# Quant Ability Study Material for MBA Exams

# Compound Interest

In compound interest, where the interest for each period is added to the principal before interest is calculated for the next period. With this method the principal grows as the interest is added to it. This method is used in investments such as savings account and bonds.

The borrower and the lender agree to fix up a certain unit of time (say yearly or half-yearly or quarterly) to settle the previous account. In such cases, the amount after the first unit of time becomes the principal for the second unit. The amount after second unit becomes the principal for the third unit and so on. After a specified period, the difference between the amount and the money borrowed is called Compound Interest for that period.

In case of compound interest, the total interest received in the present year will be added to the original principal.

For example, suppose you lend Rs. 10,000 (principal) for 3 years at 10% per annum. So, you get Rs. 1000 as interest per annum (simple interest). For three years, interest will be Rs. 3000 and thus you will get total amount of Rs. 13,000.

Compound interest involves interest on interest too, thus it gives you better amount after 3 years. While calculating the compound interest, the principal amount keeps changing year after year (if the interest is compounded annually).

After 1 year: Interest = Rs. 1,000; New Principal = Rs. 11,000

After 2 years: Interest = Rs. 1,100; New Principal = Rs. 12,100

After 3 years: Interest = Rs. 1,210; You get Rs. 13,310.

So, there is gain of Rs. 310 if you lend at compound interest.

The amount A due after t years, when a principal P is given on compound interest at the rate R% per annum is given by:

**A = P(1 + R/100) ^{t}**

**Compound interest (CI) = A - P**

Simple interest and compound interest for 1 year at a given rate of interest per annum are always equal.

If the interest is compounded **half-yearly**, then

A = P(1 + R/200)^{2t}

If the interest is compounded **quarterly**, then

A = P(1 + R/400)^{4t}

In general, if the interest is compounded n times a year, then

A = P(1 + R/(n×100))^{nt}

### Examples

**Example 1:** Mohan invested an amount of Rs. 15000 at compound interest rate 5% per annum for a period of 2 years. What amount will he receive at the end of 2 years?

Here, P = 15000, R = 5 and, t = 2

A = 15000(1 + 5/100)^{2}

A = (15000 × 21 × 21)/(20 × 20) = Rs. 16,537.50

**Example 2:** Rashi invested Rs. 16000 for two years at compound interest and received an amount of Rs. 17640 on maturity. What is the rate of interest?

Here, A = 17640, P = 16000, t = 2

17640 = 16000 × (1 + R/100)^{2}

17640 = 16000/10000 × (100 + R)^{2}

(100 + R)^{2} = 11025

100 + R = 105

R = 5

So, rate of interest is 5%.

**Example 3:** Find the compound interest on Rs. 1000 at 40% per annum compounded quarterly for 1 year?

Here, P = 1000, R = 40 and, t = 1

A = 1000 × (1 + 40/400)^{4}

A = 1000 × (11/10)^{4}

A = Rs. 1464.10

CI = A - P

CI = Rs. 1464.10 - Rs. 1000

CI = Rs. 464.10

**Example 4:** Anu invests Rs. 5000 in a bond which gives interest at 4% per annum during the first year, 5% during the second year and 10% during the third year. How much does she get at the end of the third year?

Here, P = 5000, R_{1} = 4, R_{2} = 5 and, R_{3} = 10

Amount at the end of third year = P(1 + R_{1}/100)(1 + R_{2}/100)(1 + R_{3}/100)

A = 5000 × (1 + 4/100) × (1 + 5/100) × (1 + 10/100)

A = 5000 × 26/25 × 21/20 × 11/10

A = Rs. 6006

**Example 5:** What will be the difference between simple and compound interest on a sum of Rs. 4500 put for 2 years at 5% per annum?

Here, P = 4500, R = 5, T = 2

SI = (4500 × 5 × 2)/100

SI = Rs. 450

For compound interest,

A = 4500 × (1 + 5/100)^{2}

A = (4500 × 21 × 21)/(20 × 20)

A = Rs. 4961.25

CI = A - P = 4961.25 - 4500 = 461.25

Difference between CI and SI

CI - SI = 461.25 - 450 = Rs. 11.25

**Example 6:** A certain sum of money placed at compound interest doubles in 3 years, then in how many years it will become 8 times?

Let the principle be Rs. P.

A = 2P = P(1 + R/100)^{3}

2 = (1 + R/100)^{3} [Equation 1]

Let, the sum become 8 times in t years.

8P = P(1 + R/100)^{t}

2^{3} = (1 + R/100)^{t}

2 = (1 + R/100)^{t/3} [Equation 2]

Equating both the equations,

t/3 = 3

t = 9 years

In every 3 years, the principal becomes doubles of itself. So, in 6 years it will be 4 times and in next three years it will be double of 4 times, that is 8 times of original principal. So, the required answer is 9 years.

**Example 7: **A man took Rs. 5000 at 10% simple interest and gave it to another person at 10% compound interest, which is being compounded annually. After 3 years, how much extra money he will get?

Simple Interest on Rs. 5000 = (5000 × 10 × 3)/100 = Rs. 1,500

He has to pay amount = Rs. 5000 + Rs. 1500 = Rs. 6500

Amount with compound interest = 5000(1 + 10/100)^{3} = Rs. 6655

So, the answer is 6655 - 6500 = Rs 155.