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# Compound Interest | The Complete Lesson (Part-1)

#### What is Compound Interest? Let’s explore the basic concepts for Compound Interest

Compound Interest is part of the wider topic of Interest. In the first part of this topic, we learned about Simple Interest. Well, simple interest is not the only interest mechanism used by lenders and borrowers. There is another method, one which actually runs majority of business operations and goes by the name of Compound Interest.

#### Compound Interest Definitions

Four basic definitions that you need to know in order to figure out
Principal (P): The original sum of money loaned/deposited. Also known as capital.
Interest (I): The amount of money that you pay to borrow money or the amount of money that you earn on a deposit.
Time (T): The duration for which the money is borrowed/deposited. The duration does not necessarily have to be years. The duration can be semi-annual, quarterly or any which may be deemed fit.
Rate of Interest (R): The percent of interest that you pay for money borrowed, or earn for money deposited.

#### Compound Interest Formula:

Time for the actual learning to begin and you should learn the formula for Compound Interest.

Amount Due at the end of the time period is given by

Where:
P: Principal (original amount)
R: Rate of Interest (in %)
T: Time period (yearly, half-yearly etc.)

The Compound Interest over the time period T is given by by the formula:

This can be written as:

Compound Interest= P {(1+r/100)t -1}

Useful Tip: In case of simple interest, principal and interest remain same for every year while in case of compound interest, the amount at the end of each year will be the principal for next year and interest for different years is not the same.

Example 1: Maninder took a loan of Rs. 10000 from Prashant. If the rate of interest is 5% per annum compounded annually, find the amount received by Prashant by the end of three years

Solution: The following is the data given:
Principal, P= 10000
Rate = 5%
Time =3 years
Using the formula for Compound Interest:
A = P(1+R/100)t
So A10000(1+5/100)3
A = 10000(1+1/20)3
A = 10000 x 21/20 x 21/20 x21/20 =11576.25
So, the total amount paid by Maninder at the end of third year is Rs.11576.25

Example 2: Richa gave Rs. 8100 to Bharat at a rate of 9% for 2 years compounded annually. Find the amount of money which she gained as a compound interest from Bharat at the end of second year.

Solution:
Principal value = 8100
Rate = 9%
Time = 2 years
So the total amount paid by Bharat
= 8100(1+9/100)2
=Rs. 9623.61
The question does not probe the amount, rather, it wants to know the CI paid, that the difference between the total amount and original principal.
The Compound Interest = 9623.61 – 8100 = 1523.61

#### Simple Interest vs Compound Interest: Difference between simple interest and compound interest

Simple Interest and Compound Interest are both related to the wider topic of interest. Even though these two are two types of interests on a certain sum or quantity, there is a marked difference between the two. When compounding is done on a sum of money, then the interest for the first year also gets added to the principal while calculating the interest payout for the second year. This happens for all subsequent years or the time duration when the interest is being levied (effectively, interest is being charged on interest itself).

In case the same principal P is invested in two schemes, at the same rate of interest r and for the same time period t, then in that case:
Simple Interest = (P x R x T)/100
Compound Interest = P [(1+R/100)T – 1] So, the difference between them is
= P[(1+R/100)T -1] – PRT/100
= P [(1+R/100)T -1-RT/100]

Two Useful Shortcuts based on the above concept:
Difference between CI and SI when time given is 2 years = P(R/100)2
Difference between CI and SI when time given is 3 years = P[(R/100)3 + 3(R/100)2

If difference and principal is given, you can easily calculate rate %  from this formula.

Example 3: The difference between compound interest and simple interest is 2500 for two years at 2% rate, then find the original sum.

Solution:
Given difference is = 2500
So, Simple Interest = (P X R X T)/100
Compound Interest = P [(1+R/100)t – 1] So the difference between both of them is
=  P[(1+R/100)T -1] – PRT/100
= P [(1+R/100)T -1-RT/100] 2500 = P [{(1+2/100)2-1}-4/100] On simplification this equation, the sum will be = Rs. 6250000

Let’s use the shortcut method:
When time given is 2 years, Difference = P(R/100)2
Since the difference given is Rs 2500, we have
2500 = P (2/100)2
=> 2500 = P (1/50)2
=> 2500= P (1/2500)
=> 6250000=P
So, the sum is Rs.6250000.

#### Compound Interest Exercises: Questions and Answers featuring Basic Compound Interest problems

In the following section, we solve 5 questions that are based on the basics of Compound Interest. You will solve some compound interest questions in this section and with the help of these questions, you will be able to learn the basic concepts for this topic. Remember, learning a topic such through solved problems and questions is the best approach possible.

Question 1: An amount of Rs 1000 is borrowed at CI at the rate of 2% per annum. What will be the amount to be paid after 3 years if the interest is compounded annually?

A. 926.24
B. 1248.34
C. 1061.28
D. 1678.34

Option C.

Question 2: An amount of Rs 1000 amount equates to Rs 1728 in 3 years when interest is compounded annually. What will be the rate per annum?

A. 24%
B. 20%
C. 18%
D. 16%

Option BUsing Formula

Question 3: The difference between the compound interest and the simple interest on a certain sum at 4% rate for 2 years is Rs 100. What will be the amount invested?

A. 45500
B. 52500
C. 62000
D. 62500

Option DThe difference between compound interest and simple interest for two years is given by

Question 4: The difference between the compound interest and the simple interest on a certain sum at 4% rate for 3 years is Rs 100. What will be the amount invested?

A. 20559
B. 25559
C. 16559
D. 28559

Option AThe difference between CI and SI for three years is given by

Question 5: A sum of money invested at compound interest amounts to Rs. 650 at the end of first year and Rs. 676 at the end of second year. The sum of money is:

A. Rs. 600
B. Rs. 540
C. Rs. 625
D. Rs. 560

Option CSimple Interest for one year = Compound Interest for one year

Interest on Rs. 650 for 1 year = 676 –650 = Rs. 26

### Compound Interest Examples, Calculations & Applications

#### Compound Interest Installments: Concept of Equal Installments

How does the concept of equal installments work for Compound interests? Well, in this case, the problem basically tells us that a certain sum of money is borrowed on compound interest for a certain period and it is returned with the help of equal installments.
Let us derive a formula where the amount is returned in two equal installments for a time period of two years.

Assume P to be the principal and r the rate of interest.
Step 1: P[{1+r/100}]= PI (amount of one year)

Step 2: New Principal
Now let X be the first installment. After giving the first installment, the principal value will change and the new principal will be = P1 – X

Step 3: Amount and Interest for the second year
Now the interest charged will be charged on this amount.
Amount at the end of second year is P2 = (P1 –X ){1+r/100}

Step 4: Since the installments are equal, this new amount has to be equal to X.
Hence,
[P(1+r/100)-X][1+r/100]=X
On solving, we have
P (1+R/100)2-X (1+R/100)]= X
P (1+R/100)2]= X+X (1+R/100)
Divide both sides by (1+r/100)2

So we are left with:

where X is the installment and n refers to the number of installments.

#### Compound Interest Examples: Let’s solve an example question to understand the compound interest questions.

Question-1: Richa borrowed a sum of Rs. 4800 from Ankita as a loan. She promised Ankita that she will pay it back in two equal installments.If the rate of Interest be 5% per annum compounded annually, find the amount of each installment.

Solution:
Given that principal value = 4800
Rate =5%
Two equal installments annually = 2 years
Appling the formula, P = X/(1+r/100)n…………………….X/(1+r/100)
So, we have here two equal installments.
P= X/(1+r/100)2 + X/(1+r/100)
4800=X/(1+5/100)2 + X/(1+5/100)
On simplifying
We have x= Rs. 2581.46
So, the amount of each installment is Rs 2581.46

Compound Interest Examples: Use of  Compound Interest for concepts of population growth.
We study two compound interest examples in this section where we study the concept of population growth. These are particular types of compound interest problems and we are going to explain how you approach these questions. Let’s take up one case at a time and then take up some compound interest examples questions to understand these concepts.

Case 1: When population growing in a constant rate
If the population increases with a constant rate say by r% then the population after T years will be
= P (1+R/100)t
In fact, this is nothing else but an application of the fundamentals of compound interest.
It is actually similar to finding the compound amount after time T years
Net population after T years = = P (1+R/100)t
Net population increase = P [(1+R/100)t– 1]

Example 2:The population of Chandigarh is increasing at a rate of 4% per annum. The population of Chandigarh is 450000, find the population in 3 years hence.

Solution:
P = 450000
Rate of increasing = 4%
Time =3 years
Therefore, the total population after 3 years will be
T = P (1+R/100)3
=> T = 450000(1+4/100)3
=> T = 506188

Case 2: When the population growing with different rates and for different intervals of time
If the rate growth of population increased with different rate and for different intervals of time then the population after T years will be =
P (1+R1/100)t1x  (1+R2/100)t2……………………………..  (1+RN/100)tn

Let us take an example for this concept.

Example 3:The population of Chandigarh increased at a rate of 1% for first year, the rate for second year is 2%, and for third year, it is 3%. Then what will be the population after 3 years if present population of Chandigarh is 45000?

Solution:
Since the rate growth of population increased with different rates for the three difference years, the population after T years will be =
P (1+R1/100)t1 x  (1+R2/100)t2x (1+R3/100)t3
45000(1+1/100)1 x (1+2/100)1 x (1+3/100)1=47749.77» 47750

Case 3: When the population is decreasing with rate R
Population after a time period of T years=P (1-R/100)t
Where P is the initial population
R is rate at which the population is decreasing

Example 4:The population of Chandigarh is increases at a rate of 1% for first year, it decreases at the rate of 2% for the second year and for third year it again increases at the rate of 3%. Then what will be the population after 3 years if present population of Chandigarh is 45000?

Solution:
The rate growth of population increases, then decreases for the second year and again it increases for third year.
The population after T years will be
=P (1+1/100)x  (1-2/100)x (1+3/100)
=45000(1+1/100)1 x (1-2/100)1 x (1+3/100)1=45877.23» 45877

This completest the section on population growth problems. Hope with the help of above compound interest examples, you have been able to understand how these problems need to be approached.

Tooltip 3: Negative Compound Interest

As we can see from the last case above, it is not necessary that there is always an increase in any quantity or amount. There can also be a reduction in the amount of something. This reduction is called the rate of depreciation, especially in the financial world. In this case, we do nothing else but take the interest rate to be negative. The formula for this is as follows:
Let P be the original amount.
Let P1 be the new amount at the end of t years.
P1 = P (1-R/100)t
Here R is the rate of interest (negative rate).
Always remember, the rate of depreciation is nothing else but negative rate of interest.

Example 5: Manpreet bought a new car. The value of the car is Rs. 45000. If rate of depreciation is 10% per annum then what will be the value of the car after 2 years

Solution:
Here P = 45000
Rate of depreciation = 10%
T = 2 years
Therefore, the value will be after 2 years = P (1 – R/100)t
= 45000(1-10/100)2
=Rs 36450

Question 1: A builder borrows Rs. 2550 to be paid back with compound in­terest at the rate of 4% per an­num by the end of 2 years in two equal yearly installments. How much will each installment be?

A. Rs. 1352
B. Rs. 1377
C. Rs. 1275
D. Rs. 1283

Option AAmount = Rs 2550

Rate = 4% per annum

Time = 2 years

Applying the formula

P= X/(1+r/100)n+ …………………….X/(1+r/100)

Here we have two equal installments, so

Question 2: A man buys a scooter on making a cash down payment of Rs. 16224 and promises to pay two more yearly installments of equivalent amount in next two years. If the rate of interest is 4% per annum, compounded yearly, the cash value of the scooter, is
A. Rs. 40000
B. Rs. 46824
C. Rs. 46000
D. Rs. 50000

Option BConcept used in this question is: you need to calculate principal for every year unlike simple interest where principal used to be same for every year.

Let principal (present worth) for first year be P1 and that for two years be P2.

The total payment will be ( cash down payment + installments paid )

Cash value of the scooter

= Rs. (16224 + 15600 + 15000) = Rs. 46824.

Question 3: Kamal took Rs. 6800 as a loan which along with interest is to be repaid in two equal annual installments. If the rate of interest is 25/2 %, compounded annually, then the value of each installment is
A. Rs. 4200
B. Rs. 4150
C. Rs. 4050
D. Rs. 4000

Option CLet the annual installment be Rs. z.

Now

Question 4: The population of Chandigarh is increases at a rate of 1% for first year, it decreases at the rate of 4% for the second year and for third year it again increases at the rate of 5%. Then what will be the population after 3 years if present population of Chandigarh is 50000.

A. Rs. 51006

B. Rs. 50904

C. Rs. 50836

D. Rs. 51125

Option BSince the rate growth of population is increasing first and then decreasing for the second year and again it increases for third year, then the population after T years will be

Question 5: A person bought a new machine. The value of the machine is Rs. 10000. If rate of depreciation is 5 % per annum, then what will be the value of the machine after 2 years?

A. Rs. 9025

B. Rs. 9044

C. Rs. 9110

D. Rs. 9080