CBSE Class 10 Mathematics Simple and Compound Interest Notes Set 17

Simple Interest

Simple interest is calculated on the original principal only: SI = P × R × T/100, where P is the principal, R the rate per cent per annum and T the time in years. The amount is A = P + SI = P(1 + RT/100).
Months are converted to years (9 months = 3/4 year). Equal interest on two sums means P1R1T1 = P2R2T2. If a sum becomes n times itself in T years, the rate is R = 100(n − 1)/T.

Compound Interest

In compound interest the interest of each period is added to the principal for the next period: \( A = P(1 + R/100)^n \) and CI = A − P.
If the interest is compounded half-yearly, use rate R/2 and 2n periods; if quarterly, rate R/4 and 4n periods. For different rates in successive years, A = P(1 + R1/100)(1 + R2/100)(1 + R3/100).
If a sum becomes k times in t years, it becomes k² times in 2t years and k³ times in 3t years. The same idea applies to depreciation, \( A = P(1 - R/100)^n \), and to population growth.

Difference Between CI and SI

For 2 years: CI − SI = P(R/100)². For 3 years: CI − SI = P(R/100)²(3 + R/100).
The rate can also be found from CI − SI = SI for one year × R/100.
For equal instalments x paid at the end of each of n years at compound interest, the sum borrowed = x/(1 + R/100) + x/(1 + R/100)² + ... \( + x/(1 + R/100)^n \).

Worked Examples

Illustration 1. Find the simple interest on Rs. 6000 for 2 years 6 months at 8% per annum.
Answer:
SI = 6000 × 8 × 2.5/100 = Rs. 1200.

Illustration 2. Find the compound interest on Rs. 8000 for 2 years at 5% per annum.
Answer:
Amount = 8000 × 1.05² = Rs. 8820.
CI = 8820 − 8000 = Rs. 820.

Illustration 3. The difference between CI and SI on a sum for 2 years at 10% is Rs. 30. Find the sum.
Answer:
P × (10/100)² = 30, so P = Rs. 3000.