3.1 Simple & Compound Interest
Key Takeaways
- Simple interest is charged only on the original principal: SI = (P × R × T) / 100.
- Compound interest is charged on principal plus accumulated interest: A = P(1 + R/100)^T and CI = A − P.
- For the same principal, rate and time, CI is always greater than SI for T ≥ 2 years; for 2 years the difference equals P(R/100)^2.
- When interest is compounded half-yearly, halve the rate and double the time: A = P(1 + R/200)^(2T).
- Time given in months must be converted to years (divide by 12) before substituting into any interest formula.
Why Interest Matters for RRB Group D
Interest problems are a staple of the Mathematics section, which carries 25 questions in the 100-mark CBT. They test whether you can quickly manipulate the basic formulas for simple interest (SI), compound interest (CI), and amounts. Most RRB questions are 1-2 step calculations, but they reward fast unit handling and confident formula recall.
Simple Interest (SI)
Simple interest is calculated only on the original principal for the entire period. Nothing is added back to the principal to earn further interest.
Formulas
- SI = (P × R × T) / 100
- Amount A = P + SI = P(1 + RT/100)
Where P = principal (in ₹), R = rate of interest (% per annum), T = time (in years).
Worked example 1 (basic SI)
Find the simple interest on ₹4,500 at 8% per annum for 3 years.
- SI = (4500 × 8 × 3) / 100
- SI = 108000 / 100 = ₹1,080
- Amount = 4500 + 1080 = ₹5,580
Worked example 2 (find the rate)
A sum of ₹12,000 amounts to ₹14,400 in 2 years at SI. Find the rate of interest.
- SI = 14400 − 12000 = ₹2,400
- 2400 = (12000 × R × 2) / 100 → 2400 = 240R
- R = 10% per annum
Compound Interest (CI)
Compound interest is calculated on the principal plus the interest accumulated in previous periods. Each period's interest is added back to the principal, so the base keeps growing.
Formulas
- Amount A = P(1 + R/100)^T
- CI = A − P = P[(1 + R/100)^T − 1]
Worked example 3 (CI on 2 years)
Find the compound interest on ₹10,000 at 10% per annum for 2 years, compounded annually.
- A = 10000 × (1.10)^2 = 10000 × 1.21 = ₹12,100
- CI = 12100 − 10000 = ₹2,100
For the same data, SI would have been (10000 × 10 × 2)/100 = ₹2,000. The extra ₹100 is the interest earned in year 2 on the ₹1,000 interest of year 1 — the "interest on interest" that defines compounding.
Difference between SI and CI
For the same P, R and T:
- For 2 years: CI − SI = P(R/100)^2
- For 3 years: CI − SI = P(R/100)^2 × (3 + R/100)
Worked example 4 (find principal from the difference)
The difference between CI and SI on a certain sum for 2 years at 5% per annum is ₹25. Find the sum.
- 25 = P × (5/100)^2 = P × 0.0025
- P = 25 / 0.0025 = ₹10,000
CI for Fractional Periods
When interest is compounded half-yearly, divide R by 2 and multiply T by 2:
- A = P(1 + R/200)^(2T)
When compounded quarterly: A = P(1 + R/400)^(4T).
Worked example 5 (half-yearly compounding)
Find the CI on ₹8,000 at 10% per annum for 1.5 years, compounded half-yearly.
- Rate per half-year = 10/2 = 5%; number of half-year periods = 1.5 × 2 = 3
- A = 8000 × (1.05)^3 = 8000 × 1.157625 = ₹9,261
- CI = 9261 − 8000 = ₹1,261
Formula Reference Table
| Case | Formula |
|---|---|
| Simple Interest | SI = PRT / 100 |
| SI Amount | A = P(1 + RT/100) |
| CI Amount (annual compounding) | A = P(1 + R/100)^T |
| CI Amount (half-yearly) | A = P(1 + R/200)^(2T) |
| CI Amount (quarterly) | A = P(1 + R/400)^(4T) |
| CI − SI for 2 years | P(R/100)^2 |
| CI − SI for 3 years | P(R/100)^2 × (3 + R/100) |
Finding the Missing Variable
Every SI/CI question is the same formula with one unknown. Identify which of P, R, T or A is asked, isolate it, and substitute the rest. For SI questions the algebra is linear; for CI questions it is usually a power or root.
Worked example 6 (find time from amount under SI)
In how many years will a sum of ₹6,000 become ₹7,500 at 5% per annum simple interest?
- SI = 7500 − 6000 = ₹1,500
- T = (100 × SI) / (P × R) = (100 × 1500) / (6000 × 5) = 150000 / 30000 = 5 years
Worked example 7 (find principal from amount under CI)
What sum will amount to ₹4,410 in 1 year at 10% per annum compounded half-yearly?
- Half-yearly rate = 5%, periods = 2
- 4410 = P × (1.05)^2 = P × 1.1025
- P = 4410 / 1.1025 = ₹4,000
The question is engineered so the compounding factor divides cleanly. When it does not, work backwards from the options instead of long-dividing: multiply each option by 1.1025 and see which gives the stated amount.
CI When Rates Differ Year to Year
If rates are R1%, R2%, R3% for successive years, the amount is:
- A = P × (1 + R1/100) × (1 + R2/100) × (1 + R3/100)
This form appears when a question says "10% in the first year, 12% in the second" — do not average the rates.
Quick Mental-Math Shortcuts
| Problem | Shortcut |
|---|---|
| CI for 2 years at R% | Effective rate = R + R + R²/100 (% of P) |
| CI for 3 years at R% | Effective rate = R + R + R + 3R²/100 + R³/10000 (% of P) |
| Money doubles at R% SI | Time = 100/R years; under CI the "rule of 72" gives roughly 72/R years |
| Population-growth formula | P_final = P (1 + R/100)^T — same as CI |
Common Exam Traps
- "Payable half-yearly" does NOT mean R is already the half-yearly rate — R is the annual rate; you must split it.
- "For 18 months" = 1.5 years; under half-yearly CI, treat as 3 half-year periods.
- Time given in months must be converted to years (divide by 12) before substituting into any formula.
- "Rate 5% per quarter" means the annual rate is 20%; multiply by the number of quarters in a year to get R before using the standard annual formula.
- If a question asks for the "amount", give P + interest; if it asks for the "interest", give only the interest portion.
- "Compounded annually" is the default — if the question does not say otherwise, use A = P(1 + R/100)^T, not the half-yearly version.
- Population, depreciation and bacterial-growth problems use the same CI formula; depreciation uses (1 − R/100) instead of (1 + R/100).
- The 2-year CI − SI shortcut P(R/100)^2 works ONLY for 2 years — for 3 years you must use the longer form P(R/100)^2 × (3 + R/100), not the 2-year expression twice.
Find the simple interest on ₹5,000 at 6% per annum for 2 years.
The difference between compound interest and simple interest on a sum for 2 years at 10% per annum (compounded annually) is ₹40. The sum is:
₹2,000 at 5% per annum simple interest amounts to ₹2,400 in how many years?