Installments – Various Cases and Questions including Simple and Compound Interest

Nowadays, loan has become crucial part of our life. We all have learnt living our life on credit. Whether be it a businessman taking loans to run his business or a household to buy a car, everyone has become dependent on sustaining their life and fulfilling their wishes with the help of these loans. But, when the amount has been borrowed then it has to be returned too and now not just the principal loan amount but some interest as well. Interest plays a very significant role in our life. It is a deciding factor whether or not loan has to be taken or not as higher the interest then higher the amount that has to repaid. Now, after the loan has been taken it could either be returned along with the interest in a lumpsum after some specified period of time or it can also be recovered in form of installments of some kind in which some amount of interest along with principal sum is repaid at some time intervals. Currently, all major finance lending institutions such as banks etc. recover their loans through EMI’s i.e. Equated monthly installments. Today, in this blog we will discuss the how to calculate these installments considering various different factors and cases.
Interest charged on the loan can be of any type either Simple Interest or Compound Interest. Though we have discussed regarding it but for revision’s sake.
Simple interest is a the one where interest once credited does not earn interest on it.
SI = (P * R * T)/ 100
Compound Interest is where interest earns itself interest. It is the most common form of interest that is being charged nowadays.
CI = P(1 + r/100)^n
INSTALLMENTS UNDER SIMPLE INTEREST
Suppose Ravi bought a T.V. worth ₹20000 on EMI’s and every month a fix installment has to be for next n months where interest is charged @ r% per annum on simple interest.
Now, if the loan is for n months then Ravi will pay end the of 1st month interest for (n1) months, at the end of second month he’ll pay interest for (n2) months, at the end of 3rd month he’ll pay interest for (n3) months and similarly, at the end of nth month he’ll pay no interest i.e.
Therefore, total amount paid by Ravi = [x+ (x* (n1) * r)/ 12* 100] + [x+ (x* (n2) * r)/ 12* 100] + [x+ (x* (n3) * r)/ 12* 100] … [x+ (x* 1* r)/ 12* 100] + x
This will be equal to the total interest charged for n months i.e. [P+ (P* n* r)/ 12* 100].
Thus, [P+ (P* n* r)/ 12* 100] = [x+ (x* (n1) * r)/ 12* 100] + [x+ (x* (n2) * r)/ 12* 100] + [x+ (x* (n3) * r)/ 12* 100] … [x+ (x* 1* r)/ 12* 100] + x
Simplifying and generalizing the above equation we get the following formula, x = P (1 + nr/100)/ (n + n(n1)/2 * r/100))
And instead of principal sum total amount (Principal + Interest) to be repaid is given then, x = 100A/ 100n + n(n1) r/2
INSTALLMENTS UNDER COMPOUND INTEREST
Let a person takes a loan from bank at r% and agrees to pay loan in equal installments for n years. Then, the value of each installment is given by
P (1 + r/100)^n = X (1 + r/100)^(n1) + X (1 + r/100)^(n2) + X (1 + r/100)^(n3) + …. + X (1 + r/100)
Using the Present Value Method,
P = X/ (1 + r/100)^n … X/ (1 + r/100)^2 + X/ (1 + r/100)
MISCELLANEOUS CASES OF INSTALLMENTS ON SIMPLE INTEREST AND COMPOUND INTEREST
CASE 1: To calculate the installment when interest is charged on SI
A mobile phone is available for ₹2500 or ₹520 down payment followed by 4 monthly equal installments. If the rate of interest is 24%p.a. SI, calculate the installment.
Sol: This is one basic question. You have to just use the above formula and calculate the amount of installment.
Therefore, x = P (1 + nr/100)/ (n + n(n1)/2 * r/100))
Here P = 2500 – 520 = 1980
R = 25% p.a.
T = 4 months
Hence, x = 1980(1 + 15 * 12/ 1200)/ (4 + 4* 3* 12/ 2 * 12 * 100)
= ₹520
CASE 2: To calculate the installment when interest is charged on CI
What annual payment will discharge a debt of ₹7620 due in 3 years at 50/3% p.a. compounded interest?
Sol: Again, we will use the following formula,
P (1 + r/100)^n = X (1 + r/100)^(n1) + X (1 + r/100)^(n2) + X (1 + r/100)^(n3) + …. + X (1 + r/100)
Here P = ₹7620
n = 3 years
r = 50/3% p.a.
7620(1 + 50/300)^3 = x (1 + 50/300)^2 + x (1 + 50/300) + x
12100.2778 = x (1.36111 + 1.1667 + 1)
x = ₹3430
CASE 3: To calculate loan amount when interest charged is Compound Interest
Ram borrowed money and returned it in 3 equal quarterly installments of ₹17576 each. What sum he had borrowed if the rate of interest was 16 p.a. compounded quarterly?
Sol: In this case, we will use present value method as we need to find the original sum borrowed by Ram.
Since, P = X/ (1 + r/100)^n … X/ (1 + r/100)^2 + X/ (1 + r/100)
Therefore, P = 17576/ (1 + 4/100)^3 + 17576/ (1 + 4/100)^2 + 17576/ (1 + 4/100)
= 17576 (0.8889 + 0.92455 + 0.96153)
= 17576 * 2774988
= 48773.1972
CASE 4:
Gopal borrows ₹1,00,000 from a bank at 10% p.a. simple interest and clears the debt in five years. If the installments paid at the end of the first, second, third and fourth years to clear the debt are ₹10,000, ₹20,000, ₹30,000 and ₹40,000 respectively, what amount should be paid at the end of the fifth year to clear the debt?
Sol: Total principal amount left after 5th year = 100000 – (10000 + 20000 + 30000 + 40000)
= 100000 – 100000 = 0
Therefore, only interest component has to be paid in the last installment.
Hence, Interest for the first year = (100000 * 10 * 1) /100 = ₹10000
Interest for the second year = (100000 – 10000) * 10/ 100 = ₹9000
Interest for the third year = (100000 – 10000 – 20000) * 10/ 100 = ₹7000
Interest for the fourth year = (100000 – 10000 – 20000 – 30000) * 10/ 100 = ₹4000
Thus, Amount that need to paid in the fifth installment = (10000 + 9000 + 7000 + 4000) = ₹30000
CASE 5:
An amount of ₹12820 due in 3 years, hence is fully repaid in three annual installments starting after 1 year. The first installment is 1/2 the second installment and the second installment is 2/3 of the third installment. If rate of interest is 10% p.a. Find the first installment.
Sol: Let the third installment be x.
Since, second installment is 2/3 of the third, it will be 2/3 x. And finally, 1st installment will be 1/2 * 2/3 * x
Now proceeding in the similar fashion as we did earlier and using the compound interest formula to calculate the installment amount.
P (1 + r/100)^n = X (1 + r/100)^(n1) + X (1 + r/100)^(n2) + X (1 + r/100)^(n3) + …. + X (1 + r/100)
12820 (1 + 10/100)^3 = 1/3 X (1 + 10/100)^2 + 1/2 X (1 + 10/100)^1 + X
12820(1.1)^3 = x (1/3 * (1.1)^2 + 1/2 * (1.1) + 1)
17063.42 = x * (0.40333 + 0.55 + 1)
17063.42 = x * 1.953333
x = ₹8735.53
Therefore, the amount of first installment will be 2/3 * 1/2 * x, i.e. 1/3 * 8735.53 = ₹2911.84
CASE 6:
Ravi lent out ₹9 to Sam on the condition that the amount is payable in 10 months by 10 equal installments of ₹1 each payable at the start of every month. What is the rate of interest per annum if the first installment has to paid one month from the date the loan is availed.
Sol: The value of money coming in should equal the value of the money going out for the loan to be completely paid off. Therefore,
₹9 + Interest on ₹9 for 10 months = (₹1 + interest on ₹1 for 9 months) + (₹1 + interest on ₹1 for 8 months) + (₹1 + interest on ₹1 for 7 months) + (₹1 + interest on ₹1 for 6 months) + …. + (₹1 + interest on ₹1 for 2 months) + (₹1 + interest on ₹1 for 1 month) + ₹1
₹9 + interest on ₹1 for 90 months = ₹10 + interest on ₹1 for 45 months
interest on ₹1 for 90 months – interest on ₹1 for 45 months = ₹10 – ₹9
interest on ₹1 for 45 months = ₹1 (i.e. money would double in 45 months)
Hence, the rate of interest = 100%/ 45 = 2.2222%
I hope you are now clear with the installment concept and can now numerous other these and varied type of questions!