Placement Aptitude: Simple and Compound Interest – Formulas and Concepts

1. Interest

Interest is money paid to the lender by the borrower for using lenders’ money for a specifies period or time. It is the cost of using a sum of money over a period of time. Various terms used and their representation are as follows :

  • Interest – The money paid by the borrower for using lender’s money. Often denoted by “I”.
  • Principal – The original sum that was initially borrowed. Often denoted by “P”.
  • Time – The time period for which the money is borrowed. Often denoted by “T” or “n”
  • Rate of Interest – The rate at which interest is calculated on the principal amount. It is expressed as a percentage often denoted by “R” or “r”
  • Amount – The sum of Principle and Interest, often denoted by “A”
2. Simple Interest
  • When Interest is calculated on the original principal, such interest is called Simple Interest.
  • In case of Simple Interest, the principal remains the unchanged every year. Basically the interest for any year is the same as that for any other year.
  • $SI = \frac{PRT}{100}$
  • $A = \frac{P+PRT}{100} = P{\displaystyle\LARGE[}1+\frac{RT}{100}{\displaystyle\LARGE]}$

Q: Ayesha wants to save up for a vacation. She deposits ₹10,000 in a bank account that offers a simple interest rate of 6% per annum. If she plans to use the money after 3 years, how much interest will she earn, and what will be the total amount she has after 3 years?

Sol: Principal (P) = ₹10,000

Rate of Interest (R) = 6% per annum

Time (T) = 3 years

Simple Interest (SI)=P×R×T/100I=10,000×6×3/100= ₹1,800

Total Amount after 3 years =₹10,000+₹1,800=₹11,800

3. Compound Interest
  • Under Compound Interest, the interest is added to the principal at the end of each period to arrive the new principle for the next period. Basically, the Amount at the end of 1st year becomes the Principle for the 2nd year and so on for the consecutive years.
  • $A = P{\displaystyle\LARGE[}1+\frac{R}{100}{\displaystyle\LARGE]}^n$
  • The compound Interest for the given period can be calculated as follows
  • $CI = A – P$

Q: Bob invests ₹15,000 in a bank account that offers a compound interest rate of 5% per annum, compounded annually. How much money will he have in his account after 2 years?

Solution: Principal (P) = ₹15,000

Rate of Interest (R) = 5% per annum

Time (T) = 2 years

$A = 15000*{\displaystyle\LARGE[}1+\frac{5}{100}{\displaystyle\LARGE]}^2$ = ₹16,537.50

So, Bob will have ₹16,537.50 in his account after 2 years.

Compounding at Different Rates

  • When the rate of Interest for each year is different each year we calculate the Amount for the defined period as follows
  • $A = P{\displaystyle\LARGE[}1+\frac{r_1}{100}{\displaystyle\LARGE]}{\displaystyle\LARGE[}1+\frac{r_2}{100}{\displaystyle\LARGE]}{\displaystyle\LARGE[}1+\frac{r_3}{100}{\displaystyle\LARGE]}$

Compounding more then once a year

  • Compounding can also be done more frequently than once a year. For Example, If interest is added to the principle every 6 months, we say that compounding is done every “twice a year”, If interest is added to the principle every 4 months, we say that compounding is done “thrice a year” and so on.
  • $A = P{\displaystyle\LARGE[}1+\frac{r/x}{100}{\displaystyle\LARGE]}^{nx} $; where x = times compounding is done every year.
  • when compounded, half yearly (x = 2), quarterly (x = 3), every 3 months (x = 4), monthly (x = 12)

Q. A scientist is observing the growth of a bacteria colony. The initial population of the bacteria is 500, and it doubles every 4 hours. How many bacteria will there be after 12 hours?

Sol: Initial Population (P₀) = 500 bacteria

Doubling Period = 4 hours

Time (T) = 12 hours

Formula for Exponential Growth (since population doubles every time period):

Population after T hours $(P)=P0×2^{\frac{T}{Doubling Period}}$

$P=500 * 2^{\frac{12}{4}} = 500×2^3=500×8=4,000$ bacteria

So, after 12 hours, the population of the bacteria will be 4,000.

4. Depreciation of Value
  • The value of an machine/asset is subject to wear and tear, decrease with time. This reduction is value of an asset is known as depreciation and is calculated as: $V_1 = V_0{\displaystyle\LARGE[}1-\frac{r}{100}{\displaystyle\LARGE]}^n$
  • Here, [V1] is value of asset after n years, [V0] is value of asset at the beginning of the period.

Q. Tom buys a new car for ₹8,00,000. The car’s value depreciates by 15% every year. What will be the value of the car after 3 years?

Solution: Initial Value of Car (V₀) = ₹8,00,000

Rate of Depreciation (R) = 15% per annum

Time (T) = 3 years

Value after T years = $800000{\displaystyle\LARGE[}1-\frac{15}{100}{\displaystyle\LARGE]}^3$ =₹4,91,280$

So, after 3 years, the value of the car will be ₹4,91,280.

You can also refer following videos to get more perspective on the topic:

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