Lesson No. 1: Simple Interest

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Let S.I. be Simple Interest,

P, the Principal Or the sum of money borrowed or lent or invested, T, the time period in years and R, the rate of interest p.a. (per annum)

Then, the interest to be paid on P for T years at R% rate of interest p.a. is:

Note:

From the above formula, also remember the following for:

P = (S.I. × 100)/ (T × R)

T = (S.I. × 100)/ (P × R)

R = (S.I. × 100)/ (T × P)

Unless otherwise mentioned, Amount stands for:Sum of principal and interest (simple interest or compound interest) i.e. A = P + S.I. or A = P + C.I.

Note:

  1. In this method of S.I., interest is always calculated on the original principal, i.e. the sum of money invested (or lent or borrowed). Therefore, as a consequence, interest for every year is same. Again, if interest for one length of time is known, simple interest for any other length of time can be found. For example, if S.I. is 200 for 1 year on some principal and rate of interest, then it is 500 for 5 years, 350 for 3.5 years and so on.
  2. This is unlike C.I., in which method, interest is regularly added to the original principal at agreed intervals of time to create a new and fresh principal for calculating interest for the subsequent period of time.

Example 1:

What interest will a sum of money 1000 invested at a rate of 5% p.a. yield in 2 years, if the interest is calculated on the original principal? Also, find the amount of money after the said time period.

Solution:

Since interest is calculated on principal invested at the beginning, therefore S.I. formula will work here.

Here, P = 1000, R = 5, T = 2

Apply the formula:

S.I. = (P × T × R)/100

S.I. = (1000 × 2 × 5)/100 = 100

Amount of money accruing in 2 years of time is sum of principal and interest, i.e. A = 1000 + 100 = 1100.


Example 2:

A certain sum of money (principal) amounts to 1200 in 3 years and 1300 in 5 years under S.I. Find the sum of money and the interest.

Solution:

1200 and 1300 are amounts of money (principal + interest) for 3 years and 5 years.

So, what should the difference 1300 ― 1200 represent? It stands for S.I. for the time period of 2 years.

So, S.I. is 100 for 2 years

100 in 2 years is how much in 3 years (or 5 years?)

It is 150

Therefore, we can write:

Amount in 3 years = P + S.I. for 3 years. i.e.

1200 = P + 150.

Therefore, principal P = 1200 – 150 = 1050

Now, to find the rate of interest p.a. under S.I., apply:

R = (S.I. × 100)/ (T × P)

R = (450 × 100)/ (3 × 1050) = 100/7 = 14% approximately.


Example 3:

Under S.I., in how much time will a principal invested at 8% p.a. double?

Solution:

Amount is twice P, therefore S.I. is 2P – P = P

Now, to find the time apply:

T = (S.I. × 100)/ (R × P)

T = (P × 100)/ (8 × P) = 100/8 = 12.5 years.


Example 4:

Under S.I., a principal generates 90 more interest in 1 year had it been invested at 3% more rate of interest p.a. Find the principal.

Solution:

Interest increases by 90, if rate of interest is increased by 3%. Therefore, 90 is the interest on some Principal P invested at 3% p.a.

So that we can use:

P = (S.I. × 100)/ (R × T)

P = (90 × 100)/ (3 × 1) = 3000

So, the principal invested is 3000