Core formula
A = P (1 + r/100)^n and CI = A - P. Interest is earned on interest, so CI grows faster than SI after the first year (for annual compounding, CI = SI in year 1). Target: 40 seconds.
Compounding periods
| Compounded | Rate per period | Number of periods |
|---|---|---|
| Annually | r | n |
| Half-yearly | r/2 | 2n |
| Quarterly | r/4 | 4n |
Effective rates (memorise)
| Rate | 2 years | 3 years |
|---|---|---|
| 5% | 10.25% | 15.7625% |
| 10% | 21% | 33.1% |
| 15% | 32.25% | 52.0875% |
| 20% | 44% | 72.8% |
Two-year effective rate = 2r + r²/100 (the successive-percentage formula with a = b = r).
CI minus SI shortcuts
- 2 years: CI - SI = P (r/100)²
- 3 years: CI - SI = P (r/100)² (3 + r/100) = P r² (300 + r)/100³
- Interest of year k+1 minus interest of year k = interest of year k x r/100.
Amounts in consecutive years
If the amount is A1 after n years and A2 after n + 1 years, then r = (A2 - A1)/A1 x 100. Example: 17,640 and 18,522 give r = 882/17,640 = 5%.
Multiples
If a sum becomes k times in t years, it becomes k² times in 2t years and k³ times in 3t years (8 times in 9 years means 64 times in 18 years).
Equal instalments
Loan P repaid in n equal annual instalments x at r%: P = x/(1 + r/100) + x/(1 + r/100)² + ... + x/(1 + r/100)^n.
Worked example
CI on Rs. 10,000 at 10% p.a. for 2 years = 10,000 x 21% = Rs. 2,100.
Worked example 2
Difference between CI and SI on Rs. 5,000 at 6% for 2 years = 5,000 x 0.0036 = Rs. 18.
Worked example 3
Difference between CI and SI on Rs. 8,000 at 5% for 3 years = 8,000 x 25 x 305/10,00,000 = Rs. 61.
Common traps
- Half-yearly compounding: halve the rate and double the periods.
- Different rates each year: multiply the year factors (1.10 x 1.12 x 1.15), do not add rates.
- CI is asked, not amount (or the reverse).
- For 1.5 years compounded annually, the last half year uses SI on the amount after 1 year.
Worked example 4 (multiples)
A sum doubles in 5 years at CI. It becomes 4 times in 10 years and 8 times in 15 years, because each further 5 years multiplies it by 2 again.
Worked example 5 (depreciation)
A machine worth Rs. 50,000 depreciates by 10% every year. Value after 2 years = 50,000 x 0.9 x 0.9 = Rs. 40,500. Depreciation uses (1 - r/100)^n.