SIMPLE INTEREST: Simple Interest (S.I.) is the simple method for calculating the interest amount for some principal amount of money. We often have to borrow money from banks in the form of a loan. During payback, apart from the loan principal amount, we have to payback some more money that depends on the loan amount as well as the time for which we borrowed. This is called interest and can be Simple or Compound. This term finds extensive usage in banking.
S.I. = P * R * T / 100
Where S.I. = Simple Interest
P = Principal Amount
R = Rate of interest per annum (in percentage)
T = Time in years
In order to calculate the total amount, the following formula is used:
Amount (A) = Principal (P) + Simple Interest (S.I)
Amount (A) is the total money paid back at the end of the time period for which it was borrowed.
The total amount formula in case of simple interest can also be written as:
A = P (1 + RT)
Here,
A = Total amount after the given time period
P = Principal amount
R = Rate of interest (per annum)
T = Time (in years)
Simple Interest Formula For Months
Let us see the formula to calculate the interest for months. Suppose P be the principal amount, R be the rate of interest per annum and M be the time (in months), then the formula can be written as:
Simple Interest for n months = (P × M × R)/ (12 ×100)
The list of formulas of simple interest for when the time period is given in years, months and days are tabulated below:
| Time | Simple interest Formula | Explanation |
| Years | PTR/100 | T = Number of years |
| Months | (P × M × R)/ (12 ×100) | n = Number of months |
| Days | (P × d × R)/ (365 ×100) | d = Number of days (non-leap year) |
Simple Interest Problems
Let us see some simple interest examples using the simple interest formula in maths.
Example 1:
RAM takes a loan of Rs 10000 from a bank for a period of 1 year. The rate of interest is 10% per annum. Find the interest and the amount he has to pay at the end of a year.
Solution:
Here, the loan sum = P = Rs 10000
Rate of interest per year = R = 10%
Time for which it is borrowed = T = 1 year
Thus, simple interest for a year, SI = (P × R ×T) / 100 = (10000 × 10 ×1) / 100 = Rs 1000
Amount that RAM has to pay to the bank at the end of the year = Principal + Interest = 10000 + 1000 = Rs 11,000
Example 2:
Nimmi borrowed Rs 50,000 for 3 years at the rate of 3.5% per annum. Find the interest accumulated at the end of 3 years.
Solution:
P = Rs 50,000
R = 3.5%
T = 3 years
SI = (P × R ×T) / 100 = (50,000× 3.5 ×3) / 100 = Rs 5250
At the rate of 12% per year simple interest, a sum
· of Rs.6800 will earn how much interest by the end of 4 years?
- a) Rs.3500
b) Rs.3264
c) RS.3624
d) Rs.3462
· Calculate the simple interest on Rs.5600 at 8% p.a. for 6 months.
- a) Rs.250
b) Rs.232
c) Rs.224
d) Rs.242
· Find the interest to be paid at the end of one year on Rs.3250 at 6% per annum.
- a) Rs.195
b) Rs.159
c) Ra.190
d) Rs.100
COMPOUND INTEREST
Compound interest is the interest imposed on a loan or deposit amount. The compound interest for an amount depends on both Principal and interest gained over periods. This is the main difference between compound and simple interest.
Compound interest is the interest calculated on the principal and the interest accumulated over the previous period. It is different from simple interest, where interest is not added to the principal while calculating the interest during the next period. In Mathematics, compound interest is usually denoted by C.I.
Compound Interest Formula:
A = P (1 + r/n)t
- A = amount
- P = principal
- r = rate of interest
- n = number of times interest is compounded per year
- t = time (in years)
- Alternatively, we can write the formula as given below:
- CI = A – P
| Examples 1:A town had 10,000 residents in 2000. Its population declines at a rate of 10% per annum. What will be its total population in 2005?Solution:The population of the town decreases by 10% every year. Thus, it has a new population every year. So the population for the next year is calculated on the current year population. For the decrease, we have the formula A = P(1 – R/100)nTherefore, the population at the end of 5 years = 10000(1 – 10/100)5= 10000(1 – 0.1)5 = 10000 x 0.95 = 5904 (Approx.) |
| Examples 2:The count of a certain breed of bacteria was found to increase at the rate of 2% per hour. Find the bacteria at the end of 2 hours if the count was initially 600000.Solution:Since the population of bacteria increases at the rate of 2% per hour, we use the formulaA = P(1 + R/100)nThus, the population at the end of 2 hours = 600000(1 + 2/100)2= 600000(1 + 0.02)2 = 600000(1.02)2 = 624240 |