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Arithmetic I

Interest & Growth

ALL MODULES

CAT 2025 Lesson : Interest & Growth - Present Value & EMI

3.3 Present Value of Multiple Payments

To get the present value of future payments, we discount each of these payments at the given interest rate for the time periods they correspond to. This is an extension of the formula given in the above Section 3.2.

Let

p

be the principal or present value of a loan lent at

r \%

interest rate per annum. If

A

B

and

C

are the repayments made at the end of

a

b

and

c

years, wherein the loan is fully repaid, then

p = \dfrac{A}{\left(1 + \dfrac{r}{100} \right)^{a}} + \dfrac{B}{\left(1 + \dfrac{r}{100} \right)^{b}} + \dfrac{C}{\left(1 + \dfrac{r}{100} \right)^{c}}

Example 11

Caishen wants to lend a certain amount to Jade at

20 \%

per annum compound interest. Jade agrees to repay

1000

coins at the end of each of first, third and fifth years. How many coins (rounded to the nearest integer) should Caishen lend?

Solution

Let the present value or principal be

p

p = \dfrac{1000}{\left(1 + \dfrac{20}{100} \right)^{1}} + \dfrac{1000}{\left(1 + \dfrac{20}{100} \right)^{3}} + \dfrac{1000}{\left(1 + \dfrac{20}{100} \right)^{5}}

= \dfrac{1000}{1.2} + \dfrac{1000}{1.2^{3}} + \dfrac{1000}{1.2^{5}}

Using the calculator, we get

p = 1813.91 \thicksim 1814

Answer:

1814

3.4 Equated Monthly Instalment (EMI)

For most retail loans (such as home loans, auto loans and personal loans), the borrowers repay by paying a fixed amount every month over the tenure of the loan. This fixed amount is called Equated Monthly Instalment or EMI. EMI includes interest and principal.

For instance, where Rs.

25,000

was lent on

1^{\text{st}}

t January

2019

and the interest is

12 \%

per annum, which is

1 \%

per month, and the EMI is Rs.

1250

, the following table provides the impact on the principal outstanding, i.e. the principal remaining.

Date	EMI	Interest	Principal Repaid	Principal Outstanding
$1^{\text{st}}$ January				$25,000$
$31^{\text{st}}$ January	$1,250$	$250$	$1,000$	$24,000$
$28^{\text{th}}$ February	$1,250$	$240$	$1,010$	$22,990$
$31^{\text{st}}$ March	$1,250$	$229.90$	$1,020.10$	$21,969.90$

As a certain portion of the EMI is principal repayment, the principal outstanding decreases over time. As a result, interest decreases over time and principal repayment portion of the EMI increases with time.

While the above are general observations, we need to understand how EMI is calculated. The present value of all future payments should equal the loan amount. Compound interest applies for these loans.

Let the loan amount be

\bm{p}

, the EMI be

\bm{e}

, interest rate per time period be

\bm{r \%}

and number of time periods be

\bm{n}

p = \dfrac{e}{\left(1 + \dfrac{r}{100} \right)^{1}} + \dfrac{e}{\left(1 + \dfrac{r}{100} \right)^{2}} + ... + \dfrac{e}{\left(1 + \dfrac{r}{100} \right)^{n}}

p

is the sum of

n

terms in Geometric Progression. We are not going create another formula here. You simply need to remember the sum to

n

terms of GP formula and apply here.

S_{n} = \dfrac{a(1 - r^{n})}{(1 - r)}

, where

\bm{a}

is the first term and

\bm{r}

is the common ratio.
[Note:

\bm{r}

in the above formula is common ratio and not rate of interest.]

Example 12

Jessica borrows Rs.

1,00,000

from HBSC bank at

12 \%

per annum compounded monthly. If she wishes to repay through EMI, a fixed amount every month, over

70

months, what is EMI she pays every month?
[Assume

(1.01)^{70} = 2]

(1)

2000

(2)

2200

(3)

2400

(4)

2500

Solution

Let the EMI amount be

\bm{e}

.
Rate of interest

= \bm{r} = 1 \%

per month

Present Value of Loan = Present value of EMIs

100000 = \dfrac{e}{\left(1 + \dfrac{r}{100} \right)^{1}}

+ \dfrac{e}{\left(1 + \dfrac{r}{100} \right)^{2}} + ... + \dfrac{e}{\left(1 + \dfrac{r}{100} \right)^{70}}

⇒

100000 = \dfrac{e}{(1.01)^{1}}

+ \dfrac{e}{(1.01)^{2}} + ... + \dfrac{e}{(1.01)^{70}}

The RHS is in GP where the first term

\bm{a} = \dfrac{e}{1.01}

, the common ratio

\bm{r} = \dfrac{1}{1.01}

and

\bm{n} = 70

Applying the sum of

n

terms in GP formula, i.e.

\dfrac{a(1 - r^{n})}{(1 - r)}

⇒

100000 = \dfrac{e}{1.01} \times \dfrac{\left(1 - \left(\dfrac{1}{1.01} \right)^{70} \right)}{1 - \dfrac{1}{1.01}}

= \dfrac{e}{1.01} \times \dfrac{\left(1 - \dfrac{1}{1.01^{70}} \right)}{\dfrac{0.01}{1.01}}

⇒

100000 = \dfrac{e}{0.01} \times \left(1 - \dfrac{1}{2} \right) = \dfrac{e}{0.02}

⇒

e = 2000

Answer: (1)

2000

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