Simple and Compound Interest

Simple Interest (SI): interest paid only on principal.

SI = P × R × T / 100

where P = principal, R = annual rate %, T = time in years.

Total amount: A = P + SI = P(1 + RT/100).

Compound Interest (CI): interest paid on principal AND accumulated interest.

A = P · (1 + R/100)^T, CI = A − P.

Compounded n times per year: A = P · (1 + R/(100n))^(nT).

Half-yearly: n = 2. Quarterly: n = 4.

Continuous compounding (limit): A = P · e^(RT/100). Rare in SSC; common in actuarial.

SHORTCUT 1: CI − SI for 2 years.

CI − SI = P · (R/100)²

Example: P = 10,000, R = 10%, T = 2. CI − SI = 10000 × 0.01 = ₹100. (Verify: SI = 2000, CI = 10000(1.21 − 1) = 2100 → diff = 100.)

SHORTCUT 2: CI − SI for 3 years.

CI − SI = P · (R/100)² · (3 + R/100)

SHORTCUT 3: SI doubling time.

P doubles when SI = P → T = 100/R years.

Example: at 8% SI, money doubles in 12.5 years.

SHORTCUT 4: CI doubling time (Rule of 72).

T ≈ 72/R years.

Example: at 8% CI, money doubles in ~9 years (precise: 72/8 = 9).

Rule of 70 / 69: more accurate alternatives. Rule of 69.3 used by economists.

Worked example. Find CI on ₹8,000 at 5% per annum for 2 years compounded annually.

A = 8000 × 1.05² = 8000 × 1.1025 = 8820. CI = 8820 − 8000 = ₹820.

Quick check via shortcut: SI = 8000 × 5 × 2 / 100 = 800. CI − SI = 8000 × 0.0025 = 20. So CI = 800 + 20 = 820. ✓

INTEREST = money charged for using a borrowed amount.

• Principal (P): original amount borrowed/invested.
• Rate (R): % per year (annum).
• Time (T): in years.
• Amount (A): total to be paid back = P + Interest.

SIMPLE INTEREST (SI):

SI = P × R × T / 100.

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

In SI, interest is calculated on the ORIGINAL principal only (no interest on interest).

Examples:

Q1. SI on ₹5000 @ 8% for 3 years.
SI = 5000 × 8 × 3 / 100 = ₹1200.
A = 5000 + 1200 = ₹6200.

Q2. In how many years does ₹1000 double at 10% SI?
SI = P → 1000 × 10 × T / 100 = 1000 → T = 10 years.

COMPOUND INTEREST (CI):

Interest is added to principal periodically; subsequent interest is calculated on the new amount.

A = P (1 + R/100)^T (yearly compounding).

CI = A − P.

COMPOUNDING FREQUENCIES:

EXAMPLES:

Q3. CI on ₹10000 @ 10% for 2 years.
A = 10000 × (1.1)² = 10000 × 1.21 = ₹12100.
CI = 12100 − 10000 = ₹2100.

Q4. SI on ₹10000 @ 10% for 2 years = ₹2000.

So CI > SI by ₹100. (Always CI > SI for periods > 1.)

KEY DIFFERENCES (SI vs CI):

For 2 years:

• SI₂ = P × 2R/100 = 2RP/100.
• CI₂ = P[(1 + R/100)² − 1] = P[2R/100 + R²/10000].
• CI − SI = PR²/10000 = P(R/100)².

For 3 years:

• CI − SI = PR²(R + 300) / 10⁶ = P(R/100)² × (3 + R/100).

These shortcuts save time.

RULE OF 72:

Quick estimate: years to double at rate r% (CI) ≈ 72/r.

E.g., at 8% CI, money doubles in ~9 years.
(Exact: log 2 / log 1.08 ≈ 9.006.)

ADVANCED PROBLEMS:

Q5. Difference between CI and SI on ₹10000 @ 5% for 2 years.

CI − SI = P(R/100)² = 10000 × (5/100)² = 10000 × 0.0025 = ₹25.

Q6. A sum doubles in 5 years at SI. How long to triple?

Doubling: P + SI = 2P → SI = P → P × R × 5 / 100 = P → R = 20%.

Tripling: SI = 2P → P × 20 × T / 100 = 2P → T = 10 years.

Q7. A sum becomes 4P in 6 years at CI. Find rate.

A = P(1 + R/100)^6 = 4P.
(1 + R/100)^6 = 4.
1 + R/100 = 4^(1/6) = 2^(1/3) ≈ 1.26.
R ≈ 26%.

Q8. What is CI on ₹4000 for 2 years at 5%, compounded half-yearly?

R/2 = 2.5%, periods = 4.
A = 4000 × (1.025)⁴ ≈ 4000 × 1.1038 ≈ ₹4415.20.
CI ≈ ₹415.20.

(Compare with annual compounding @ 5% for 2 years: A = 4000 × 1.1025 = 4410; CI = 410.)

INSTALLMENTS (CI):

If a sum P is to be paid in n equal annual installments X, at rate R (CI):

P = X × [(1 − (1+R/100)^(-n)) / (R/100)].

Use case: EMI on loans.

EXAM HOOKS:

• SI ∝ T (linear); CI grows exponentially.
• CI > SI for T > 1.
• CI − SI for 2 years = P(R/100)².
• Rule of 72: doubling time ≈ 72/r.
• Compounding more frequently → slightly higher CI.
• For very short times, CI ≈ SI.