1. Amount to be repaid after N years if simple interest is applied = P + (P x N x R) = P (1 + N x R)

2. Simple Interest = P x N x R

3. Amount to be repaid after N years if interest is compounded = P [(1 + R)^N]

4. Compound Interest = [P x (1 + R)^N] - P

Question: Simple interest on a certain sum of money for 3 years at 8% per annum is half the compound interest on Rs.4000 for 2 years at 10% per annum. Calculate the principal placed on simple interest.

Solution:

Let the Principal be Rs. P

Then, SI = (P x R x T) = 0.24P

Given CI = 4000(1 + 0.1)^{2} – 4000 = 4000(1.21 – 1) =4000 x 0.21

According to the question,

0.24P = 2000 x 0.21

=> P = 2000 x 0.21 / 0.24 = 2000 x 7 / 8 = Rs. 1750

Tip #2: If the interest rate is applied on a half-yearly, quarterly or monthly basis, the effective annual rate is calculated by compounding the interest

Question: What is the effective annual rate of interest corresponding to a nominal rate of 6% per annum payable half-yearly?

Solution:

Let the Principal be Rs. P.

Now, the rate is 6% per annum but the interest has to be paid twice a year (i.e.) an interest of 3% is applied every 6 months.

Amount to be repaid after 1 year = P x 1.03 x 1.03 = 1.0609P

=> Effective Annual Rate of interest = 6.09%

Tip #3: Use logarithms to find the time when compound rates are applied

1. log 2 = 0.301

2. log 3 = 0.477

3. log 4 = 0.602

4. log 5 = 0.699

5. log 6 = 0.778

6. log 7 = 0.845

Question: At 3% annual interest compounded monthly, how long will it take to double your money?

Solution:

Let the number of months be n and the Principal be Rs. P.

Then, P(1 + 0.03)^{n} = 2P

=> (1 + 0.03)^{n} = 2

=> n log ( 1.03) = log 2

=> n = log 2 / log 1.03 = 0.301 / 0.128 = 23.5

Thus. It’ll take 1 year and 11.5 months.

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Simple and Compound Interest

Aptitude Test on Simple & Compound Distance: Tips and Aptitude Questions