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Simple and Compound Interest
Simple Interest: Linear Growth
Simple Interest is the most straightforward way to calculate earnings on money. The interest is computed only on the original principal and never on accumulated interest, producing steady, predictable growth.
$$SI = \frac{P \times R \times T}{100}$$
Simple interest is the product of principal, rate, and time divided by 100.
$SI$=Simple interest earned (or charged)(currency)
$P$=Principal — the original amount of money deposited or borrowed(currency)
$R$=Annual [rate of interest] (in percentage)(% per year)
$T$=[Time period] for which the money is invested or borrowed(years)
$T$ is in months
→Convert to years: $T = \frac{\text{months}}{12}$
$T$ is in days
→Convert to years: $T = \frac{\text{days}}{365}$
Flat Growth: Interest each year is the same: $\frac{P \times R}{100}$. After $T$ years, total interest is simply $T$ times the yearly interest.
Total Amount: The amount at the end of $T$ years is $A = P + SI$.
Directly Proportional: SI is directly proportional to $P$, $R$, and $T$ individually — doubling any one of them doubles the interest.
The amount under simple interest combines the original principal with the total interest earned. This formula gives the final value you receive (or owe) after $T$ years.
$$A = P\left(1 + \frac{RT}{100}\right)$$
The amount grows linearly with time — it's a straight-line function of $T$.
$A$=Total amount after $T$ years (principal + interest)(currency)
$1 + \frac{RT}{100}$=Growth factor — the multiplier applied to the principal(unitless)
$T = 1$
→$A = P(1 + \frac{R}{100})$ — the one-year multiplier
Linear Relationship: Plotting $A$ vs $T$ gives a straight line with slope $\frac{PR}{100}$ and y-intercept $P$.
Finding Principal from Amount: Rearrange to $P = \frac{A \times 100}{100 + RT}$ — useful when you know the final amount and need the original investment.
Compound Interest: Exponential Growth
Compound Interest is calculated on both the original principal and any previously earned interest. This reinvestment of interest creates a snowball effect — the money grows faster each year because the base keeps getting larger.
$$A = P\left(1 + \frac{R}{n \times 100}\right)^{nT}$$
The amount after compounding depends on how frequently interest is added to the principal.
$A$=Total amount after $T$ years(currency)
$P$=Original principal(currency)
$R$=Annual [rate of interest] (percentage)(% per year)
$n$=[Compounding frequency] — number of times interest is compounded per year(per year)
$T$=Number of years(years)
$n = 1$ (annual)
→$A = P(1 + \frac{R}{100})^T$ — interest added once per year
$n = 2$ (semi-annual)
→$A = P(1 + \frac{R}{200})^{2T}$ — rate halved, periods doubled
$n = 4$ (quarterly)
→$A = P(1 + \frac{R}{400})^{4T}$ — rate quartered, periods quadrupled
$n = 12$ (monthly)
→$A = P(1 + \frac{R}{1200})^{12T}$ — rate divided by 12, periods multiplied by 12
Exponential Nature: CI follows $A = P(1 + r)^t$ where $r = \frac{R}{100}$ — this is an exponential function, not linear like SI.
Compounding Effect: The more frequently you compound, the larger the final amount — but with diminishing returns. Annual → semi-annual gives a noticeable jump; monthly → daily gives very little extra.
CI Calculation Shortcut: $CI = A - P$. Always compute $A$ first, then subtract $P$.
Greater Than SI: Compound interest is always greater than or equal to simple interest for the same $P$, $R$, and $T$ (equal only when $T \leq 1$).
CI for Non-Integer Years
When the time period includes a fraction of a year (e.g., 2.5 years), split the calculation into a whole number of compounding periods plus a simple interest portion for the remaining fraction.
$$A = P\left(1 + \frac{R}{100}\right)^t \times \left(1 + \frac{R \times f}{100}\right)$$
Compound the full years, then apply simple interest for the fractional part.
$t$=Whole number of years (integer part of $T$)(years)
$f$=Fractional part of a year ($f = T - t$)(years)
$T$ is already an integer
→This reduces to the standard compound interest formula — no fractional adjustment needed
Mixed Approach: For 3 years 4 months with annual compounding: compound for 3 full years, then add SI for $\frac{4}{12} = \frac{1}{3}$ year on the accumulated amount.
Why Not Just Use $T$ Directly?: The exponent $T$ in $(1 + R/100)^T$ only produces clean results when $T$ is an integer. For fractional exponents, this mixed approach is the conventional method.
Effective Annual Rate
The effective interest rate is the rate that, when compounded annually, would produce the same final amount as a nominal rate compounded more frequently. It reveals the true cost of borrowing or the true return on investment.
$$R_{\text{eff}} = \left(1 + \frac{R}{n \times 100}\right)^n - 1$$
Converts any compounding frequency into an equivalent annual rate for fair comparison.
$R_{\text{eff}}$=Effective annual rate (as a decimal, multiply by 100 for percentage)(decimal)
$R$=Nominal (stated) annual rate in percentage(% per year)
$n$=Number of compounding periods per year(per year)
$n = 1$
→$R_{\text{eff}} = R$ — nominal and effective rates are identical
$n \to \infty$ (continuous)
→$R_{\text{eff}} = e^{R/100} - 1$ — the maximum possible effective rate
Always Greater Than Nominal: For $n > 1$, $R_{\text{eff}} > R$. The more frequent the compounding, the higher the effective rate.
Comparison Tool: Use this to compare two investments with different compounding frequencies — the one with the higher effective rate is always better, regardless of how the nominal rate is stated.
Practical Example: 12% compounded quarterly gives $R_{\text{eff}} = (1 + 0.03)^4 - 1 = 0.1255 = 12.55\%$. You earn more than the stated 12%.
Simple vs Compound Interest: The Difference
The gap between compound interest and simple interest grows wider with time. In the first year they are equal, but from year two onward, compound interest pulls ahead because it earns interest on previously accumulated interest.
$$CI - SI = P\left(1 + \frac{R}{100}\right)^T - P\left(1 + \frac{RT}{100}\right)$$
The difference is the compound amount minus the simple amount, both starting from the same principal.
$CI - SI$=Excess of compound interest over simple interest(currency)
Year 1: No Difference: When $T = 1$, $CI = SI = \frac{PRT}{100}$. The gap only appears from year 2 onward.
Accelerating Gap: The difference grows faster each year because CI is exponential while SI is linear.
Shortcut for 2 Years: $CI - SI = P\left(\frac{R}{100}\right)^2$ — the difference equals the interest on one year's SI.
Shortcut for 3 Years: $CI - SI = P\left(\frac{R}{100}\right)^2 \times \left(3 + \frac{R}{100}\right)$.
For a two-year period, there is a clean shortcut to find the difference between compound and simple interest without computing each separately.
$$CI - SI = P\left(\frac{R}{100}\right)^2$$
The 2-year difference equals the principal multiplied by the square of the rate fraction.
$\frac{R}{100}$=Rate expressed as a decimal(decimal)
Difference and rate are given
→Find principal: $P = \frac{\text{Difference}}{(R/100)^2}$
Why It Works: In year 2, CI earns interest on year 1's interest too. The extra amount is exactly $\frac{R}{100}$ of year 1's SI = $\frac{R}{100} \times \frac{PR}{100} = P(R/100)^2$.
Inverse Problem: If $CI - SI = 80$ and $R = 10\%$, then $P = \frac{80}{0.01} = 8000$.
Installments and the Time Value of Money
An installment is a fixed periodic payment made to repay a loan or purchase over time. Because of the time value of money, each installment is worth less in present-value terms than the final amount — money received sooner is worth more than money received later.
$$\text{Present Value} = \frac{X}{\left(1 + \frac{R}{100}\right)^t}$$
Discounts a future payment back to its value today at the given interest rate.
$X$=Future payment (installment amount)(currency)
$R$=Interest rate per period(% per period)
$t$=Number of periods from now until the payment is made(periods)
Equal Installments Principle: If $n$ equal annual installments of $X$ repay a loan of $P$ at rate $R\%$, then $P = \frac{X}{(1+R/100)} + \frac{X}{(1+R/100)^2} + \dots + \frac{X}{(1+R/100)^n}$.
Each Installment's Time Value: The first installment (paid after 1 year) is discounted by 1 period, the second by 2 periods, and so on.
Total Repayment: Total money paid = $n \times X$, which is always greater than $P$ because of interest.
Application Strategy
Worked Examples
Problem Statement
A small business owner borrows Rs. 50,000 at 8% simple interest per year for 3 years. How much total interest does she pay, and what is the total amount to be repaid?
Simple InterestAmount
Strategic Toolkit
$$SI = \frac{P \times R \times T}{100}, \quad A = P + SI$$
Step 1
Identify the given values
Principal $P = 50{,}000$, Rate $R = 8\%$, Time $T = 3$ years.
Step 2
Calculate the simple interest
Substitute into the SI formula.
$$SI = \frac{50{,}000 \times 8 \times 3}{100} = \frac{1{,}200{,}000}{100} = 12{,}000$$
Step 3
Calculate the total amount
Add the interest to the principal.
$$A = P + SI = 50{,}000 + 12{,}000 = 62{,}000$$
Step 4
State the final answer
The business owner pays Rs. 12,000 in interest and repays a total of Rs. 62,000 after 3 years.
Problem Statement
Ahmad deposits Rs. 20,000 in a savings account that pays 10% interest compounded annually. What is the balance after 2 years, and how much interest did he earn?
Compound InterestAmount
Strategic Toolkit
$$A = P\left(1 + \frac{R}{100}\right)^T$$
Step 1
Identify the given values
Principal $P = 20{,}000$, Rate $R = 10\%$, Time $T = 2$ years, $n = 1$ (annual).
Step 2
Apply the compound interest formula
Substitute into the annual compounding formula.
$$A = 20{,}000 \times \left(1 + \frac{10}{100}\right)^2 = 20{,}000 \times (1.1)^2$$
Step 3
Compute the result
Evaluate $(1.1)^2$ and multiply.
$$A = 20{,}000 \times 1.21 = 24{,}200$$
Step 4
Find the compound interest earned
Subtract the principal from the amount.
$$CI = A - P = 24{,}200 - 20{,}000 = 4{,}200$$
Step 5
Verify against simple interest
SI for 2 years would be $\frac{20{,}000 \times 10 \times 2}{100} = 4{,}000$. Compound interest gives Rs. 200 more — exactly $P(R/100)^2 = 20{,}000 \times 0.01 = 200$.
Problem Statement
A bank offers 12% interest compounded quarterly on a deposit of Rs. 15,000 for 1 year. Find the amount at maturity and the effective annual rate.
Compound InterestCompounding FrequencyEffective Interest Rate
Strategic Toolkit
$$A = P\left(1 + \frac{R}{400}\right)^{4T}, \quad R_{\text{eff}} = \left(1 + \frac{R}{400}\right)^4 - 1$$
Step 1
Set up for quarterly compounding
Quarterly means $n = 4$. Rate per quarter = $\frac{12}{4} = 3\%$. Number of quarters = $4 \times 1 = 4$.
Step 2
Calculate the amount
Apply the formula with $n = 4$.
$$A = 15{,}000 \times \left(1 + \frac{12}{400}\right)^4 = 15{,}000 \times (1.03)^4$$
Step 3
Evaluate
$(1.03)^4 = 1.12550881$.
$$A = 15{,}000 \times 1.1255 \approx 16{,}882.63$$
Step 4
Calculate the effective annual rate
Convert the quarterly rate to an equivalent annual rate.
$$R_{\text{eff}} = (1.03)^4 - 1 = 1.1255 - 1 = 0.1255 = 12.55\%$$
Step 5
Interpret
Although the stated rate is 12%, the quarterly compounding makes the effective return 12.55%. The investor earns Rs. 1,882.63 in interest.
Problem Statement
The difference between compound interest and simple interest on a sum at 10% per annum for 2 years is Rs. 250. Find the original sum (principal).
Simple InterestCompound Interest
Strategic Toolkit
$$CI - SI = P\left(\frac{R}{100}\right)^2$$
Step 1
Identify the given values
Difference = Rs. 250, $R = 10\%$, $T = 2$ years.
Step 2
Use the 2-year difference formula
For 2 years, $CI - SI = P(R/100)^2$.
$$250 = P\left(\frac{10}{100}\right)^2 = P(0.1)^2 = 0.01P$$
Step 3
Solve for P
Rearrange to isolate the principal.
$$P = \frac{250}{0.01} = 25{,}000$$
Step 4
Verify
SI = $\frac{25{,}000 \times 10 \times 2}{100} = 5{,}000$. CI = $25{,}000 \times (1.1)^2 - 25{,}000 = 30{,}250 - 25{,}000 = 5{,}250$. Difference = $5{,}250 - 5{,}000 = 250$. ✓
Problem Statement
A laptop costs Rs. 40,000 cash. A store offers it on equal annual installments at 10% simple interest. If Zara pays Rs. 15,200 per year for 3 years, verify that this fairly repays the cost.
InstallmentsTime Value of MoneySimple Interest
Strategic Toolkit
$$P = \frac{X}{(1+R/100)} + \frac{X}{(1+R/100)^2} + \frac{X}{(1+R/100)^3}$$
Step 1
Understand the setup
Each installment of Rs. 15,200 is paid at the end of each year. We need to find the present value (PV) of all three installments and check if it equals Rs. 40,000.
Step 2
Discount each installment to present value
At 10%, the discount factor for year $t$ is $\frac{1}{(1.1)^t}$.
$$\text{PV} = \frac{15{,}200}{1.1} + \frac{15{,}200}{(1.1)^2} + \frac{15{,}200}{(1.1)^3}$$
Step 3
Evaluate each term
Compute each discounted payment.
$$\frac{15{,}200}{1.1} = 13{,}818.18, \quad \frac{15{,}200}{1.21} = 12{,}561.98, \quad \frac{15{,}200}{1.331} = 11{,}419.99$$
Step 4
Sum the present values
Add all three discounted amounts.
$$\text{PV} = 13{,}818.18 + 12{,}561.98 + 11{,}419.99 \approx 37{,}800$$
Step 5
Interpret
The present value of three installments of Rs. 15,200 each is approximately Rs. 37,800, which is less than the Rs. 40,000 cash price. This means Zara effectively pays less in present-value terms — a reasonable installment deal.
Memorize formulas
Simple Interest
Use when interest is only charged on the original principal — flat, linear growth.
Formula
$SI = \dfrac{P \times R \times T}{100}$
Amount under Simple Interest
Use to find the total repayment or maturity value under simple interest.
Formula
$A = P\left(1 + \dfrac{RT}{100}\right)$
Compound Interest (Annual)
Use when interest is reinvested annually — exponential growth.
Formula
$A = P\left(1 + \dfrac{R}{100}\right)^T$
Compound Interest (General)
Use for any compounding frequency — semi-annual, quarterly, monthly.
Formula
$A = P\left(1 + \dfrac{R}{n \times 100}\right)^{nT}$
Effective Annual Rate
Use to compare rates with different compounding frequencies.
Formula
$R_{\text{eff}} = \left(1 + \dfrac{R}{n \times 100}\right)^n - 1$
Review
1
A small business borrows Rs. 50,000 at 8% per annum simple interest for 3 years. What is the total interest paid over the loan period?
Ans
Rs. 12,000
Review
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Apply the simple interest formula
Substitute the principal, rate, and time directly into the formula.
$$SI = \dfrac{50000 \times 8 \times 3}{100} = 12000$$
2
An investment of Rs. 10,000 earns 10% per annum compound interest for 2 years. What is the total amount at maturity?
Ans
Rs. 12,100
Review
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Apply the compound interest formula
Compute the amount by raising the growth factor to the power of the time period.
$$A = 10000\left(1 + \dfrac{10}{100}\right)^2 = 10000(1.1)^2$$
Calculate the final amount
Evaluate $(1.1)^2 = 1.21$ and multiply by the principal.
$$A = 10000 \times 1.21 = 12100$$
3
A sum of Rs. 25,000 amounts to Rs. 35,000 in 4 years at simple interest. What is the annual rate of interest?
Ans
10%
Review
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Find the simple interest earned
Subtract the principal from the maturity amount to get the interest.
$$SI = 35000 - 25000 = 10000$$
Solve for the rate
Rearrange the SI formula to isolate $R$.
$$R = \dfrac{10000 \times 100}{25000 \times 4} = 10\%$$
4
A sum of money doubles itself in 6 years at compound interest. How many years will it take for the same sum to become 8 times its original value?
Ans
18 years
Review
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Identify the doubling time
The money doubles every 6 years, meaning the amount multiplies by 2 for each 6-year interval.
$$A = P \times 2^{T/6}$$
Set the target and solve for time
Set $P \times 2^{T/6} = 8P$ and solve for $T$.
$$2^{T/6} = 8 = 2^3 \implies T/6 = 3 \implies T = 18 \text{ years}$$
5
Rs. 20,000 is invested at 20% per annum compound interest for 2.5 years. If the fractional year is computed using simple interest on the compounded amount, what is the total amount?
Ans
Rs. 31,680
Review
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Compound for the full 2 years
Apply the compound interest formula for the integer part of the time period.
$$P(1.2)^2 = 20000 \times 1.44 = 28800$$
Apply simple interest for the fractional 0.5 year
Multiply the compounded amount by $(1 + f \times R/100)$ for the remaining half year.
$$28800 \times (1 + 0.5 \times 0.2) = 28800 \times 1.1 = 31680$$
6
A bank offers 12% interest per annum compounded quarterly. What is the effective annual rate (rounded to two decimal places)?
Ans
12.55%
Review
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Identify the compounding parameters
Quarterly compounding means $n = 4$ periods per year, with each period earning $R/4 = 3\%$.
$$\dfrac{R}{n \times 100} = \dfrac{12}{4 \times 100} = 0.03$$
Apply the effective annual rate formula
Raise the per-period growth factor to the number of periods and subtract 1.
$$R_{\text{eff}} = (1.03)^4 - 1 = 1.125509 - 1 \approx 12.55\%$$
7
Rs. 8,000 is invested at 5% per annum for 3 years. How much more does compound interest yield compared to simple interest?
Ans
Rs. 61
Review
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Calculate simple interest
Use the SI formula with the given values.
$$SI = \dfrac{8000 \times 5 \times 3}{100} = 1200$$
Calculate compound interest
Use the CI formula to find the maturity amount, then subtract the principal.
$$A = 8000(1.05)^3 = 9261 \implies CI = 1261$$
Find the difference
Subtract the simple interest from the compound interest.
$$CI - SI = 1261 - 1200 = 61$$
8
The difference between compound interest and simple interest on a certain sum at 10% per annum for 2 years is Rs. 250. What is the principal?
Ans
Rs. 25,000
Review
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Use the CI–SI difference formula for 2 years
For 2 years, the difference between compound and simple interest equals $P(R/100)^2$.
$$250 = P\left(\dfrac{10}{100}\right)^2 = P(0.01)$$
Solve for the principal
Divide both sides by 0.01 to isolate the principal.
$$P = \dfrac{250}{0.01} = 25000$$
9
A laptop costs Rs. 24,000 cash, or Rs. 6,000 down payment plus 3 equal monthly installments. If simple interest at 20% per annum is charged on the remaining balance, what is each monthly installment?
Ans
Rs. 6,300
Review
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Calculate the balance after down payment
Subtract the down payment from the total price.
$$\text{Balance} = 24000 - 6000 = 18000$$
Compute the simple interest on the balance
Apply SI formula with $T = 3/12 = 0.25$ years at 20%.
$$SI = \dfrac{18000 \times 20 \times 0.25}{100} = 900$$
Divide the total repayment into equal installments
Add the interest to the balance and split across 3 equal payments.
$$\text{Each installment} = \dfrac{18000 + 900}{3} = 6300$$
10
Bank A offers 8% per annum compounded quarterly. Bank B offers 8.1% per annum compounded semi-annually. Which bank gives a higher effective annual rate, and by approximately how much?
Ans
Bank B by about 0.02%
Review
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Calculate Bank A's effective rate
Quarterly compounding at 8% with $n = 4$.
$$R_{\text{eff}, A} = (1.02)^4 - 1 \approx 8.24\%$$
Calculate Bank B's effective rate
Semi-annual compounding at 8.1% with $n = 2$.
$$R_{\text{eff}, B} = (1.0405)^2 - 1 \approx 8.26\%$$
Compare the two rates
Subtract the smaller effective rate from the larger one.
$$\text{Difference} \approx 8.26\% - 8.24\% = 0.02\%$$
11
What is the compound interest earned on Rs. 15,000 at 16% per annum compounded semi-annually for 1 year?
Ans
Rs. 2,496
Review
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Identify the per-period rate
Semi-annual compounding means $n = 2$, so each period earns $16/2 = 8\%$.
$$\dfrac{R}{n \times 100} = \dfrac{16}{2 \times 100} = 0.08$$
Calculate the maturity amount
Raise the per-period factor to the total number of compounding periods ($nT = 2$).
$$A = 15000(1.08)^2 = 15000 \times 1.1664 = 17496$$
Subtract the principal to get CI
The compound interest is the amount minus the original principal.
$$CI = 17496 - 15000 = 2496$$
12
A vendor is offered Rs. 10,000 cash now or Rs. 13,000 after 2 years. If the vendor can invest money at 10% compound interest per annum, which option is financially better and by how much?
Ans
Option 2 (later) is better by Rs. 900
Review
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Find the future value of the immediate payment
Invest Rs. 10,000 at 10% compound interest for 2 years to compare both options at the same point in time.
$$A = 10000(1.1)^2 = 12100$$
Compare with the deferred payment
Subtract the future value of Option 1 from the Option 2 amount to determine which is better.
$$13000 - 12100 = 900 \text{ in favor of Option 2}$$
Question 1 of 12
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1
A small business borrows Rs. 50,000 at 8% per annum simple interest for 3 years. What is the total interest paid over the loan period?