Core formula
SI = P x R x T / 100, Amount A = P + SI. Interest is the same every year because it is always charged on the original principal. Target: 30 seconds.
Key formulas
| Find | Formula |
|---|---|
| Principal | P = 100 x SI / (R x T) |
| Rate | R = 100 x SI / (P x T) |
| Time | T = 100 x SI / (P x R) |
| Sum becomes n times in T years | R = (n - 1) x 100 / T |
| Amount A1 after T1 years, A2 after T2 years | yearly SI = (A2 - A1)/(T2 - T1) |
| Different rates for different periods | SI = P x (R1T1 + R2T2 + ...)/100 |
Doubling and multiples
- A sum doubles in 8 years: R = 12.5%. It becomes 3 times in 16 years, 4 times in 24 years (interest grows linearly).
- Becomes n times: total interest = (n - 1) x P.
Instalments and division of a sum
- If SI on two parts must be equal: P1 x R1 x T1 = P2 x R2 x T2, so P1 : P2 = R2T2 : R1T1.
- Lending at two rates with a known total interest: use alligation on rates. Average rate = total interest/total sum x 100.
Worked example
Find the SI on Rs. 8,000 at 7.5% p.a. for 4 years.
SI = 8,000 x 7.5 x 4 / 100 = Rs. 2,400.
Worked example 2
A sum amounts to Rs. 7,200 in 2 years and Rs. 8,100 in 3.5 years at SI. Find the principal and rate.
SI for 1.5 years = 900, so 1 year = 600. P = 7,200 - 2 x 600 = 6,000. R = 600/6,000 x 100 = 10%.
Worked example 3 (two rates)
Rs. 20,000 is lent partly at 6% and partly at 10%. Total annual interest is Rs. 1,600. Average rate = 8%, so by alligation the parts are in ratio (10 - 8) : (8 - 6) = 1 : 1, i.e. Rs. 10,000 each.
Common traps
- Time in months: divide by 12 (9 months = 3/4 year).
- "Becomes 4 times" means interest = 3P, not 4P.
- Mixing up amount and interest.
- Days: for exam questions use 365 days unless told otherwise; count the date of deposit or withdrawal only once.
Worked example 4
In how many years will Rs. 6,000 become Rs. 8,400 at 8% p.a. simple interest?
SI needed = 2,400. One year's SI = 6,000 x 8% = 480. Time = 2,400/480 = 5 years.