Simple & Compound Interest (RRB)

Simple Interest (SI) is interest on the original principal only.

Formula: SI = P × R × T / 100
where P = principal (₹), R = rate per annum (%), T = time in years.

Amount under SI: A = P + SI = P(1 + RT/100).

Compound Interest (CI) adds the previous interest to the principal each period.

Formula (compounded annually): A = P(1 + R/100)ⁿ where n = number of years. CI = A − P.

Compounded k times per year: A = P(1 + R/(100·k))ⁿᵏ. Common cases: half-yearly k=2, quarterly k=4.

Key facts:

• SI is the same each year. CI grows each year because the base grows.
• CI for year 1 = SI for year 1, always. The difference starts from year 2.
• 2-year CI−SI difference: P · (R/100)². This is the most-asked RRB shortcut.
• 3-year CI−SI difference: P · (R/100)² · (R + 300)/100, or use P·R²(R+300)/100³.

Doubling under CI — uses Rule of 72: time to double ≈ 72 / R years. So at 8% pa, money doubles in about 9 years.

Quick conversion when R is small enough that R/100 simplifies:

• 5% → 1/20, 10% → 1/10, 12.5% → 1/8, 20% → 1/5, 25% → 1/4.
  This lets you compute (1 + R/100) as a clean fraction: 1 + 1/20 = 21/20.

Example: P = ₹8000 at 5% for 2 years (CI annually).
A = 8000 × (21/20)² = 8000 × 441/400 = 20 × 441 = ₹8820. So CI = ₹820.

Example 1 — SI on doubling:
If a sum doubles in 8 years at SI, in how many years will it triple?
Method: "Doubling" means SI = P; so P = P·R·8/100 ⟹ R = 12.5%. "Tripling" means SI = 2P; so 2P = P·12.5·T/100 ⟹ T = 16 years.

Shortcut: at SI, time to multiply by n is (n−1) × (time to double). So time to triple = 2 × 8 = 16.

Example 2 — CI−SI 2-year difference:
Find the difference between CI and SI on ₹5000 at 8% for 2 years.
Method: Diff = P(R/100)² = 5000 × (8/100)² = 5000 × 64/10000 = ₹32.

Example 3 — CI half-yearly:
₹4000 at 10% pa compounded half-yearly for 1 year.
Method: effective rate per half-year = 5%, n=2. A = 4000 × (21/20)² = 4000 × 441/400 = ₹4410. CI = ₹410.

Example 4 — Population growth (CI logic):
A town's population is 64,000 and grows at 5% pa. Population after 2 years = 64000 × (21/20)² = 64000 × 441/400 = 70,560.

Example 5 — Depreciation (negative CI):
A machine worth ₹12,000 depreciates 10% pa. Value after 3 years = 12000 × (9/10)³ = 12000 × 729/1000 = ₹8,748.

Common traps:

1. "Per annum compounded half-yearly" — always halve rate AND double time.
2. CI questions saying "rate of interest" without specifying compounding period — assume annual.
3. CI − SI for time = 1 year is always 0 (one period: both methods give the same interest).