Simple and compound interest
Simple interest is the same every year; compound interest adds each year's interest to the principal, so it earns interest on interest.
Formula sheet
| Item | Formula |
|---|---|
| Simple interest | SI = P × R × T ÷ 100 |
| Amount (simple) | A = P + SI = P(1 + RT/100) |
| Compound amount, yearly | A = P(1 + R/100)ⁿ |
| Compounded half-yearly | A = P(1 + R/200)^(2n) |
| Compounded quarterly | A = P(1 + R/400)^(4n) |
| Different rates in different years | A = P(1 + R₁/100)(1 + R₂/100)… |
| CI − SI for 2 years | P(R/100)² |
| CI − SI for 3 years | P(R/100)²(3 + R/100) |
| Sum doubles under SI in | 100 ÷ R years |
| Sum becomes k times under SI in | (k − 1) × 100 ÷ R years |
Simple and compound interest
For one year compounded yearly, SI and CI are equal. The gap appears from the second year.
Worked example 1: simple interest
Meena borrows ₹12,000 from a self-help group at 9% simple interest per annum. How much must she repay after 2 years 6 months?
- Standard method: T = 2.5 years. SI = 12,000 × 9 × 2.5 ÷ 100 = ₹2,700. Amount = 12,000 + 2,700 = ₹14,700.
- Shortcut (total per cent): 9% a year for 2.5 years = 22.5% in all. 10% of 12,000 = 1,200, so 22.5% = 1,200 × 2.25 = ₹2,700.
- Check: 2,700 ÷ 2.5 = ₹1,080 a year, which is 9% of 12,000.
₹14,700 (interest ₹2,700).
Worked example 2: compound interest and the SI gap
Find the compound interest on ₹15,000 for 2 years at 10% per annum, compounded yearly, and its difference from the simple interest.
- Standard method: A = 15,000 × 1.1 × 1.1 = 15,000 × 1.21 = ₹18,150, so CI = ₹3,150.
- Shortcut (year by year): year 1 interest = 10% of 15,000 = 1,500; year 2 = 1,500 + 10% of 1,500 = 1,650. CI = 1,500 + 1,650 = ₹3,150 (an effective 21% for 2 years).
- SI = 15,000 × 10 × 2 ÷ 100 = ₹3,000, so the gap is ₹150.
- Check the gap: P(R/100)² = 15,000 × 0.01 = ₹150.
CI ₹3,150; it exceeds SI by ₹150.
Worked example 3: find the sum from the CI-SI gap
The difference between the compound and simple interest on a sum for 2 years at 8% per annum is ₹96. Find the sum.
- Standard method: let the sum be P. SI = 0.16P; CI = P(1.08² − 1) = 0.1664P. Difference = 0.0064P = 96, so P = ₹15,000.
- Shortcut: P = gap ÷ (R/100)² = 96 ÷ 0.0064 = ₹15,000.
- Check: SI = ₹2,400; CI = 15,000 × 1.1664 − 15,000 = ₹2,496; the gap is ₹96.
₹15,000.
Bank and insurance papers love CI − SI gaps, half-yearly compounding and 'in how many years will it double'; SSC and railway papers keep to straight SI and two-year CI. Placement tests often give the amount and ask for the rate or the principal.
2 marks: “At what rate of simple interest will a sum double itself in 8 years?” Interest must equal the principal: R = 100 ÷ 8 = 12.5% per annum.
5 marks: “Find the compound interest on ₹15,000 for 2 years at 10% per annum compounded yearly, and its difference from the simple interest.” A = 15,000 × 1.21 = ₹18,150; CI = ₹3,150; SI = ₹3,000; gap ₹150 = P(R/100)².
10 marks: “(a) Find the SI on ₹12,000 at 9% for 2½ years. (b) Find the CI on ₹8,000 for 1 year at 10% per annum compounded half-yearly. (c) The CI − SI gap on a sum for 2 years at 8% is ₹96; find the sum.” (a) ₹2,700. (b) 8,000 × 1.05² = ₹8,820, so CI = ₹820. (c) ₹15,000. Show the formula, the working and a check for each.
SI = PRT/100, the same every year. CI = P(1 + R/100)ⁿ − P. The two-year gap is P(R/100)². Half-yearly: halve the rate, double the time.