# EXAMPLE PROBLEMS ON SIMPLE INTEREST

What is simple interest ?

If the interest on a sum borrowed for a certain period is reckoned uniformly, then it is called simple interest.

By definitionsimple interest is the interest amount for a particular principal amount of money at some rate of interest.

Simple interest = (PRT/100)

P = Principal, R = Rate of interest per annum and T = number of years

R can be converted as decimal from percentage, then don't have to use the denominator as 100.

To find amount,

Amount = Simple interest over a period of time + Principal

Find the simple interest charged when :

Example 1 :

\$5000 is borrowed for 1 year at 12% per annum simple interest

Solution :

Principal P = \$5000, R = 12%, T= 1 year

Simple interest I = (P × T × R)/100

= (5000 × 12)/100

= 60000/100

= 600

Simple interest for 1 year is \$600.

Example 2 :

\$2500 is borrowed for 2 years at 8% per annum simple interest

Solution :

P = \$2500, R = 8%, T= 2 years

Simple interest I = (P × T × R)/100

= (2500 × 2 × 8)/100

= 40000/100

= 400

Simple interest for 2 years is \$400.

Example 3 :

\$40000 is borrowed for 5 years at 11% per annum simple interest

Solution :

P= \$40000, R = 11%, T = 5 years

Simple interest I = (P × T × R)/100

= (40000 × 5 × 11)/100

= 2200000/100

= 22000

Simple interest for 5 years is \$22000.

Example 4 :

\$250000 is borrowed for 9 months at 20% per annum simple interest.

Solution :

P = \$250000, Rate R = 20%, Time T= 9 months

1 year = 12 months

Time T = 9/12 ==> 3/4

Simple interest I = (P × T × R)/100

= (250000 × 3/4 × 20)/100

= 3750000/100

= 37500

Simple interest for 9 months is \$37500.

## Example Problems on Finding Amount on Simple Interest

Find the total amount needed to repay a loan of :

Example 5 :

\$2400 borrowed for 3 years at 10% per annum simple interest.

Solution :

P = \$2400, R = 10%, T = 3 years

Simple interest I = (P × T × R)/100

= (2400 × 3 × 10)/100

= 72000/100

= 720

Simple interest for 3 years is \$720.

Total amount = P + I

= 2400 + 720

= 3120

So, total amount is \$3120.

Example 6 :

\$8000 borrowed for 7 years at 12% per annum simple interest

Solution :

P = \$8000, R = 12%, T = 7 years

Simple interest I = (P × T × R)/100

= (8000 × 7 × 12)/100

= 672000/100

= 6720

Simple interest for 7 years is \$6720.

Total amount = P + I

= 8000 + 6720

= 14720

So, total amount is \$14720.

Example 7 :

\$7500 borrowed for 2 1/2 years at 8% per annum simple interest.

Solution :

P = \$7500, R = 8%, T = 5/2 years

Simple interest I = (P × T × R)/100

= (7500 × 5/2 × 8)/100

= 150000/100

= 1500

Simple interest for 2 1/2 years is \$1500.

Total amount = P + I

= 7500 + 1500

= 9000

So, total amount is \$9000.

Example 8 :

\$23000 borrowed for 4 months at 15% per annum simple interest.

Solution :

P = \$23000, R = 15%, T= 4 months

1 year = 12 months, T = 4/12 ==> 1/3

Simple interest I = (P × T × R)/100

= (23000 × 1/3 × 15)/100

= 115000/100

= 1150

Simple interest for 9 months is \$1150.

Total amount = P + I

= 23000 + 1150

= 24150

So, total amount is \$24150.

Example 9 :

Kyle borrows \$25000 at 6% p.a. Simple interest for 4 years.

(a)  Find the total amount needed to repay the loan.

(b)  Calculate the monthly repayment required to pay this loan off in 48 equal instalments.

Solution :

P = \$25000, R = 6%, T = 4 years

Simple interest I = (P × T × R)/100

= (25000 × 4 × 6)/100

= 600000/100

= 6000

Simple interest for 4 years is \$6000.

Total amount = P + I

= 25000 + 6000

= 31000

So, total amount is \$31000.

(b) Amount is being paid for 48 equal monthly instalments.

\$31000 = 48 months

= 31000 ÷ 48

Instalment amount for a month = \$645.83

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