= \textsf{₹ } 1500$

Compound Interest (CI) $= \textsf{₹ } 1655$

For a period of 3 years, the Compound Interest ($\textsf{₹ } 1655$) is greater than the Simple Interest ($\textsf{₹ } 1500$). The difference (CI - SI) is $\textsf{₹ } 1655 - \textsf{₹ } 1500 = \textsf{₹ } 155$. This difference is the interest earned on the interest from the previous years (Interest on $\textsf{₹ } 500$ from Year 1 for 2 years and interest on $\textsf{₹ } 550$ from Year 2 for 1 year, using SI logic on previous interests is complicated, the direct formula is better!).

The extra interest in CI comes from the interest earned on the first year's interest ($\textsf{₹ } 500$) and the second year's interest ($\textsf{₹ } 550$ includes interest on original principal + interest on first year's interest).


Competitive Exam Notes:

Understanding the difference between SI and CI is crucial. CI problems are more frequent and often involve calculations over multiple periods.

  • Key Difference: SI is on original Principal; CI is on Principal + Accumulated Interest.
  • Growth Pattern: SI grows linearly; CI grows exponentially.
  • Equality: SI = CI for Time = 1 year (when compounded annually).
  • CI > SI: For Time > 1 year, CI is always greater than SI (for positive rate).
  • Calculation Methods:
    • Step-by-step: Useful for understanding, especially with short periods.
    • Formula: Essential for longer periods or when compounding is not annual.
  • Difference between CI and SI: Often asked. For 2 years, CI - SI = Interest on 1st year's SI. For longer periods, the formula method for both CI and SI is needed, then find the difference.


Compound Interest Formula and Calculation (Annual Compounding)

Calculating Compound Interest (CI) step-by-step for each period can be tedious, especially for longer time durations. Fortunately, there is a general formula that directly calculates the total Amount after compounding, from which the Compound Interest can be found.


Derivation of the Formula for Amount (Annual Compounding)

Let the Principal amount be P, the rate of interest per annum be R% (meaning $\frac{R}{100}$ as a fraction), and the number of years be $n$. We assume the interest is compounded annually (once a year).

At the end of the 1st year:

The interest for the first year is calculated on the original principal, P. Using the Simple Interest formula for 1 year:

Interest (Year 1) $= \frac{\text{P} \times \text{R} \times 1}{100} = \text{P} \frac{\text{R}}{100}$

The Amount at the end of the 1st year ($A_1$) is the sum of the principal and the interest earned in the first year:

$\boldsymbol{A_1 = \text{P} + \text{P} \frac{\text{R}}{100} = \text{P} \left(1 + \frac{\text{R}}{100}\right)}$

... (i)

At the end of the 2nd year:

For the second year, the principal is the amount at the end of the 1st year, which is $A_1 = \text{P} \left(1 + \frac{\text{R}}{100}\right)$. The interest for the second year is calculated on this new principal $A_1$ at the rate R% for 1 year.

Interest (Year 2) $= \frac{A_1 \times \text{R} \times 1}{100} = A_1 \frac{\text{R}}{100}$

The Amount at the end of the 2nd year ($A_2$) is the sum of the principal for the 2nd year ($A_1$) and the interest earned in the 2nd year:

$\boldsymbol{A_2 = A_1 + A_1 \frac{\text{R}}{100} = A_1 \left(1 + \frac{\text{R}}{100}\right)}$

... (ii)

Now, substitute the expression for $A_1$ from equation (i) into equation (ii):

$\boldsymbol{A_2 = \text{P} \left(1 + \frac{\text{R}}{100}\right) \times \left(1 + \frac{\text{R}}{100}\right) = \text{P} \left(1 + \frac{\text{R}}{100}\right)^2}$

... (iii)

At the end of the 3rd year:

The principal for the third year is the amount at the end of the 2nd year, $A_2 = \text{P} \left(1 + \frac{\text{R}}{100}\right)^2$. The interest for the third year is calculated on $A_2$ at rate R% for 1 year.

Interest (Year 3) $= A_2 \frac{\text{R}}{100}$

The Amount at the end of the 3rd year ($A_3$) is $A_2 + \text{Interest (Year 3)}$:

$\boldsymbol{A_3 = A_2 \left(1 + \frac{\text{R}}{100}\right)}$

... (iv)

Substitute the expression for $A_2$ from equation (iii) into equation (iv):

$\boldsymbol{A_3 = \text{P} \left(1 + \frac{\text{R}}{100}\right)^2 \times \left(1 + \frac{\text{R}}{100}\right) = \text{P} \left(1 + \frac{\text{R}}{100}\right)^3}$

... (v)

Observing the pattern in equations (i), (iii), and (v), we can generalize the formula for the Amount (A) at the end of 'n' years, when compounded annually:

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

... (vi)

Where:

This formula is for situations where interest is compounded annually.

Formula for Compound Interest (CI)

The Compound Interest (CI) earned is the difference between the total Amount (A) at the end of the period and the original Principal (P). Using the definition of Amount from equation (i) in the previous section:

$\boldsymbol{\text{CI} = \text{A} - \text{P}}$

... (vii)

Substituting the formula for A (vi) into equation (vii):

$\boldsymbol{\text{CI} = \text{P} \left(1 + \frac{\text{R}}{100}\right)^n - \text{P}}$

... (viii)

We can factor out P:

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

... (ix)


Example 1. Calculate the Compound Interest on $\textsf{₹ } 15,000$ at 8% per annum for 2 years, compounded annually. Also, find the amount.

Answer:

Given: Principal (P) $= \textsf{₹ } 15000$, Rate (R) $= 8\%$ p.a., Time (n) $= 2$ years.

Interest is compounded annually.

Calculate the Amount (A):

Using the formula for Amount (vi): $\text{A} = \text{P} \left(1 + \frac{\text{R}}{100}\right)^n$

$\text{A} = 15000 \left(1 + \frac{8}{100}\right)^2$

Simplify the term inside the bracket:

$1 + \frac{8}{100} = 1 + \frac{2}{25} = \frac{25+2}{25} = \frac{27}{25}$

$\text{A} = 15000 \left(\frac{27}{25}\right)^2 = 15000 \times \frac{27}{25} \times \frac{27}{25}$

Calculate the amount:

$\text{A} = \cancel{15000}^{\normalsize 600} \times \frac{27}{\cancel{25}^{\normalsize 1}} \times \frac{27}{25}$

$\text{A} = \cancel{600}^{\normalsize 24} \times \frac{27}{1} \times \frac{27}{\cancel{25}^{\normalsize 1}}$

A $= 24 \times 27 \times 27 = 24 \times (27)^2 = 24 \times 729$

Perform the multiplication:

$\begin{array}{cccc}& & & 7 & 2 & 9 \\ \times & & & & 2 & 4 \\ \hline & & 2 & 9 & 1 & 6 \\ & 1 4 & 5 & 8 & \times \\ \hline 1 & 7 & 4 & 9 & 6 \\ \hline \end{array}$

$\boldsymbol{\text{A} = \textsf{₹ } 17496}$

The Amount after 2 years is $\boldsymbol{\textsf{₹ } 17496}$.

Calculate the Compound Interest (CI):

Using the formula (vii): $\text{CI} = \text{A} - \text{P}$

CI $= \textsf{₹ } 17496 - \textsf{₹ } 15000$

$\boldsymbol{\text{CI} = \textsf{₹ } 2496}$

The Compound Interest is $\boldsymbol{\textsf{₹ } 2496}$.


Competitive Exam Notes:

The compound interest formulas for Amount and CI are essential tools. Practice using them efficiently, especially with simplifying fractions or using decimal equivalents.

  • Amount Formula: $\text{A} = \text{P} \left(1 + \frac{\text{R}}{100}\right)^n$. This gives the total amount including principal and accumulated interest.
  • CI Formula: $\text{CI} = \text{A} - \text{P}$ or $\text{CI} = \text{P} \left[ \left(1 + \frac{\text{R}}{100}\right)^n - 1 \right]$. Use the first one after calculating A.
  • R and n Units: Ensure R is the rate per compounding period, and n is the total number of compounding periods. For annual compounding, R is the annual rate and n is the number of years.
  • Calculation Steps: Calculate $(1 + \frac{R}{100})$, raise it to the power of n, then multiply by P for A. Subtract P to get CI. Be comfortable with squares and cubes of common numbers.
  • Fractional Approach: Converting $\left(1 + \frac{R}{100}\right)$ into a single fraction can sometimes make multiplication easier, especially if P is a multiple of the denominator. E.g., $10\% = \frac{1}{10}$, so $1 + \frac{1}{10} = \frac{11}{10}$. $8\% = \frac{8}{100} = \frac{2}{25}$, so $1 + \frac{2}{25} = \frac{27}{25}$.


Compound Interest for Different Compounding Periods (Half-yearly, Quarterly, etc.)

Interest is not always compounded annually. The frequency of compounding can be different, such as half-yearly, quarterly, monthly, or even daily. When the interest is compounded more frequently than once a year, the standard annual compounding formula needs to be adjusted to reflect the shorter compounding intervals.


Adjustments to Rate and Time for Different Compounding Periods

Let the annual rate of interest be R% per annum, and the total time period be $n$ years.

If interest is compounded k times in a year (e.g., $k=2$ for half-yearly, $k=4$ for quarterly, $k=12$ for monthly):

We use these adjusted values for the rate and number of periods in the compound interest formula derived for annual compounding (Amount $= \text{P} \left(1 + \frac{\text{Rate}}{100}\right)^{\text{Number of periods}}$).

Substituting the adjusted rate and number of periods:

$\boldsymbol{\text{Amount (A)} = \text{P} \left(1 + \frac{\text{R}/k}{100}\right)^{nk}}$

... (i)

This can also be written as:

$\boldsymbol{\text{A} = \text{P} \left(1 + \frac{\text{R}}{100k}\right)^{nk}}$

... (ii)

Where:

The Compound Interest (CI) is still given by $\text{CI} = \text{A} - \text{P}$.

Specific Compounding Frequencies:


Example 1. Calculate the Compound Interest on $\textsf{₹ } 20,000$ at 16% per annum for 9 months, compounded quarterly.

Answer:

Given: Principal (P) $= \textsf{₹ } 20000$, Annual Rate (R) $= 16\%$ p.a., Time $= 9$ months.

Compounding is quarterly ($k=4$).

Convert time into years: 9 months $= \frac{9}{12}$ years $= 0.75$ years.

Adjust the rate for the compounding period:

Rate per quarter $= \frac{\text{Annual Rate}}{4} = \frac{16\%}{4} = 4\%$

Adjust the time into the number of compounding periods:

Number of quarters in 9 months $= 9 \text{ months} \div 3 \text{ months/quarter} = 3$ quarters.

(Alternatively, Number of periods $= n \times k = 0.75 \times 4 = 3$ periods)

Using the Amount formula with adjusted values: $\text{A} = \text{P} \left(1 + \frac{\text{Rate per period}}{100}\right)^{\text{Number of periods}}$

$\text{A} = 20000 \left(1 + \frac{4}{100}\right)^3$

Simplify the term inside the bracket:

$1 + \frac{4}{100} = 1 + \frac{1}{25} = \frac{25+1}{25} = \frac{26}{25}$

$\text{A} = 20000 \left(\frac{26}{25}\right)^3 = 20000 \times \frac{26}{25} \times \frac{26}{25} \times \frac{26}{25}$

Calculate the amount:

$\text{A} = \cancel{20000}^{\normalsize 800} \times \frac{26}{\cancel{25}^{\normalsize 1}} \times \frac{26}{25} \times \frac{26}{25}$

$\text{A} = \cancel{800}^{\normalsize 32} \times \frac{26}{1} \times \frac{26}{\cancel{25}^{\normalsize 1}} \times \frac{26}{25}$

$\text{A} = 32 \times 26 \times 26 \times \frac{26}{25} = 32 \times (26)^3 \times \frac{1}{25} = 32 \times 17576 \times \frac{1}{25}$

$\text{A} = 20000 \times \frac{26}{25} \times \frac{26}{25} \times \frac{26}{25} = \frac{20000 \times 26^3}{25^3} = \frac{20000 \times 17576}{15625}$

$\frac{20000}{15625} = \frac{20000 \div 625}{15625 \div 625} = \frac{32}{25}$

$\text{A} = \frac{32}{25} \times 17576 = \frac{562432}{25}$

$\begin{array}{r} 22497.28 \\ 25{\overline{\smash{\big)}\,562432.00}} \\ \underline{-~\phantom{(}50\phantom{2432)}} \\ 62\phantom{432.00)} \\ \underline{-~\phantom{()}\,50\phantom{432.00)}} \\ 124\phantom{32.00)} \\ \underline{-~\phantom{()}\,100\phantom{32.00)}} \\ 243\phantom{2.00)} \\ \underline{-~\phantom{()}\,225\phantom{2.00)}} \\ 182\phantom{.00)} \\ \underline{-~\phantom{()}\,175\phantom{.00)}} \\ 7.00\phantom{)} \\ \underline{-~\phantom{()}\,5.00\phantom{)}} \\ 2.000 \\ \underline{-~\phantom{()}\,2.000} \\ 0 \end{array}$

$\boldsymbol{\text{A} = \textsf{₹ } 22497.28}$

Amount after 9 months (3 quarters) is $\textsf{₹ } 22497.28$.

Compound Interest (CI) = A - P

CI $= \textsf{₹ } 22497.28 - \textsf{₹ } 20000$

$\boldsymbol{\text{CI} = \textsf{₹ } 2497.28}$

The Compound Interest is $\boldsymbol{\textsf{₹ } 2497.28}$.


When Time is a Fraction or Mixed Number (Annual Compounding)

If the total time period is not a whole number of years, say it is $n$ whole years and a fraction $\frac{a}{b}$ of a year (i.e., time $= n \frac{a}{b}$ years), and the interest is compounded annually, the formula needs a slight modification. The amount is compounded for the whole number of years $n$, and then simple interest is calculated on the amount accumulated at the end of $n$ years for the remaining fractional part of the year $\frac{a}{b}$.

Amount at the end of $n$ whole years $= \text{P} \left(1 + \frac{\text{R}}{100}\right)^n$. Let this amount be $A_n$.

Interest for the fractional part $\frac{a}{b}$ of a year is calculated as Simple Interest on $A_n$ for $\frac{a}{b}$ years at rate R% p.a.

Interest for fractional part $= \frac{A_n \times \text{R} \times (a/b)}{100} = A_n \times \frac{(a/b)R}{100}$

The total Amount at the end of $n \frac{a}{b}$ years is $A_n + \text{Interest for fractional part}$.

$\text{A} = A_n \left(1 + \frac{(a/b)R}{100}\right)$

Substitute the expression for $A_n$:

$\boldsymbol{\text{A} = \text{P} \left(1 + \frac{\text{R}}{100}\right)^n \left(1 + \frac{\frac{a}{b} \times \text{R}}{100}\right)}$

... (iii)

The Compound Interest (CI) is $A - P$.

Example 2. Find the Compound Interest on $\textsf{₹ } 12000$ for $2\frac{1}{2}$ years at 10% per annum, compounded annually.

Answer:

Given: P $= \textsf{₹ } 12000$, R $= 10\%$ p.a., Time $= 2\frac{1}{2}$ years.

Here, $n=2$ (whole years) and the fractional part is $\frac{a}{b} = \frac{1}{2}$ year.

Compounding is annual.

Using the formula (iii) for fractional time:

$\text{A} = \text{P} \left(1 + \frac{\text{R}}{100}\right)^n \left(1 + \frac{\frac{a}{b} \times \text{R}}{100}\right)$

$\text{A} = 12000 \left(1 + \frac{10}{100}\right)^2 \left(1 + \frac{\frac{1}{2} \times 10}{100}\right)$

Simplify the terms:

$1 + \frac{10}{100} = 1 + \frac{1}{10} = \frac{11}{10}$

$1 + \frac{\frac{1}{2} \times 10}{100} = 1 + \frac{5}{100} = 1 + \frac{1}{20} = \frac{20+1}{20} = \frac{21}{20}$

$\text{A} = 12000 \left(\frac{11}{10}\right)^2 \left(\frac{21}{20}\right)$

$\text{A} = 12000 \times \frac{121}{100} \times \frac{21}{20}$

Calculate the amount:

$\text{A} = \cancel{12000}^{\normalsize 120} \times \frac{121}{\cancel{100}^{\normalsize 1}} \times \frac{21}{20}$

$\text{A} = \cancel{120}^{\normalsize 6} \times 121 \times \frac{21}{\cancel{20}^{\normalsize 1}}$

A $= 6 \times 121 \times 21$

$121 \times 6 = 726$

$\begin{array}{cccc}& & 7 & 2 & 6 \\ \times & & & 2 & 1 \\ \hline & & 7 & 2 & 6 \\ 1 & 4 & 5 & 2 & \times \\ \hline 1 & 5 & 2 & 4 & 6 \\ \hline \end{array}$

$\boldsymbol{\text{A} = \textsf{₹ } 15246}$

The Amount after $2\frac{1}{2}$ years is $\boldsymbol{\textsf{₹ } 15246}$.

Compound Interest (CI) = A - P

CI $= \textsf{₹ } 15246 - \textsf{₹ } 12000$

$\boldsymbol{\text{CI} = \textsf{₹ } 3246}$

The Compound Interest is $\boldsymbol{\textsf{₹ } 3246}$.


Competitive Exam Notes:

Compounding frequency is a key factor in CI problems. Always check if compounding is annual, half-yearly, quarterly, etc.

  • Adjustments: For $k$ compoundings per year, the rate used in the formula is $\frac{R}{k}$, and the number of periods is $n \times k$.
  • Formula: $\text{A} = \text{P} \left(1 + \frac{\text{R}_{\text{annual}}}{100k}\right)^{nk}$.
  • Half-yearly: $k=2$, use $\frac{R}{2}$ and $2n$.
  • Quarterly: $k=4$, use $\frac{R}{4}$ and $4n$.
  • Monthly: $k=12$, use $\frac{R}{12}$ and $12n$.
  • Fractional Time (Annual Compounding): For $n\frac{a}{b}$ years compounded annually, amount $= \text{P} \left(1 + \frac{\text{R}}{100}\right)^n \left(1 + \frac{(a/b)R}{100}\right)$. This is equivalent to compounding for $n$ years and calculating simple interest on that amount for the remaining fraction.
  • Calculation Steps: Pay careful attention to simplifying the term $(1 + \frac{\text{R}}{100k})$ before raising it to the power. Use fractional forms where appropriate.


Applications of Compound Interest (Growth and Depreciation)

The compound interest concept is a specific instance of a broader mathematical model used to describe exponential growth or decay. This model is applicable in various real-world scenarios where a quantity changes by a fixed percentage over regular intervals. Common applications include population growth, the appreciation (increase in value) of assets like property, and the depreciation (decrease in value) of assets like machinery or vehicles.


Growth Applications (Compound Growth)

If a quantity increases at a constant percentage rate over a fixed period, its growth follows the pattern of compound interest. This is often referred to as compound growth.

Let $P_0$ be the initial quantity (or value) at the start, R be the rate of growth per period (expressed as a percentage), and $n$ be the number of periods over which the growth occurs.

The formula for the quantity after $n$ periods of growth is analogous to the Amount formula in compound interest:

$\boldsymbol{\text{Quantity after n periods} = P_0 \left(1 + \frac{\text{R}}{100}\right)^n}$

... (i)

Here, $P_0$ is the initial value (like the Principal), R is the growth rate per period (like the annual interest rate), and $n$ is the number of growth periods (like the number of years). The $(1 + \frac{R}{100})$ factor represents the multiplier for each period's growth.

Examples where this applies include:


Depreciation Applications (Exponential Decay)

Depreciation is the decrease in the value of an asset over time, usually due to usage, age, or obsolescence. If an asset depreciates at a fixed percentage rate per period, its value decreases exponentially. This is similar to negative compound interest or exponential decay.

Let $V_0$ be the initial value of the asset, R be the rate of depreciation per period (expressed as a percentage), and $n$ be the number of periods over which the depreciation occurs.

The formula for the value of the asset after $n$ periods of depreciation is:

$\boldsymbol{\text{Value after n periods} = V_0 \left(1 - \frac{\text{R}}{100}\right)^n}$

... (ii)

The minus sign in the formula signifies that the value is decreasing. The $(1 - \frac{R}{100})$ factor represents the multiplier for each period's depreciation, resulting in a value less than the previous period's value.

Examples where this applies include:


Example 1. The population of a city is 1,50,000. If it increases at a rate of 5% per annum, what will be its population after 3 years?

Answer:

Given: Initial Population ($P_0$) $= 150000$, Annual Rate of Growth (R) $= 5\%$, Time (n) $= 3$ years.

This is a growth application, similar to compound interest.

Using the growth formula (i):

$\text{Population after 3 years} = P_0 \left(1 + \frac{\text{R}}{100}\right)^n$

$= 150000 \left(1 + \frac{5}{100}\right)^3$

Simplify the term inside the bracket:

$1 + \frac{5}{100} = 1 + \frac{1}{20} = \frac{20+1}{20} = \frac{21}{20}$

$= 150000 \left(\frac{21}{20}\right)^3 = 150000 \times \frac{21}{20} \times \frac{21}{20} \times \frac{21}{20}$

$= 150000 \times \frac{9261}{8000}$

Calculate the population:

$= \cancel{150000}^{\normalsize 150} \times \frac{9261}{\cancel{8000}^{\normalsize 8}}$

$= \frac{150 \times 9261}{8}$

$= \frac{\cancel{150}^{\normalsize 75} \times 9261}{\cancel{8}^{\normalsize 4}} = \frac{75 \times 9261}{4}$

$= \frac{694575}{4} = 173643.75$

$\boldsymbol{\text{Population after 3 years} = 173643.75}$

In practical population problems, the result is often rounded to the nearest whole number as population consists of discrete individuals. However, based strictly on the formula for exponential growth, the calculated value is 173643.75.

Assuming the problem implies calculating the exact value based on the compound growth model, the population after 3 years will be $\boldsymbol{173643.75}$. If a whole number is required, it would typically be stated to round to the nearest integer.


Example 2. The value of a car depreciates at the rate of 12% per annum. If its present value is $\textsf{₹ } 8,00,000$, what will be its value after 2 years?

Answer:

Given: Present Value ($V_0$) $= \textsf{₹ } 800000$, Annual Rate of Depreciation (R) $= 12\%$, Time (n) $= 2$ years.

This is a depreciation application, using the exponential decay formula.

Using the depreciation formula (ii):

$\text{Value after 2 years} = V_0 \left(1 - \frac{\text{R}}{100}\right)^n$

$= 800000 \left(1 - \frac{12}{100}\right)^2$

Simplify the term inside the bracket:

$1 - \frac{12}{100} = 1 - \frac{3}{25} = \frac{25-3}{25} = \frac{22}{25}$

$= 800000 \left(\frac{22}{25}\right)^2 = 800000 \times \frac{22}{25} \times \frac{22}{25}$

$= 800000 \times \frac{484}{625}$

Calculate the value after depreciation:

$= \cancel{800000}^{\normalsize 32000} \times \frac{484}{\cancel{25}^{\normalsize 1}}$

$= \cancel{32000}^{\normalsize 1280} \times 484 \times \frac{1}{\cancel{25}^{\normalsize 1}}$

Value after 2 years $= 1280 \times 484$

Perform the multiplication:

$\begin{array}{ccccccc}& & & 1 & 2 & 8 & 0 \\ \times & & & & 4 & 8 & 4 \\ \hline & & & 5 & 1 & 2 & 0 \\ & & 10 & 2 & 4 & 0 & \times \\ & 5 & 1 & 2 & 0 & \times & \times \\ \hline 6 & 1 & 9 & 5 & 2 & 0 & 0 \\ \hline \end{array}$

Value after 2 years $= \boldsymbol{\textsf{₹ } 619520}$

The value of the car after 2 years will be $\boldsymbol{\textsf{₹ } 6,19,520}$.


Competitive Exam Notes:

Growth and depreciation problems are direct applications of compound interest principles. The key is to use the correct sign (+ for growth, - for depreciation) in the formula.

  • Growth Formula: Final Value $= \text{Initial Value} \times \left(1 + \frac{\text{Rate}}{100}\right)^n$. Rate is positive.
  • Depreciation Formula: Final Value $= \text{Initial Value} \times \left(1 - \frac{\text{Rate}}{100}\right)^n$. Rate is positive, but used with a minus sign.
  • Identifying Parameters: Clearly identify the initial value, the rate per period, and the number of periods from the problem statement. Ensure units (like years) are consistent.
  • Rate per Period: If the rate is given annually but the growth/depreciation happens over different periods (e.g., monthly), adjust the rate and number of periods as done for compound interest (Rate per period = R/k, Number of periods = nk). However, most simple growth/depreciation problems use annual rates and whole years.
  • Back Calculation: Sometimes you might be asked to find the original value given the final value and the rate/time. Rearrange the formulas accordingly: $\text{Initial Value} = \frac{\text{Final Value}}{\left(1 \pm \frac{\text{R}}{100}\right)^n}$.


Solving Problems on Simple and Compound Interest

This section focuses on applying the formulas and concepts of Simple Interest (SI) and Compound Interest (CI) to solve various types of problems. It includes finding interest or amount, determining principal, rate, or time, and dealing with the difference between CI and SI.


Example 1. At what rate percent per annum Simple Interest will a sum of money double itself in 8 years?

Answer:

Let the Principal amount be P.

If the sum of money doubles itself, the Amount (A) at the end of the period will be twice the Principal, i.e., $A = 2P$.

The Time period (T) is 8 years.

We need to find the Rate of Interest (R) per annum under Simple Interest.

The Simple Interest (SI) earned is the difference between the Amount and the Principal:

SI $= A - P = 2P - P = P$

So, the interest earned is equal to the original principal amount.

Using the formula for Simple Interest: $\text{SI} = \frac{\text{P} \times \text{R} \times \text{T}}{100}$

Substitute the known values (SI = P, Time = 8):

$\boldsymbol{P = \frac{\text{P} \times \text{R} \times 8}{100}}$

... (i)

We can cancel P from both sides of equation (i), assuming $P \neq 0$:

$\boldsymbol{1 = \frac{\text{R} \times 8}{100}}$

Now, solve for R:

$\text{R} \times 8 = 100 \times 1$

$\boldsymbol{\text{R} = \frac{100}{8}}$

Simplify the fraction:

$\frac{100}{8} = \frac{50}{4} = \frac{25}{2} = 12.5$

$\boldsymbol{\text{R} = 12.5}$

The rate of interest is $\boldsymbol{12.5\%}$ per annum.

Verification:

Let P = $\textsf{₹ } 1000$. R = 12.5%, T = 8 years.

SI $= \frac{1000 \times 12.5 \times 8}{100} = \frac{1000 \times 100}{100} = 1000$

Amount $= P + SI = 1000 + 1000 = 2000$. The sum doubles itself.


Example 2. The difference between Compound Interest and Simple Interest on a certain sum of money for 2 years at 8% per annum is $\textsf{₹ } 32$. Find the sum of money.

Answer:

Let the sum of money (Principal) be P.

Time (n) = 2 years, Rate (R) $= 8\%$ p.a.

Given: Difference between CI and SI = $\textsf{₹ } 32$.

We need to find the Principal (P).

Method 1: Calculate SI and CI separately.

Calculate Simple Interest (SI) for 2 years:

$\text{SI} = \frac{\text{P} \times \text{R} \times \text{n}}{100} = \frac{\text{P} \times 8 \times 2}{100} = \frac{16\text{P}}{100} = \frac{4\text{P}}{25}$

Calculate Compound Interest (CI) for 2 years (compounded annually):

Amount under CI: $A = P \left(1 + \frac{\text{R}}{100}\right)^n = P \left(1 + \frac{8}{100}\right)^2 = P \left(1 + \frac{2}{25}\right)^2 = P \left(\frac{27}{25}\right)^2 = P \times \frac{729}{625}$.

CI $= A - P = P \frac{729}{625} - P = P \left(\frac{729}{625} - 1\right) = P \left(\frac{729 - 625}{625}\right) = \frac{104\text{P}}{625}$

Given that CI - SI = $\textsf{₹ } 32$.

$\frac{104\text{P}}{625} - \frac{4\text{P}}{25} = 32$

Find a common denominator (625). $\frac{4P}{25} = \frac{4P \times 25}{25 \times 25} = \frac{100P}{625}$.

$\frac{104\text{P}}{625} - \frac{100\text{P}}{625} = 32$

$\frac{(104 - 100)\text{P}}{625} = 32$

$\frac{4\text{P}}{625} = 32$

... (ii)

Solve for P from equation (ii):

$\boldsymbol{4\text{P} = 32 \times 625}$

$\boldsymbol{\text{P} = \frac{32 \times 625}{4}}$

$\text{P} = \cancel{32}^{\normalsize 8} \times \frac{625}{\cancel{4}^{\normalsize 1}} = 8 \times 625$

Perform the multiplication:

$\begin{array}{cccc}& & & 6 & 2 & 5 \\ \times & & & & 8 \\ \hline & 5 & 0 & 0 & 0 \\ \hline \end{array}$

$\boldsymbol{\text{P} = \textsf{₹ } 5000}$

The sum of money (Principal) is $\textsf{₹ } 5000$.

Alternative Method (Formula for Difference for 2 years):

For 2 years, the difference between CI and SI is given by the formula: $\text{CI} - \text{SI} = \text{P} \left(\frac{\text{R}}{100}\right)^2$.

Given: CI - SI $= \textsf{₹ } 32$, Rate (R) $= 8\%$.

Substitute the values into the formula:

$\boldsymbol{32 = \text{P} \left(\frac{8}{100}\right)^2}$

... (iii)

Simplify the term inside the bracket:

$\frac{8}{100} = \frac{2}{25}$

$\left(\frac{8}{100}\right)^2 = \left(\frac{2}{25}\right)^2 = \frac{4}{625}$

Substitute this back into equation (iii):

$\boldsymbol{32 = \text{P} \times \frac{4}{625}}$

Solve for P:

$\boldsymbol{\text{P} = 32 \times \frac{625}{4}}$

$\text{P} = \cancel{32}^{\normalsize 8} \times \frac{625}{\cancel{4}^{\normalsize 1}} = 8 \times 625$

$\boldsymbol{\text{P} = \textsf{₹ } 5000}$

The sum of money is $\textsf{₹ } 5000$. This formula method is significantly faster for 2-year difference problems.


Example 3. A sum of money invested at Compound Interest doubles itself in 5 years. In how many years will it become 8 times itself at the same rate?

Answer:

Let the Principal be P and the annual rate be R%. Let the time be $n$ years.

Relationship for Compound Interest Amount: $A = P \left(1 + \frac{R}{100}\right)^n$.

Given that the sum doubles itself in 5 years. So, when $n=5$, $A=2P$.

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

Assuming $P \neq 0$, divide both sides by P:

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

... (iv)

Now, we want to find the number of years ($N$) it takes for the sum to become 8 times itself. So, when $n=N$, $A=8P$.

$\boldsymbol{8P = P \left(1 + \frac{R}{100}\right)^N}$

Assuming $P \neq 0$, divide both sides by P:

$\boldsymbol{8 = \left(1 + \frac{R}{100}\right)^N}$

... (v)

We need to find $N$. We can express 8 as a power of 2, since $8 = 2^3$.

Substitute the expression for 2 from equation (iv) into the equation $8 = \left(1 + \frac{R}{100}\right)^N$:

$\boldsymbol{2^3 = \left(1 + \frac{R}{100}\right)^N}$

$\boldsymbol{\left[\left(1 + \frac{R}{100}\right)^5\right]^3 = \left(1 + \frac{R}{100}\right)^N}$

Using the property of exponents $(a^m)^n = a^{mn}$:

$\boldsymbol{\left(1 + \frac{R}{100}\right)^{5 \times 3} = \left(1 + \frac{R}{100}\right)^N}$

$\boldsymbol{\left(1 + \frac{R}{100}\right)^{15} = \left(1 + \frac{R}{100}\right)^N}$

Since the bases are equal, the exponents must be equal:

$\boldsymbol{N = 15}$

It will take $\boldsymbol{15}$ years for the sum to become 8 times itself at the same rate compounded annually.

General Principle for Doubling/Tripling/etc. in CI:

If a sum of money at CI becomes $x$ times itself in $n$ years, then it will become $x^k$ times itself in $nk$ years. In this example, the sum doubled (became 2 times) in 5 years. It needed to become 8 times, and $8 = 2^3$. Here $x=2$, $n=5$, $k=3$. So the time is $n \times k = 5 \times 3 = 15$ years.


Competitive Exam Notes:

Problems combining concepts or requiring rearrangement of formulas are common. Be prepared to handle situations like differences between CI and SI, or finding time/rate for money to grow by a certain factor.

  • Rate/Time for Multiples of Principal (SI): If money becomes $m$ times in $T$ years at SI, then $(m-1)P = PRT/100 \implies (m-1)100 = RT$. E.g., doubles ($m=2$): $100=RT$. Triples ($m=3$): $200=RT$.
  • Rate/Time for Multiples of Principal (CI): If money becomes $x$ times in $n$ years at CI, then $x = (1+R/100)^n$. If it becomes $x^k$ times, the time taken is $nk$ years. This is a very quick method for such problems.
  • Difference between CI and SI: For 2 years, CI - SI $= P \left(\frac{R}{100}\right)^2$. For 3 years, CI - SI $= P \left(\frac{R}{100}\right)^2 \left(3 + \frac{R}{100}\right)$. Memorizing these formulas saves significant time compared to calculating SI and CI separately.
  • Effective Rate: If two banks offer different nominal rates but with different compounding frequencies, compare them using the Effective Annual Rate (EAR) formula $\text{EAR} = \left[\left(1 + \frac{\text{R}_{\text{nominal}}}{100k}\right)^k - 1\right] \times 100\%$.
  • Combined Problems: Some problems might ask for the time taken for the difference between CI and SI to reach a certain amount, or relate CI calculations to profit/loss (e.g., a person borrows at SI and lends at CI). Break these down into steps using the formulas.
  • Approximation: For small rates and short periods, $(1+r)^n \approx 1+nr$. For 2 years, $(1+r)^2 = 1+2r+r^2$. CI = $P[(1+r)^2-1] = P(2r+r^2)$. SI = $P(2r)$. CI - SI = $Pr^2 = P(R/100)^2$. This justifies the 2-year difference formula. Approximations for longer periods or specific rates are less common in general aptitude but might appear.