Loans and Installment: Definition, Formula, Examples, Calculator

Loans and instalments are essential financial tools that allow individuals and businesses to borrow money for various purposes, such as purchasing a home, financing education, or expanding a business. Understanding how to solve an instalment loan example, whether instalment loan interest rates are simple or compound, is crucial for effective financial planning and management. This article will delve into the instalment loan meaning and formulas. Additionally, we will explore the use of an instalment loan calculator, the concept of instalment loans for bad credit, and how loan instalment payments work.

Formula for simple interest and compound interest

Where,

P = Principal

R = Annual interest rate (in percentage)

T = Time (in years)

n = Number of compounding periods per year

SI = Simple interest

CI = Compound interest

How do you find the value of the annual instalment if the interest accrued is simple interest?

To find the value of the annual instalment with simple interest, we use the following formula:

Annual instalment (AI) = $\frac{100A}{100T+\frac{RT(T-1)}{2}}$

Where A is the total amount to be paid in T years with R% annual interest.

Example:

What annual instalment amount will clear a debt of Rs.10,750 due in 4 years at 5% p.a. simple interest?

Sol: Here, the amount (A) = Rs.10,750, Interest rate (R) = 5% p.a. and time (T) = 4 years

So, the annual instalment = $\frac{100×10,750}{100×4+\frac{5×4(4-1)}{2}}$

= $\frac{10,75,000}{400+30}$

= $\frac{1,07,500}{43}$

= Rs.2500

How do you find the value of the monthly instalment if the interest accrued is simple?

To find the value of the monthly instalment with simple interest, we use the following formula:

Monthly instalment (MI) = $\frac{100A}{1200T+\frac{RT(12T-1)}{2}}$

Where A is the total amount to be paid in T years with R% annual interest.

Example:

What monthly instalment amount will clear a debt of Rs.44,850 with a simple interest of 10% per annum for 5 years?

Sol: Here, the amount (A) = Rs.44,850, Interest rate (R) = 10% p.a. and time (T) = 5 years

So, the monthly instalment = $\frac{44,850×100}{1200×5+\frac{10×5(12×5-1)}{2}}$

= $\frac{44,85,000}{6000+1475}$

= $\frac{44,85,000}{7475}$

= Rs.600

How do you find the value of the annual instalment if the interest accrued is compound interest?

To find the value of the annual instalment with compound interest, we use the following formula:

Annual instalment (AI) = $\frac{P}{1-(\frac{100}{100+R})^n}×\frac{R}{100}$

Where A is the total amount that is to be paid in n years with R% annual interest.

Example:

A sum of Rs.5500 is borrowed at 20% p.a. compound interest and paid back in 2 equal annual instalments. Find the amount of each annual instalment.

Sol: Here, the amount (A) = Rs.5500, Interest rate (R) = 20% p.a. and time (n) = 2 years

So, the annual instalment = $\frac{5500}{1-(\frac{100}{100+20})^2}×\frac{20}{100}$

= $\frac{1100}{1-(\frac{25}{36})}$

= $\frac{1100×36}{11}$

= Rs.3600

How do you find the value of the monthly instalment if the interest accrued is compound interest?

To find the value of the monthly instalment with compound interest, we use the following formula:

Monthly instalment (MI) = $\frac{AR(1+R)^n}{(1+R)^n - 1}$

Where A is the total amount to be paid in n months with a monthly interest rate of R (decimal equivalent of percentage).

Example:

Find the monthly instalment amount of a loan of Rs.8,00,000 with a compound interest of 12% per annum for 3 years.

⇒ Here, the amount (A) = Rs.8,00,000, monthly interest rate (R) = $\frac{12}{12}$ = 1% = 0.01 and time (T) = 3 years = (3 × 12) = 36 months

So, the monthly instalment = $\frac{8,00,000×0.01(1+0.01)^{36}}{(1+0.01)^{36} - 1}$

= $\frac{8,00,000×0.01(1+0.01)^{36}}{(1+0.01)^{36} - 1}$

= $\frac{11,446.15}{0.431}$

= Rs.26,557.20

Table of formulas used

Tips and Tricks

• Have a clear understanding of the key terms, for example, principal, interest, time, rate etc.

• Identify the type of interest given, whether it is a simple interest or a compound interest.

• Convert the interest rates and time into the specific format used in the formula.

• Check if the instalment is monthly or yearly, then use the specific formula for that.

Practice Questions/Solved Examples

Q.1.

A sum of Rs. 210 was taken as a loan. This is to be paid back in two equal instalments. If the rate of interest is 10% compounded annually, then the value of each instalment is:

1. Rs.127

2. Rs.121

3. Rs.210

4. Rs.225

Hint: Let $x$ be the equal instalment at the end of one year.

First, find the interest at the end of the first year and then at the end of the second year.

Substitute these values in the relation: Total instalment = Amount taken as loan + Interest for the first year + Interest of the second year to solve this.

Solution:

Let $x$ be the equal instalment at the end of one year.

In the first year,

Sum, $P$ = Rs. 210

Interest = $\frac{PnR}{100}$

Interest at the end of one year at a 10% rate of interest = 210 × 0.1 = 21

At the beginning of the second year after paying '$x$',

The sum for the second year becomes $P = 210+21-x$

⇒ $P = 231 - x$

Interest at the end of the second year at 10% rate of interest $= (231-x) × 0.1 = 23.1-0.1x$

Now, total instalment = Amount taken as loan + Interest for the first year + Interest of second year

So, the total value of instalments, $2x = 210+21+23.1–0.1x$

⇒ $x = \frac{254.1}{2.1}$

⇒ $x = 121$

Hence, the correct answer is option (2).

Q.2.

A loan of Rs. 12300 at 5% per annum Compound Interest be repaid in two equal annual instalments at the end of every year. Find the amount of each instalment.

1. Rs.6651

2. Rs.6615

3. Rs.6516

4. Rs.6156

Hint: $\frac{x}{(1+\frac{R}{100})} +\frac{x}{(1+\frac{R}{100})^2} +...........+\frac{x}{(1+\frac{R}{100})^n} =A$

where $x$ is the instalment value, 'A' is the total amount that was created in instalments and n is the number of instalments.

Solution:

Let the instalment value be Rs. $P$.

According to the concept,

$\frac{x}{(1+\frac{R}{100})} +\frac{x}{(1+\frac{R}{100})^2} +...........+\frac{x}{(1+\frac{R}{100})^n} =A$

where $x$ is the instalment value, 'A' is the total amount that was created in instalments and n is the number of instalments.

$P(1+\frac{5}{100})+P=12300(1+\frac{5}{100})^{2}$

⇒ $1.05P+P=12300×\frac{21}{20}×\frac{21}{20}$

⇒ $2.05P=13560.75$

⇒ $P=\frac{13560.75}{2.05}$ = 6615

Hence, the correct answer is option (2).

Q.3.

What is the amount (in INR) of debt that will be discharged in 6 equal instalments of INR 800 each, if the debt is due in 6 years at 5% per annum?

1. 6600

2. 7500

3. 8000

4. 5400

Hint: Use the given formula.

Installment = $\frac{\text{Debt×100}}{100×t+\frac{r×t(t-1)}{2}}$

Where, $t$ = time, $r$ =rate

Solution:

Given:

Installment = INR $800$

Rate, $r$ = $5$%

Time, $t$ = $6$ years

Let the debt be $x$.

Installment $=\frac{\text{Debt×100}}{100×t+\frac{r×t(t-1)}{2}}$

⇒ $800=\frac{100x}{100×6+\frac{5×6(6-1)}{2}}$

⇒ $800=\frac{100x}{675}$

$\therefore x = 5400 $

Hence, the correct answer is option (4).

Q.4.

A man borrows Rs. 4,000 from a bank at 10% per annum simple interest and clears the debt in three years. If the instalments paid at the end of the first, and second year to clear the debt are Rs. 1,500, and Rs. 2,500, respectively, what amount (in Rs.) should be paid at the end of the third year to clear the debt?

1. 700

2. 650

3. 500

4. 550

Hint: Use the formula, to find the simple interest separately for two years from the known values.

Simple interest (S.I.) = $\frac{\text{P} \times \text{R} \times \text{T}}{100}$

Where P is the principal amount, R is the rate of interest per annum, and T is the time in years.

Solution:

For the first year,

Principal, P = Rs. 4000

Rate, R = 10%

Time, T = 1 year

Simple interest = $\frac{\text{P} \times \text{R} \times \text{T}}{100}$ = $\frac{4000 \times 10 \times 1}{100}$ = Rs. 400

Remaining amount after the installment is paid = Rs. 4000 – Rs. 1500 = Rs. 2500

For the second year,

Principal, P = Rs. 2500

Rate, R = 10%

Time, T = 1 year

Simple interest = $\frac{\text{P} \times \text{R} \times \text{T}}{100}$ = $\frac{2500 \times 10 \times 1}{100}$ = Rs. 250

Remaining amount after the installment is paid = Rs. 2500 – Rs. 2500 = Rs. 0

The amount left to be paid for the third year = Total simple interest in two years

= Rs. 400 + Rs. 250 = Rs. 650

Hence, the correct answer is option (2).

Q.5.

A computer is available for INR 39,000 on cash payment or INR 19,000 as cash payment followed by five monthly instalments of INR 4,200 each. What is the rate of interest per annum under the instalment plan?

1. $20 \frac{19}{29} \%$

2. $20 \frac{17}{29} \%$

3. $20 \frac{20}{29} \%$

4. $20 \frac{18}{29} \%$

Hint: Simple interest = $\frac{\text{Principal × Rate × Time}}{100}$

The sum of the amounts of these instalments = The amount of Rs. 20000 for 5 months.

Solution:

Total cost of the computer = Rs. 39000

Down payment = Rs. 19000

Balance = Rs. 20000

Let the rate of interest be $r$

Here simple interest is denoted by SI.

Hence, the amount of Rs. 20000 for 5 months

= $20000+ \frac{20000\times r\times \frac{5}{12}}{100}$

= $20000+ \frac{250r}{3}$

The customer pays the shopkeeper Rs. 4200 after 1 month, Rs. 4200 after 2 months, ...... and Rs. 4200 after 5 months.

Thus, the shopkeeper keeps Rs. 4200 for 4 months, Rs. 4200 for 3 months, Rs. 4200 for 2 months, Rs. 4200 for 1 month and Rs. 4200 at the end.

The sum of the amounts of these instalments

= (Rs. 4200 + S.I. on Rs 4200 for 4 months) + (Rs. 4200 + S.I. on Rs. 4200 for 3 months) + ...... + (Rs. 4200 + S.I. on Rs. 4200 for 1 month) + Rs. 4200

= Rs. (4200 × 5) + S.I. on Rs. 4200 for (4 + 3 + 2 + 1) months

= Rs. 21000 + S.I. on Rs. 4200 for 10 months

= $21000+4200\times r\times \frac{10}{12}\times \frac{1}{100}=21000+35r$

Hence, $20000+ \frac{250r}{3}=21000+35r$

$⇒\frac{145r}{3}=1000$

$⇒r=\frac{600}{29}$

$⇒r=20\frac{20}{29}\%$

Hence, the correct answer is option (3).

Q.6.

A car priced INR 6,50,000 is bought by making a down payment. On the balance, a simple interest of 10% is charged in a lump sum and the money is to be paid in 20 equal annual instalments of INR 25,000. How much is the down payment?

1. INR 1,55,945

2. INR 1,95,455

3. INR 1,94,555

4. INR 1,45,955

Hint: Find the interest charged on the balance amount and then equate it with the instalments. Use this information to solve the question.

Solution:

Let the downpayment be of INR $x$.

Hence from the above data, we get the following relation,

$(650000 - x)\times (1 + \frac{10}{100}) = 20\times 25000$

$⇒650000 - x = 20\times 25000 \times \frac{10}{11} ≈ 454545$

$⇒x = 650000 - 454545 = 1,95,455$

$⇒x ≈ 195455$

∴ The downpayment was INR 1,95,455.

Hence, the correct answer is option (2).

Q.7.

A person borrows Rs.1,00,000 from a bank at 10% per annum simple interest and clears the debt in five years. If the instalment paid at the end of the first, second, third, and fourth years to clear the debt are Rs.10,000, Rs. 20,000, Rs. 30,000, and Rs. 40,000, respectively, what amount should be paid at the end of the fifth year to clear the debt?

1. Rs. 38,250

2. Rs. 39,490

3. Rs. 40,450

4. Rs. 36,450

Hint: Use the formula below.

Simple Interest = $\frac{\text{Principle×Rate×Time}}{100}$

Solution:

Given,

A person borrows Rs.1,00,000 from a bank at 10% per annum.

1st year interest = 100000 × 10% = Rs. 10,000

The instalment paid at the end of the first year = Rs. 10,000

Principal amount for 2nd year = 100,000 – 10000 + 10000 = Rs. 1,00,000

So 2nd year interest = 1,00,000 × 10% = Rs. 10,000

The instalment paid at the end of the second year = Rs. 20,000

Principal amount for 3rd year = 1,00,000 – 20,000 + 10,000 = Rs. 90,000

So 3rd year interest = 90,000 × 10% = Rs. 9,000

The instalment paid at the end of the third year = Rs. 30,000

Principal amount for 4th year = 90,000 – 30,000 + 9000 = Rs. 69,000

So 4th year interest = 69,000 × 10% = Rs. 6,900

The instalment paid at the end of the fourth year = Rs. 40,000

Principal amount for 5th year = 69000 – 40000 + 6900 = Rs. 35900

So 5th year interest = 35,900 × 10% = Rs. 3590

In 5th year he has to pay the total amount (Principal + Interest for the fifth) as the remaining Amount

= 35900 + 3590 = 39,490

Hence, the correct answer is option (2).

Q.8.

A sum of Rs.10 is lent by a child to his friend to be returned in 11 monthly instalments of Re.1 each, the interest being simple. The rate of interest is:

1. $11 \frac{9}{11} \%$

2. $21 \frac{9}{11} \%$

3. $10 \frac{2}{11} \%$

4. $9 \frac{1}{11} \%$

Hint: Use the formula below.

Simple Interest = $\frac{\text{Principle×Rate×Time}}{100}$

Solution:

Let the rate of interest be $r\%$ per annum.

Amount to be paid = $10 + \frac{10 \times r \times\frac{11}{12}}{100}= 10 + \frac{11r}{120}$

Total effective payment = (Re. 1 + interest on Re. 1 for 10 months) + (Re. 1 + interest on Re. 1 for 9 months) + ... +(Re. 1 + interest on Re. 1 for 1 months) + Re. 1

= $(1 + \frac{1 × r × \frac{10}{12}}{100}) + (1 + \frac{1 × r × \frac{9}{12}}{100}) + .... + (1 + \frac{1 × r × \frac{1}{12}}{100})+ 1$

= $(1 + \frac{10r}{1200}) + (1 + \frac{9r}{1200}) + .... + (1 + \frac{r}{1200}) + 1$

= $11 + \frac{(1 + 2 + .... + 10)r}{1200}$

= $11 + \frac{(10 × \frac{11}{2})r}{1200}$

= $11 + \frac{11r}{240}$

Now we have,

$10 + \frac{11r}{120} = 11 + \frac{11r}{240}$

⇒ $\frac{11r}{240} = 1$

⇒ $r = \frac{240}{11} = 21\frac{9}{11}\%$

Hence, the correct answer is option (2).

Q.9.

A person borrowed INR 2,000 at 5% annual simple interest repayable in 3 equal annual instalments. What will be the annual instalment?

1. INR $730\frac{10}{63}$

2. INR $840\frac{10}{63}$

3. INR $640\frac{11}{63}$

4. INR $250\frac{10}{63}$

Hint: Instalment = $\frac{A\times 100}{N\times 100 + ((N-1) + (N-2) + .....1)\times R }$

where, $A$ = Amount, $R$ = Rate, $N$ = Number of Years

Solution:

Principal(P) = INR 2,000

Rate = 5%

Time = 3 years

Instalment = $\frac{A\times 100}{N\times 100 + ((N-1) + (N-2) + .....1)\times R }$

where, $A$ = Amount, $R$ = Rate, $N$ = Number of Years

Simple Interest = $\frac{P\times R\times T}{100} = \frac{2000\times 5\times 3}{100} = 300$

Amount = 2000 + 300 = Rs. 2300

Instalment = $\frac{2300\times 100}{3\times 100 + (2+1)\times 5}$

$= \frac{46000}{63} = 730\frac{10}{63}$

Hence, the correct answer is option (1).

Q.10.

Rajnish borrowed INR 1,500 from a bank and repaid the entire amount with interest in two equal annual instalments, the first instalment being paid a year after Rajnish borrowed from the bank. If the rate of interest was 40% per annum, compounded annually, then what was the value (in INR) of each instalment paid by Rajnish?

1. 1125

2. 1470

3. 1225

4. 1350

Hint: Use the formula,

When compounded annually, Amount = Principal(1 + $\frac{\text{Rate}}{100})^{\text{Time}}$

Solution:

Let the two equal instalments be $P_1$ and $P_2$

Given: $P_1 + P_2 = 1500$ and rate of interest $(R)= 40 \%$

Since annual instalments are equal,

So, $P_1(1+\frac{R}{100})=P_2(1+\frac{R}{100})(1+\frac{R}{100})$

⇒ $\frac{P_1}{P_2}=1+\frac{R}{100}=1+\frac{40}{100}=1+\frac{2}{5}=\frac{7}{5}$

⇒ $P_2 = \frac{5P_1}{7}$

Now, $P_1 + P_2 = 1500$

⇒ $P_1+\frac{5P_1}{7}= 1500$

⇒ $\frac{12P_1}{7}= 1500$

⇒ $P_1 = \frac{1500×7}{12}=875$

Installments = $P_1(1+\frac{R}{100})=875(1+\frac{40}{100})=875×\frac{7}{5}=1225$

Hence, the correct answer is option (3).