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Loans: Financial Concepts Equated Monthly Installment (EMI): Definition and Components Calculation of EMI using Formula
Amortization Schedule (Implicit) Problems based on EMI Calculation


Loans and Equated Monthly Installments (EMI)



Loans: Financial Concepts

Definition of a Loan

A loan is a financial arrangement where funds are temporarily transferred from one party (the lender) to another (the borrower). The borrower receives a specific amount of money, known as the principal, and agrees to repay this principal amount over a set period (the tenure). In return for the use of the money, the borrower also pays an additional amount, called interest, to the lender. The terms of this agreement, including the principal amount, interest rate, repayment frequency, and tenure, are legally documented.

Loans are essential tools for managing finances, enabling large purchases (like property or vehicles), funding business operations, or covering unexpected expenses. They are a fundamental form of debt.


Key Financial Concepts in Loans

Several key concepts are central to understanding how loans are structured and managed:

The calculation and structure of many loans, particularly those with equal periodic payments (like EMIs), are directly based on the principles of the Time Value of Money, specifically the concept of the Present Value of an Annuity.

Summary for Competitive Exams

Loan: Borrower gets Principal from Lender and repays it with Interest over a specified Tenure via a Repayment Schedule.

Key Terms:

  • Principal (P): Initial amount borrowed.
  • Interest Rate (r% p.a.): Cost of borrowing (Fixed or Floating).
  • Tenure (t): Loan duration (years).
  • Installment: Regular payment covering Principal + Interest.
  • Amortisation: Process of gradually repaying principal.
  • Secured/Unsecured: Based on presence of Collateral.

Equated Monthly Installments (EMIs) are common loan payments structured as Ordinary Annuities.


Equated Monthly Installment (EMI): Definition and Components

Definition of EMI

An Equated Monthly Installment (EMI) is a fixed payment amount that a borrower pays to the lender every month on a predetermined date. EMIs are the standard method of repayment for most retail term loans, such as home loans, car loans, personal loans, and consumer durable loans.

The term 'Equated' means that the installment amount is constant throughout the loan's entire tenure, providing budgeting predictability for the borrower. 'Monthly' specifies the payment frequency, although the concept can be adapted for other frequencies. The EMI amount is calculated to cover both the interest due for the period and a portion of the principal amount, ensuring that the loan is fully repaid by the end of the agreed tenure.


Components of an EMI

Each EMI payment is composed of two distinct parts:

  1. Interest Component: This portion of the EMI covers the interest that has accrued on the outstanding principal balance of the loan during the specific payment period (usually the preceding month).

  2. Principal Component (or Principal Repayment): This portion of the EMI is applied directly towards reducing the original principal amount of the loan. With each EMI, the outstanding principal balance decreases.

Dynamic Change in EMI Components Over Time

Although the total EMI amount remains constant for every payment, the proportion of interest and principal within each EMI changes significantly over the loan's life. This is a crucial aspect of how amortizing loans work:

This shifting balance ensures that the loan is systematically paid off. By the final EMI, the principal component covers the last remaining bit of the principal, and the interest component covers the interest on that amount, bringing the loan balance to zero.

This breakdown is detailed in an amortisation schedule, a table that shows for each EMI, how much goes towards interest, how much towards principal, and the remaining outstanding balance.

Stacked bar chart showing the composition of a fixed EMI over time. Lower part (e.g., red) represents interest, upper part (e.g., blue) represents principal. The interest part shrinks over time, and the principal part grows, while the total bar height (EMI) remains constant.

The visual representation shows how the interest portion dominates early EMIs, while the principal portion dominates later EMIs, despite the fixed total payment amount.

EMI as an Ordinary Annuity: The Foundation for Calculation

From the standpoint of financial mathematics and calculation, a series of EMIs perfectly aligns with the definition of an ordinary annuity. This is because:

The initial loan amount received by the borrower today is financially equivalent to the sum of the present values of all the future EMI payments, discounted at the monthly interest rate. Therefore, the loan principal amount is the Present Value (PV) of the EMI annuity stream.

Summary for Competitive Exams

EMI: Fixed monthly payment to repay a loan.

Composition: Each EMI = Principal Repayment + Interest Payment.

Shift in Composition: Early EMIs are mostly Interest; Later EMIs are mostly Principal (as outstanding balance decreases). Total EMI is constant.

Financial Model: EMI series is an Ordinary Annuity.

PV Relationship: Loan Principal (P) = Present Value (PV) of the EMI annuity stream, discounted at the monthly rate.

The Amortisation Schedule illustrates the component breakdown of each EMI.


Calculation of EMI using Formula

Concept: Leveraging the Present Value of an Ordinary Annuity

As established, the stream of Equated Monthly Installments (EMIs) for an amortizing loan functions as an ordinary annuity. The fundamental relationship is that the amount of money received by the borrower today (the Loan Principal, $P$) is equivalent to the value today (the Present Value) of the future stream of EMI payments.

Therefore, we can use the formula for the Present Value of an Ordinary Annuity, substituting the loan principal for the PV and the EMI for the regular payment ($R$), and then rearrange the formula to solve for the EMI.

Formula Derivation

Let:

First, calculate the periodic interest rate ($i$) and the total number of periods ($n$) that match the EMI frequency (monthly):

Now, we use the formula for the Present Value (PV) of an Ordinary Annuity:

$PV = R \left[ \frac{1 - (1+i)^{-n}}{i} \right]$

Substitute $P$ for $PV$ and $EMI$ for $R$:

$P = EMI \left[ \frac{1 - (1+i)^{-n}}{i} \right]$

We need to find the EMI. Rearrange the formula by dividing both sides by the term in the square brackets:

$\frac{P}{\left[ \frac{1 - (1+i)^{-n}}{i} \right]} = EMI$

Multiply by the reciprocal of the fractional term to isolate EMI:

$EMI = P \times \frac{1}{\left[ \frac{1 - (1+i)^{-n}}{i} \right]}$

$EMI = P \times \left[ \frac{i}{1 - (1+i)^{-n}} \right]$

This is one common form of the EMI calculation formula.

An equivalent form is also widely used. Recall that $(1+i)^{-n} = \frac{1}{(1+i)^n}$. Substitute this into the denominator:

$EMI = P \left[ \frac{i}{1 - \frac{1}{(1+i)^n}} \right]$

Find a common denominator within the denominator:

$1 - \frac{1}{(1+i)^n} = \frac{(1+i)^n - 1}{(1+i)^n}$

Substitute this back into the EMI formula:

$EMI = P \left[ \frac{i}{\frac{(1+i)^n - 1}{(1+i)^n}} \right]$

To perform the division, multiply by the reciprocal of the denominator:

$EMI = P \left[ i \times \frac{(1+i)^n}{(1+i)^n - 1} \right]$

This gives the other common form of the EMI formula:

$EMI = P \times i \times \frac{(1 + i)^n}{(1 + i)^n - 1}$

Formula for Equated Monthly Installment (EMI)

The formula to calculate the fixed monthly payment (EMI) for an amortizing loan is:

$\mathbf{EMI = P \left[ \frac{i}{1 - (1+i)^{-n}} \right]}$

or equivalently,

$\mathbf{EMI = P \times i \times \frac{(1 + i)^n}{(1 + i)^n - 1}}$

Where:

This formula calculates the constant monthly payment required to fully repay the loan principal and all accrued interest over the specified tenure. Using this formula requires calculating $(1+i)^n$ or $(1+i)^{-n}$, which usually involves a financial calculator, spreadsheet software, or logarithm tables for large $n$ values.

Worked Example

Example 1. Calculate the Equated Monthly Installment (EMI) for a personal loan of $\textsf{₹}\$ 5,00,000$ taken for a tenure of 5 years at an interest rate of 11% per annum.

Answer:

Given:

  • Principal Loan Amount (P) = $\textsf{₹}\$ 5,00,000$.
  • Nominal Annual Interest Rate (R) = 11% per annum. Convert to decimal: $r = \frac{11}{100} = 0.11$.
  • Loan Tenure (t) = 5 years.
  • Payment Frequency: Monthly.

To Find:

  • Equated Monthly Installment (EMI).

Calculate Periodic (Monthly) Rate (i) and Total Number of Installments (n):

  • Periodic monthly rate $i = \frac{r}{12} = \frac{0.11}{12}$. (This is the decimal monthly rate. We can keep it as a fraction or use its decimal approximation $0.0091666...$)
  • Total number of installments $n = t \times 12 = 5 \times 12 = 60$. (60 months).

Formula:

We will use the formula $EMI = P \left[ \frac{i}{1 - (1+i)^{-n}} \right]$.

Solution:

Substitute the values of P, i, and n into the formula:

$EMI = 500000 \left[ \frac{0.11/12}{1 - (1+0.11/12)^{-60}} \right]$

Calculate $(1+0.11/12)^{-60} = (1 + 0.0091666... )^{-60}$. Using a financial calculator or spreadsheet function with the exact fraction 0.11/12:

$(1 + 0.11/12)^{-60} \approx (1.00916667)^{-60} \approx 0.576691$

So,

$EMI \approx 500000 \left[ \frac{0.11/12}{1 - 0.576691} \right]$

$EMI \approx 500000 \left[ \frac{0.00916667}{0.423309} \right]$

Calculate the value inside the brackets:

$\frac{0.00916667}{0.423309} \approx 0.02165455$

Now calculate EMI:

$EMI = 500000 \times 0.02165455$

$EMI \approx 10827.275$

Rounding to the nearest paisa, the Equated Monthly Installment (EMI) is approximately $\textsf{₹}\$ 10,827.28$.

This means the borrower will pay $\textsf{₹}\$ 10,827.28$ each month for 60 months to repay the $\textsf{₹}\$ 5,00,000$ loan with interest at 11% p.a.

Summary for Competitive Exams

EMI Calculation: Finding the fixed periodic payment amount.

Concept: Loan Principal (P) = PV of EMI stream (Ordinary Annuity).

Formula: $\mathbf{EMI = P \left[ \frac{i}{1 - (1+i)^{-n}} \right]}$

Or $\mathbf{EMI = P \times i \times \frac{(1 + i)^n}{(1 + i)^n - 1}}$

  • P: Principal Loan Amount.
  • i: Monthly Interest Rate (decimal) = Annual Rate (decimal) / 12.
  • n: Total Number of Months = Tenure in Years $\times$ 12.

Key: Use the monthly rate for $i$ and total number of months for $n$. This is a direct application of the PV of Ordinary Annuity formula solved for the payment amount (R).



Amortization Schedule (Implicit)

Definition of an Amortization Schedule

An Amortization Schedule (also commonly referred to as a Loan Amortization Table or merely an Amortization Table) is a tabular representation that provides a detailed breakdown of every single periodic payment made over the entire tenure of an amortizing loan. For each installment payment (such as an EMI), the schedule clearly shows three key pieces of information:

  1. The portion of the payment that is applied towards covering the interest accrued on the outstanding loan balance since the previous payment.
  2. The portion of the payment that is applied towards reducing the original principal amount borrowed.
  3. The remaining outstanding principal balance of the loan after the payment has been successfully made and processed.

This schedule serves as a roadmap for the loan repayment process, explicitly demonstrating how a loan's principal amount is gradually reduced (amortized) over time through regular, often fixed, payments that combine both interest and principal components.


Purpose and Significance

Amortization schedules hold significant importance for both borrowers and lenders due to the multiple benefits they provide:

In essence, the amortization schedule is a vital document that clarifies the mechanics of loan repayment and the true financial burden over the loan's life.


Structure and Implicit Calculation Process

An amortization schedule is typically presented in a tabular format. Each row in the table corresponds to one periodic payment (e.g., one monthly EMI). The columns detail the key figures and calculations for that period. Assuming a standard fixed-rate loan with constant EMI payments and interest compounded at the same frequency as payments (e.g., monthly compounding for monthly EMIs), the structure is as follows:

Let $P$ be the original Loan Principal, $EMI$ be the fixed periodic payment amount, $i$ be the periodic interest rate (e.g., monthly rate), and $n$ be the total number of periods (payments).

  1. Payment Number (Period): This column simply lists the sequential number of each payment, starting from 1 and continuing up to the total number of payments, $n$.

  2. Beginning Balance: This is the outstanding principal balance of the loan at the very start of the current payment period. For the first payment (Period 1), this is equal to the original Loan Principal ($P$). For any subsequent payment (Period $k > 1$), the Beginning Balance is the same as the Ending Balance from the immediately preceding period (Period $k-1$).

  3. Scheduled Payment (EMI): This column lists the fixed, constant amount of the regular installment payment calculated using the EMI formula. This amount typically remains the same for every payment throughout the loan's tenure, except for potentially a minor adjustment in the final payment due to cumulative rounding over many periods.

  4. Interest Paid: This is the amount of interest that has accrued on the outstanding principal balance during the current period. It is calculated by multiplying the Beginning Balance of the current period by the periodic interest rate ($i$). It is crucial that the periodic rate $i$ matches the frequency of the payment period (e.g., if payments are monthly, $i$ must be the monthly interest rate, obtained by dividing the nominal annual rate by 12).

    $Interest\$ Paid_{Period\_k} = Beginning\$ Balance_{Period\_k} \times Periodic\$ Interest\$ Rate (i)$

  5. Principal Repaid: This column shows the portion of the Scheduled Payment (EMI) that is actually applied towards reducing the outstanding principal amount of the loan. It is calculated by subtracting the calculated Interest Paid for the period from the fixed Scheduled Payment amount.

    $Principal\$ Repaid_{Period\_k} = Scheduled\$ Payment (EMI) - Interest\$ Paid_{Period\_k}$

    Since the EMI is constant and the Interest Paid component changes, the Principal Repaid component within each EMI also changes with each payment.

  6. Ending Balance: This is the remaining outstanding principal balance of the loan after the current payment has been made and the principal repayment portion has been applied. It is calculated by subtracting the Principal Repaid amount for the period from the Beginning Balance of that same period.

    $Ending\$ Balance_{Period\_k} = Beginning\$ Balance_{Period\_k} - Principal\$ Repaid_{Period\_k}$

This calculation process is carried out sequentially for each payment period. The Ending Balance of one period becomes the Beginning Balance for the next period. The ultimate goal is that the Ending Balance after the very last scheduled payment (Payment Number $n$) should be exactly zero, signifying that the loan has been fully repaid. In practice, due to minor rounding during calculations, the final balance might be a very small positive or negative value, requiring a slight adjustment to the final payment.

Key Observations Highlighted by the Amortization Schedule

Examining the values calculated in the amortization schedule provides critical insights into how loans are repaid:

This confirms the concept that early EMIs disproportionately cover interest, while later EMIs are more effective at reducing the principal balance. The amortization schedule makes this dynamic clearly visible.

Implicit Calculation of Outstanding Balance at Any Point

While the amortization schedule provides a period-by-period tracking of the balance, it also implicitly demonstrates how to calculate the outstanding loan balance at any intermediate point in time without having to build the entire table up to that point. The outstanding loan balance after $k$ payments have been made is equal to the Present Value of all the *remaining* future EMI payments (specifically, payments from $k+1$ up to $n$), discounted back to the time of the $k$-th payment using the periodic interest rate $i$.

Outstanding Balance After $k$ payments $= EMI \times \left[ \frac{1 - (1+i)^{-(n-k)}}{i} \right]$

Where $(n-k)$ represents the number of payments remaining after the $k$-th payment.

This formula will yield the exact same outstanding balance as calculated iteratively in the amortization schedule for the end of period $k$.

Understanding the step-by-step calculation within the amortization schedule, even if not performing it manually for every single payment of a long loan, provides a fundamental understanding of how loan principal is reduced and the mechanism of loan repayment over time.

Example 1. A loan of $\textsf{₹}\$ 5,00,000$ is taken at 10% per annum compounded monthly for 2 years. The calculated EMI is $\textsf{₹}\$ 22,915.99$. Show the amortization schedule for the first 4 months.

Answer:

Given:

  • Principal (P) = $\textsf{₹}\$ 5,00,000$.
  • Nominal Annual Rate (r) = 10% = 0.10.
  • Monthly Rate (i) = $0.10 / 12 \approx 0.00833333$.
  • Tenure (t) = 2 years = 24 months.
  • Total Payments (n) = 24.
  • EMI = $\textsf{₹}\$ 22,915.99$.

To Show:

  • Amortization Schedule for the first 4 months.

Calculation Steps for each period (using $i \approx 0.00833333$ for calculation clarity, results might slightly vary with exact fraction):

Period 1 (Month 1):

  • Beginning Balance = $\textsf{₹}\$ 5,00,000.00$
  • Scheduled Payment = $\textsf{₹}\$ 22,915.99$
  • Interest Paid = Beginning Balance $\times$ i = $500000.00 \times (0.10/12) \approx \textsf{₹}\$ 4,166.67$
  • Principal Repaid = Scheduled Payment - Interest Paid = $22915.99 - 4166.67 = \textsf{₹}\$ 18,749.32$
  • Ending Balance = Beginning Balance - Principal Repaid = $500000.00 - 18749.32 = \textsf{₹}\$ 4,81,250.68$

Period 2 (Month 2):

  • Beginning Balance = $\textsf{₹}\$ 4,81,250.68$
  • Scheduled Payment = $\textsf{₹}\$ 22,915.99$
  • Interest Paid = Beginning Balance $\times$ i = $481250.68 \times (0.10/12) \approx \textsf{₹}\$ 4,010.42$
  • Principal Repaid = Scheduled Payment - Interest Paid = $22915.99 - 4010.42 = \textsf{₹}\$ 18,905.57$
  • Ending Balance = Beginning Balance - Principal Repaid = $481250.68 - 18905.57 = \textsf{₹}\$ 4,62,345.11$

Period 3 (Month 3):

  • Beginning Balance = $\textsf{₹}\$ 4,62,345.11$
  • Scheduled Payment = $\textsf{₹}\$ 22,915.99$
  • Interest Paid = Beginning Balance $\times$ i = $462345.11 \times (0.10/12) \approx \textsf{₹}\$ 3,852.88$
  • Principal Repaid = Scheduled Payment - Interest Paid = $22915.99 - 3852.88 = \textsf{₹}\$ 19,063.11$
  • Ending Balance = Beginning Balance - Principal Repaid = $462345.11 - 19063.11 = \textsf{₹}\$ 4,43,282.00$

Period 4 (Month 4):

  • Beginning Balance = $\textsf{₹}\$ 4,43,282.00$
  • Scheduled Payment = $\textsf{₹}\$ 22,915.99$
  • Interest Paid = Beginning Balance $\times$ i = $443282.00 \times (0.10/12) \approx \textsf{₹}\$ 3,694.02$
  • Principal Repaid = Scheduled Payment - Interest Paid = $22915.99 - 3694.02 = \textsf{₹}\$ 19,221.97$
  • Ending Balance = Beginning Balance - Principal Repaid = $443282.00 - 19221.97 = \textsf{₹}\$ 4,24,060.03$

Amortization Schedule (First 4 Months):

Payment No. Beginning Balance ($\textsf{₹}$) Scheduled Payment ($\textsf{₹}$) Interest Paid ($\textsf{₹}$) Principal Repaid ($\textsf{₹}$) Ending Balance ($\textsf{₹}$)
1 5,00,000.00 22,915.99 4,166.67 18,749.32 4,81,250.68
2 4,81,250.68 22,915.99 4,010.42 18,905.57 4,62,345.11
3 4,62,345.11 22,915.99 3,852.88 19,063.11 4,43,282.00
4 4,43,282.00 22,915.99 3,694.02 19,221.97 4,24,060.03

The table clearly shows the pattern: the Interest Paid column decreases with each payment, and the Principal Repaid column increases, while the Scheduled Payment (EMI) remains constant. The sum of Interest Paid and Principal Repaid in each row equals the EMI $\textsf{₹}\$ 22,915.99$ (allowing for minor rounding). The Ending Balance becomes the next period's Beginning Balance, progressively reducing the loan amount.

Summary for Competitive Exams

Amortization Schedule: Table detailing how each EMI payment is split between Interest and Principal and tracks the remaining balance.

Calculation Per Period: Interest = Beginning Balance $\times$ Periodic Rate (i); Principal Repaid = EMI - Interest Paid; Ending Balance = Beginning Balance - Principal Repaid.

Key Observation: Interest paid decreases over time; Principal repaid increases over time within the fixed EMI.

Outstanding Balance after k payments: Equivalent to the PV of the remaining $(n-k)$ EMIs.

Understanding the amortization process is vital for grasping how loans are paid off and the true cost of borrowing over time.


Problems based on EMI Calculation

Problems involving EMI calculations are common applications of the Present Value of an Ordinary Annuity formula. These problems typically involve calculating the EMI itself, or working backward to find the principal loan amount, the interest rate, or sometimes the tenure, given other parameters. The core relationship is that the loan principal today is the present value of the stream of future EMI payments.

The primary formula used is $\mathbf{EMI = P \left[ \frac{i}{1 - (1+i)^{-n}} \right]}$ or its rearrangement $\mathbf{P = EMI \left[ \frac{1 - (1+i)^{-n}}{i} \right]}$. Remember that $i$ is the periodic (monthly) interest rate and $n$ is the total number of payments (months).

Worked Examples

Example 1. A person takes a car loan of $\textsf{₹}\$ 6,00,000$ at an annual interest rate of 10% for a tenure of 5 years. Calculate the EMI.

Answer:

Given:

  • Principal Loan Amount (P) = $\textsf{₹}\$ 6,00,000$.
  • Nominal Annual Interest Rate (R) = 10% per annum. Convert to decimal: $r = \frac{10}{100} = 0.10$.
  • Loan Tenure (t) = 5 years.
  • Payment Frequency: Monthly (implied by EMI).

To Find:

  • Equated Monthly Installment (EMI).

Calculate Periodic (Monthly) Rate (i) and Total Number of Installments (n):

  • Periodic monthly rate $i = \frac{r}{12} = \frac{0.10}{12}$. We will use the fractional form for calculation.
  • Total number of installments $n = t \times 12 = 5 \times 12 = 60$. (60 months).

Formula:

The formula to calculate the Equated Monthly Installment (EMI) is:

$EMI = P \left[ \frac{i}{1 - (1+i)^{-n}} \right]$

Solution:

Substitute the values of P, i, and n into the formula:

$EMI = 600000 \left[ \frac{0.10/12}{1 - (1+0.10/12)^{-60}} \right]$

Calculate $(1+0.10/12)^{-60}$. Using a financial calculator or spreadsheet function with the exact fraction 0.10/12:

$(1 + 0.10/12)^{-60} \approx (1.00833333)^{-60} \approx 0.60751315$

So,

$EMI \approx 600000 \left[ \frac{0.10/12}{1 - 0.60751315} \right]$

$EMI \approx 600000 \left[ \frac{0.00833333}{0.39248685} \right]$

Calculate the value inside the brackets:

$\frac{0.00833333}{0.39248685} \approx 0.02122377$

Now calculate EMI:

$EMI = 600000 \times 0.02122377$

$EMI \approx 12734.262$

Rounding to two decimal places, the Equated Monthly Installment (EMI) is approximately $\textsf{₹}\$ 12,734.26$.

This means the person will pay $\textsf{₹}\$ 12,734.26$ each month for 60 months to repay the $\textsf{₹}\$ 6,00,000$ loan with interest at 10% p.a.


Example 2. A bank offers a personal loan with an EMI of $\textsf{₹}\$ 15,000$ per month for a tenure of 3 years. If the interest rate is 12% per annum compounded monthly, what is the maximum principal loan amount the bank is offering?

Answer:

Given:

  • Equated Monthly Installment (EMI) = $\textsf{₹}\$ 15,000$.
  • Loan Tenure (t) = 3 years.
  • Nominal Annual Interest Rate (R) = 12% per annum. Convert to decimal: $r = \frac{12}{100} = 0.12$.
  • Payment Frequency: Monthly. Compounding: Monthly ($m=12$).

To Find:

  • Maximum Principal Loan Amount (P). (This is the Present Value of the EMI stream).

Calculate Periodic (Monthly) Rate (i) and Total Number of Installments (n):

  • Periodic monthly rate $i = \frac{r}{12} = \frac{0.12}{12} = 0.01$. (This is 1% per month).
  • Total number of installments $n = t \times 12 = 3 \times 12 = 36$. (36 months).

Formula:

We use the formula relating Principal (PV), EMI (R), i, and n:

$P = EMI \left[ \frac{1 - (1+i)^{-n}}{i} \right]$

Solution:

Substitute the given values into the formula:

$P = 15000 \left[ \frac{1 - (1+0.01)^{-36}}{0.01} \right]$

$P = 15000 \left[ \frac{1 - (1.01)^{-36}}{0.01} \right]$

Calculate $(1.01)^{-36} = \frac{1}{(1.01)^{36}}$. Using a financial calculator or spreadsheet function:

$(1.01)^{36} \approx 1.43076876

$(1.01)^{-36} \approx \frac{1}{1.43076876} \approx 0.69892473

So,

$P = 15000 \left[ \frac{1 - 0.69892473}{0.01} \right]$

$P = 15000 \left[ \frac{0.30107527}{0.01} \right]$

Calculate the value inside the brackets (the PVIFA):

$\frac{0.30107527}{0.01} = 30.107527$

Now calculate P:

$P = 15000 \times 30.107527$

$P \approx 451612.905$

Rounding to two decimal places, the maximum principal loan amount is approximately $\textsf{₹}\$ 4,51,612.91$.


Example 3. A loan of $\textsf{₹}\$ 8,00,000$ is repaid with an EMI of $\textsf{₹}\$ 19,000$ per month. If the interest rate is 10.5% per annum compounded monthly, what is the approximate tenure of the loan in years?

Answer:

Given:

  • Principal Loan Amount (P) = $\textsf{₹}\$ 8,00,000$.
  • Equated Monthly Installment (EMI) = $\textsf{₹}\$ 19,000$.
  • Nominal Annual Interest Rate (R) = 10.5% per annum. Convert to decimal: $r = \frac{10.5}{100} = 0.105$.
  • Payment Frequency: Monthly. Compounding: Monthly ($m=12$).

To Find:

  • Loan Tenure (t) in years. (First find $n$, the total number of months).

Calculate Periodic (Monthly) Rate (i):

  • Periodic monthly rate $i = \frac{r}{12} = \frac{0.105}{12} = 0.00875$. (This is 0.875% per month).

Formula:

We use the formula $P = EMI \left[ \frac{1 - (1+i)^{-n}}{i} \right]$ and solve for $n$.

Solution:

Substitute the given values into the formula:

$800000 = 19000 \left[ \frac{1 - (1+0.00875)^{-n}}{0.00875} \right]$

$800000 = 19000 \left[ \frac{1 - (1.00875)^{-n}}{0.00875} \right]$

Divide both sides by 19000:

$\frac{800000}{19000} = \left[ \frac{1 - (1.00875)^{-n}}{0.00875} \right]$

$\frac{800}{19} \approx 42.105263$

So, $42.105263 \approx \frac{1 - (1.00875)^{-n}}{0.00875}$

Multiply both sides by 0.00875:

$42.105263 \times 0.00875 \approx 1 - (1.00875)^{-n}$

$0.368421 \approx 1 - (1.00875)^{-n}$

Rearrange to isolate the term with $n$:

$(1.00875)^{-n} \approx 1 - 0.368421$

$(1.00875)^{-n} \approx 0.631579$

To solve for $n$, we can take the logarithm of both sides. Using the property $\log(a^b) = b \log(a)$:

$\log((1.00875)^{-n}) \approx \log(0.631579)$

$-n \log(1.00875) \approx \log(0.631579)$

Divide by $-\log(1.00875)$:

$n \approx \frac{\log(0.631579)}{-\log(1.00875)}$

Using a calculator for the logarithms:

$\log(0.631579) \approx -0.199756$

$\log(1.00875) \approx 0.003794$

$n \approx \frac{-0.199756}{-0.003794}$

$n \approx 52.65$

The total number of months ($n$) is approximately 52.65. To find the tenure in years ($t$), divide $n$ by 12:

$t = \frac{n}{12} \approx \frac{52.65}{12} \approx 4.3875$ years.

The approximate tenure of the loan is 52.65 months, or about 4.39 years.

(Note: In real-world loans, the tenure is usually rounded up to the next whole month if the calculation results in a partial month. However, for exam purposes, providing the calculated decimal value is often acceptable unless otherwise specified).

Summary for Competitive Exams

EMI Problems: Typically involve finding P, EMI, i, or n for an amortizing loan.

Core Formulae (Ordinary Annuity assumption):

  • To find EMI: $\mathbf{EMI = P \left[ \frac{i}{1 - (1+i)^{-n}} \right]}$
  • To find P: $\mathbf{P = EMI \left[ \frac{1 - (1+i)^{-n}}{i} \right]}$

Where:

  • P = Principal Loan Amount
  • EMI = Equated Monthly Installment
  • i = Periodic (Monthly) Interest Rate (decimal) = Annual Rate (decimal) / 12
  • n = Total Number of Months = Tenure in Years $\times$ 12

Finding i or n: Requires algebraic manipulation (often using logarithms for n) or financial calculator/software.

Key Step: Always convert the annual rate to a periodic rate (monthly $i$) and tenure to total periods (total months $n$) before using the formulae.