Do you know, If someone wants to buy a Motor Cycle, or a costly mobile or if your father wants to purchase a new flat, they generally take loan from a bank or from a financier. They need to pay monthly installments. Today, across the world, all the EMI's (Equated monthly Installments) are being calculated on compound interest. So after reading this article, you must be in a position to calculate the EMI you are going to pay for your new scooter purchase!!
Before that, let us try to understand the meaning of compounding and discounting. We know that daily prices of the goods increase at a rate. Assume, A product which costs Rs.100 costs Rs.110 next year and Rs.121 that next year. Suppose, You lend Rs.100 to your friend and he promised to give you Rs.100 after 1 year. Do you accept it? Out of friendship we accept. But think like a financier. If you receive Rs.100 after 1 year the value of this 100 rupee may not buy the same amount of goods it would have purchased today as the prices went up to Rs.100. So we need to charge some interest. So how much is the interest?
We use the formula of compound interest.
$P\left( {1 + \displaystyle\frac{R}{{100}}} \right) = A \Rightarrow 100\left( {1 + \displaystyle\frac{R}{{100}}} \right) = 110$
It is clear that interest rate is 10%
That means, if the interest rate is 10%, Rs.100 today is equal to Rs.110 earned after 1 year and Rs.121 after 2 years.
Assume we have taken a loan at period 0 and have to pay installments at the end of 1, 2, 3, 4th periods of x each.
Now loan amount plus the interest on the total loan amount P at R% rate for 4 periods is equal to all the EMI's and interests earned for the remaining period. That is the EMI in the period 2 earns 2 periods interest and EMI in the 3rd period earns only 1 period interest.
$P\left( {1 + \displaystyle\frac{R}{{100}}} \right)^4$ = $x\left( {1 + \displaystyle\frac{R}{{100}}} \right)^3$ + $x\left( {1 + \displaystyle\frac{R}{{100}}} \right)^2$ + $x\left( {1 + \displaystyle\frac{R}{{100}}} \right)^1$ + $x$
We already learnt that Rs.110 earned after 1 year is equal to Rs.100 earned today if the interest rate is 10%
So Rs.110 discounted at 10% gives you Rs.100. Don't worry how to do this. This is quite simple. You have to calculate P from the compound interest formula.
$P = \displaystyle\frac{A}{{\left( {1 + \displaystyle\frac{R}{{100}}} \right)^n }}$
Here A = Rs.110, R = 10%, n = 1.
Before that, let us try to understand the meaning of compounding and discounting. We know that daily prices of the goods increase at a rate. Assume, A product which costs Rs.100 costs Rs.110 next year and Rs.121 that next year. Suppose, You lend Rs.100 to your friend and he promised to give you Rs.100 after 1 year. Do you accept it? Out of friendship we accept. But think like a financier. If you receive Rs.100 after 1 year the value of this 100 rupee may not buy the same amount of goods it would have purchased today as the prices went up to Rs.100. So we need to charge some interest. So how much is the interest?
We use the formula of compound interest.
$P\left( {1 + \displaystyle\frac{R}{{100}}} \right) = A \Rightarrow 100\left( {1 + \displaystyle\frac{R}{{100}}} \right) = 110$
It is clear that interest rate is 10%
That means, if the interest rate is 10%, Rs.100 today is equal to Rs.110 earned after 1 year and Rs.121 after 2 years.
Formula for installments in Compound Interest:
If a buyer sells a product to you at full payment get some interest on your amount for n periods. This total amount should equal to sum of all EMI's and the interests accrued on each EMI for the remaining period. Then only the seller did not get any loss.Now loan amount plus the interest on the total loan amount P at R% rate for 4 periods is equal to all the EMI's and interests earned for the remaining period. That is the EMI in the period 2 earns 2 periods interest and EMI in the 3rd period earns only 1 period interest.
$P\left( {1 + \displaystyle\frac{R}{{100}}} \right)^4$ = $x\left( {1 + \displaystyle\frac{R}{{100}}} \right)^3$ + $x\left( {1 + \displaystyle\frac{R}{{100}}} \right)^2$ + $x\left( {1 + \displaystyle\frac{R}{{100}}} \right)^1$ + $x$
Present Value method:
We should also solve this problem by Present Value (PV method). You learn this concept in any finance text book!!We already learnt that Rs.110 earned after 1 year is equal to Rs.100 earned today if the interest rate is 10%
So Rs.110 discounted at 10% gives you Rs.100. Don't worry how to do this. This is quite simple. You have to calculate P from the compound interest formula.
$P = \displaystyle\frac{A}{{\left( {1 + \displaystyle\frac{R}{{100}}} \right)^n }}$
Here A = Rs.110, R = 10%, n = 1.
From the above diagram, if we have taken loan for 4 periods, the last installment should discount for 4 periods and so on.
So $P$ = $\displaystyle\frac{x}{{\left( {1 + \displaystyle\frac{R}{{100}}} \right)^1 }}$ + $\displaystyle\frac{x}{{\left( {1 + \displaystyle\frac{R}{{100}}} \right)^2 }}$ + $\displaystyle\frac{x}{{\left( {1 + \displaystyle\frac{R}{{100}}} \right)^3 }}$ + $\dfrac{x}{{\left( {1 + \displaystyle\frac{R}{{100}}} \right)^4 }}$
b. Rs 2,413
c. Rs 2,314
d. Rs 2,662
Answer: b
Explanation:
Sum borrowed = Rs 6000
Interest = 10% at compounded annually and time = 3 years
Let x be the amount paid at the end of each of the 3 years.
$ \Rightarrow 6000\left( {1 + \displaystyle\frac{{10}}{{100}}} \right)^3$ = $x\left( {1 + \displaystyle\frac{{10}}{{100}}} \right)^2$ + $x\left( {1 + \displaystyle\frac{{10}}{{100}}} \right) + x$
$ \Rightarrow 6000(1.1)^3 = 1.21x + 1.1x + x$
$ \Rightarrow x = \displaystyle\frac{{6000(1.331)}}{{3.31}} = 2412.68$
So Installment amount is Rs.2413
Solved Examples
1. A person borrowed a sum of Rs 6000 at 10% p.a., interest compounded annually. If the money is to be repaid in three equal annual installment, each payable at the end of the year, then what is the value of each installment?
a. Rs. 2,000b. Rs 2,413
c. Rs 2,314
d. Rs 2,662
Answer: b
Explanation:
Sum borrowed = Rs 6000
Interest = 10% at compounded annually and time = 3 years
Let x be the amount paid at the end of each of the 3 years.
$ \Rightarrow 6000\left( {1 + \displaystyle\frac{{10}}{{100}}} \right)^3$ = $x\left( {1 + \displaystyle\frac{{10}}{{100}}} \right)^2$ + $x\left( {1 + \displaystyle\frac{{10}}{{100}}} \right) + x$
$ \Rightarrow 6000(1.1)^3 = 1.21x + 1.1x + x$
$ \Rightarrow x = \displaystyle\frac{{6000(1.331)}}{{3.31}} = 2412.68$
So Installment amount is Rs.2413
2. Three equal installment, each of Rs 200, were paid at the end of year on a sum borrowed at 20% compound interest compounded annually. Find the sum.
b. Rs 400
c. Rs 421.30
d. Rs 432.10
Answer: c
Explanation:
Installments each of Rs 200 for 3 years.
$ \Rightarrow x\left( {1 + \displaystyle\frac{{20}}{{100}}} \right)^3$ = $200\left( {1 + \displaystyle\frac{{20}}{{100}}} \right)^2$ + $200\left( {1 + \displaystyle\frac{{20}}{{100}}} \right)$ + $200\left( {1 + \displaystyle\frac{{20}}{{100}}} \right)$
$ \Rightarrow x(1.2)^3 = 200(1.44) + 200(1.2) + 200$
$ \Rightarrow x(1.718) = 200(3.64)$
$ \Rightarrow x = \displaystyle\frac{{200(3.64)}}{{1.718}} = 421.3$
Alternate Method:
we use discounting method. We discount the installments for the present value.
$P = \displaystyle\frac{{200}}{{\left( {1 + \displaystyle\frac{{20}}{{100}}} \right)^1 }}$ + $\displaystyle\frac{{200}}{{\left( {1 + \displaystyle\frac{{20}}{{100}}} \right)^2 }}$ + $\displaystyle\frac{{200}}{{\left( {1 + \displaystyle\frac{{20}}{{100}}} \right)^3 }}$
$P = \displaystyle\frac{{200}}{{1.2}} + \displaystyle\frac{{200}}{{1.44}} + \displaystyle\frac{{200}}{{1.718}}$
P = 421.3
3. A man borrows a certain sum of money and pays it back in 2 years in two equal installments. If C.I. is reckoned at 5% per annum and he pays back annually Rs. 441, what sum did he borrow?
a. Rs. 820b. Rs. 800
c. Rs. 882
d. Rs. 850
Answer: a
Explanation:
Money borrowed which becomes Rs. 441 after one year =$\displaystyle\frac{{441}}{{\left( {1 + \displaystyle\frac{5}{{100}}} \right)}}$ = \(441 \times \left( {\dfrac{{20}}{{21}}} \right) = 420\)
Money borrowed which become Rs. 441 after two years = $\displaystyle\frac{{441}}{{\left( {1 + \displaystyle\frac{5}{{100}}} \right)}}$ = \(441 \times {\left( {\dfrac{{20}}{{21}}} \right)^2} = 400\)
Total money borrowed = Rs. 820.
4. A man borrows Rs. 2100 and undertakes to pay back with compound interest @ 10% p.a. in 2 equal yearly installments at the end of first and second year. What is the amount of each installment?
a. 1200
b. 1210
c. 1300
d. 1310
Answer: b
Explanation:
Let the installment be \(x\).
The value of first installment before 1 year = \(\dfrac{x}{{\left( {1 + \dfrac{{10}}{{100}}} \right)}} = \dfrac{{10x}}{{11}}\)
The value of second installment before 2 years = \(\dfrac{x}{{{{\left( {1 + \dfrac{{10}}{{100}}} \right)}^2}}} = \dfrac{{100x}}{{121}}\)
The sum of the above values should equal to principal.
\( \Rightarrow \dfrac{{10x}}{{11}} + \dfrac{{100x}}{{121}} = 2100\)
\( \Rightarrow \dfrac{{110x + 100x}}{{121}} = 2100\)
\( \Rightarrow \dfrac{{210x}}{{121}} = 2100\)
\( \Rightarrow x = 1210\)
Alternative method:
The value of the principal after 2 years = The value of 1st installment after 1 year + the value of 2nd installment.
\( \Rightarrow \) \(2100{\left( {1 + \dfrac{{10}}{{100}}} \right)^2}\) = \(x\left( {1 + \dfrac{{10}}{{100}}} \right) + x\)
\(2100\left( {\dfrac{{121}}{{100}}} \right) = x\left( {\dfrac{{11}}{{10}}} \right) + x\)
\(2100\left( {\dfrac{{121}}{{100}}} \right) = \dfrac{{21x}}{{10}}\)
\( \Rightarrow x = 1210\)
Alternative method:
Here, (1 + r) = 1 + $\displaystyle\frac{1}{{10}} = \displaystyle\frac{{11}}{{10}}$
Ratio of principals of two installments = 1 : $\displaystyle\frac{{10}}{{11}}$ = 11 : 10
Sum of ratios = 11 + 10 = 21
Therefore, Principal of first instalment = 2100 × $\displaystyle\frac{{11}}{{21}}$ = Rs. 1100
Therefore, Instalment = Principal of first instalment × (1 + r)
= 1100 × $\displaystyle\frac{{11}}{{10}}$ = Rs. 1210
5. A man borrows Rs. 820 and undertakes to pay back with compound interest @ 5% p.a. in 2 equal yearly installments at the end of first and second year. What is the amount of each installment?
a. 400
b. 420
c. 441
d. 410
Answer: c
Explanation:
The value of 820 after two years = The value of the installment after 1 year + The value of second installment
\( \Rightarrow 820{\left( {1 + \dfrac{5}{{100}}} \right)^2}\) = \(x\left( {1 + \dfrac{5}{{100}}} \right)\) + \(x\)
\( \Rightarrow 820 \times \dfrac{{441}}{{400}} = \dfrac{{21}}{{20}}x + x\)
\( \Rightarrow 820 \times \left( {\dfrac{{441}}{{400}}} \right) = \dfrac{{41}}{{20}}x\)
\( \Rightarrow x = 441\)
Alternative Method:
Here, (1 + r) = 1 + $\displaystyle\frac{1}{{20}} = \displaystyle\frac{{21}}{{20}}$
Ratio of principals of two instalments = 1 : $\displaystyle\frac{{20}}{{21}}$ = 21 : 20
Sum of ratios = 21 + 20 = 41
Therefore, Principal of first installment = $\displaystyle\frac{{21}}{{41}}$ × 820 = Rs. 420
Therefore, Installment = Principal of first installment × (1 + r) = 420 × $\displaystyle\frac{{21}}{{20}}$ = Rs. 441
6. A man borrows Rs. 1820 and undertakes to pay back with compound interest @ 20% p.a. in 3 equal yearly installments at the end of first, second and third years. What is the amount of each installment?
a. 864
b. 850
c. 820
d. 900
Answer: a
Explanation:
The value of 1820 after three years = The value of the first installment after 2 years + The value of second installment after 1 year + the value of the third installment
\( \Rightarrow 1820{\left( {1 + \dfrac{{20}}{{100}}} \right)^3}\) = \(x{\left( {1 + \dfrac{{20}}{{100}}} \right)^2}\) + \(x\left( {1 + \dfrac{{20}}{{100}}} \right) + x\)
\( \Rightarrow 1820\left( {\dfrac{{216}}{{125}}} \right)\) = \(x\left( {\dfrac{{36}}{{25}}} \right) + x\left( {\dfrac{6}{5}} \right) + x\)
\( \Rightarrow 1820\left( {\dfrac{{216}}{{125}}} \right)\) = \(x\left( {\dfrac{{36 + 30 + 25}}{{25}}} \right)\)
\( \Rightarrow 1820\left( {\dfrac{{216}}{{125}}} \right) = x\left( {\dfrac{{91}}{{25}}} \right)\)
\( \Rightarrow x = 864\)
Alternative Method:
Here, (1 + r) = 1 + $\displaystyle\frac{1}{5} = \displaystyle\frac{6}{5}$
Ratio of principals for three years = 1 : $\displaystyle\frac{5}{6}:\left( {\displaystyle\frac{5}{6}} \right)^2 $
= $6^2 $ : 6 × 5 : $5^2 $ (On multiplying each ratio by $6^2 $)
= 36 : 30 : 25
Sum of the ratios = 36 + 30 + 25 = 91
Therefore, Principal of first installment = $\displaystyle\frac{{36}}{{91}}$ × 1820 = Rs. 720
Therefore, Installment = Principal of first installment × (1 + r) = 720 × $\displaystyle\frac{6}{5}$ = Rs. 864
General Discussion On EMI's:
Suppose you have taken a loan of Rs.100000 to buy a house at 12% rate to be paid EMI's for 60 months.
They use this formula EMI = $\displaystyle\frac{{P \times r \times (1 + r)^n }}{{\left( {1 + r} \right)^n  1}}$
Here r = Rate / 1200
n = periods (5 years = 60 periods)
Monthly interest is 1% per month. Now after 1 month, interest accrued is 1000. Of the total EMI of Rs.2224, Rs.1000 used for interest and the remaining Rs.1224 for principal reduction. Now Balance of Rs.98776 becomes the principal for next month. So interest is 987.7 or Rs.988.
Suppose you have taken a loan of Rs.100000 to buy a house at 12% rate to be paid EMI's for 60 months.
They use this formula EMI = $\displaystyle\frac{{P \times r \times (1 + r)^n }}{{\left( {1 + r} \right)^n  1}}$
Here r = Rate / 1200
n = periods (5 years = 60 periods)
Sample Loan Schedule:
CarWale EMI Schedule
Following Schedule Is For : 100000 to repay in 60 months.
All calculations are based on EMI in Arrears(i.e. Rear Ended EMI's).
All calculations are based on EMI in Arrears(i.e. Rear Ended EMI's).
EMI Number

EMI Amount

Interest Amount

Principal Reduction

Balance Due

1.

Rs.2224

Rs.1000

Rs.1224

Rs.98776

2.

Rs.2224

Rs.988

Rs.1237

Rs.97539

3.

Rs.2224

Rs.975

Rs.1249

Rs.96290

4.

Rs.2224

Rs.963

Rs.1262

Rs.95028

5.

Rs.2224

Rs.950

Rs.1274

Rs.93754

Monthly interest is 1% per month. Now after 1 month, interest accrued is 1000. Of the total EMI of Rs.2224, Rs.1000 used for interest and the remaining Rs.1224 for principal reduction. Now Balance of Rs.98776 becomes the principal for next month. So interest is 987.7 or Rs.988.