Compound Interest (चक्रवृद्धि ब्याज)
Learning Goals
By the end of this chapter you will be able to:
- Find the amount and compound interest for any number of years.
- Handle half-yearly and quarterly compounding.
- Find the difference between compound and simple interest in seconds.
- Find the rate or time when the amount is given.
- Apply the same formula to population growth and depreciation.
The Hook: Interest on Interest
Put ₹10,000 in a bank at 10% a year. With simple interest you get ₹1,000 every year — ₹2,000 in two years. With compound interest, the first year's ₹1,000 is added to your money, so in the second year you earn 10% of ₹11,000 = ₹1,100. Total: ₹2,100. That extra ₹100 is interest on interest — and over many years it grows like a snowball.
Why Students Struggle
- They use the simple-interest formula out of habit.
- They forget to halve the rate and double the time for half-yearly compounding.
- They do long multiplications instead of using shortcuts for 2 and 3 years.
🧠 Active Recall: In the example above, why is the second year's interest ₹1,100 and not ₹1,000?
Chunk 1: The Main Formula
Amount A = P (1 + R/100)ⁿ and CI = A − P
where P = principal, R = rate per cent per year, n = number of years.
Example: P = ₹8,000, R = 5%, n = 2 years.
A = 8000 × (21/20)² = 8000 × 441/400 = ₹8,820. CI = ₹820.
Chunk 2: Compounding More Than Once a Year
| Compounding | Rate per period | Number of periods |
|---|---|---|
| Yearly | R | n |
| Half-yearly | R/2 | 2n |
| Quarterly | R/4 | 4n |
Example: ₹10,000 for 1 year at 20% compounded half-yearly → 10% for 2 periods → A = 10000 × (1.1)² = ₹12,100, CI = ₹2,100 (not ₹2,000).
🧠 Active Recall: For quarterly compounding at 12% for 1 year, what are the rate per period and the number of periods?
Chunk 3: CI − SI Shortcuts
- 2 years: CI − SI = P (R/100)²
- 3 years: CI − SI = P (R/100)² (3 + R/100)
Example: P = ₹5,000, R = 10%, 2 years → 5000 × (1/10)² = ₹50.
Chunk 4: Finding Rate or Time
If A/P is a perfect square for 2 years, take the square root: ₹6,250 becomes ₹7,840 in 2 years → (1 + R/100)² = 7840/6250 = 784/625 → 1 + R/100 = 28/25 → R = 12%.
A sum that doubles in n years at compound interest becomes 4 times in 2n years and 8 times in 3n years.
Chunk 5: Growth and Depreciation
- Population after n years = P (1 + R/100)ⁿ
- Value of a machine after n years = P (1 − R/100)ⁿ
Mnemonic — "PRN-A": Principal, Rate, n years → Amount. Growth adds R, depreciation subtracts R.
Summary Table
| Situation | Formula |
|---|---|
| Amount | P (1 + R/100)ⁿ |
| Compound interest | A − P |
| Half-yearly | rate R/2, periods 2n |
| CI − SI, 2 years | P (R/100)² |
| CI − SI, 3 years | P (R/100)² (3 + R/100) |
| Depreciation | P (1 − R/100)ⁿ |
📅 Review this chapter after 2 days, then solve one practice set with a timer.
⚠️ Caution: Examiner's Trap
- Simple interest in disguise: the option equal to simple interest is always present. Compound interest is larger after the first year.
- Half-yearly confusion: using the yearly rate for half-yearly compounding gives a smaller answer that is placed in the options.
- Amount vs interest: the question asks for interest, but the amount appears as an option — subtract the principal.