Simple and Compound Interest

Simple and Compound Interest

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Simple Interest (SI) is computed only on the original Principal (P) at a fixed Rate (R) per annum over Time (T) in years, using SI = (P × R × T) / 100, with the Amount A = P + SI. Compound Interest (CI) adds each year’s interest back into the principal before recalculating, so A = P(1 + R/100)^T and CI = A − P. For NAT-I Quantitative Reasoning, expect 1–2 questions (≈3% weight) that swap P, R, T, A, or CI in the formula and ask you to solve. Remember: for T = 1 year, SI = CI; for T ≥ 2 years, CI > SI. Convert months to years (6 months = 0.5 yr) before substituting, and apply SI on the compounded amount for the leftover fraction.

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🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Core Formulas

• Simple Interest: SI = (P × R × T) / 100, so A = P(1 + RT/100)
• Compound Interest (annual): A = P(1 + R/100)^T, and CI = P(1 + R/100)^T − P
• Compounding n times per year: A = P(1 + R/(100n))^(nT)
• 2-year CI − SI shortcut: Difference = P(R/100)^2
• Effective annual rate: (1 + R/(100n))^n − 1

How the Two Differ

In SI, the yearly interest is constant: every year earns (P·R/100), so the total grows linearly with time. In CI, year-1 interest is added to P, year-2 interest is calculated on the new balance, and so on — the amount grows geometrically. This is why a Rs 10,000 deposit at 10% for 3 years yields SI = Rs 3,000 but CI ≈ Rs 3,310.

Worked Relationship

Let P = 10,000, R = 10%, T = 3 years.

• SI = (10000 × 10 × 3)/100 = Rs 3,000 → A = 13,000
• CI = 10000 × (1.1)^3 − 10000 = 10000 × 1.331 − 10000 = Rs 3,310 → A = 13,310
• Difference = 310 = P(R/100)^2 × (something) — the shortcut P(R/100)^2 gives the 2-year difference only.

Typical NAT-I Question Types

1. Direct substitution — given three of {P, R, T, A, CI}, find the fourth.
2. Rate/time swap — “at what rate will Rs 5,000 become Rs 6,050 in 2 years compounded annually?”
3. Fractional periods — compound for whole years, then apply SI on the resulting amount for the remaining months.
4. Population/depreciation — replace R with a growth or decay percentage in the CI formula.

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Edge Cases and Mechanisms

Fractional time (e.g., 2 years 6 months) is the most common trap. The standard NAT-I convention is: compound for the whole years first (using the CI formula), then apply simple interest on the resulting amount for the leftover fraction. A shortcut: compute A = P(1 + R/100)^2, then A_final = A[1 + (R/100) × 0.5]. Plugging 2.5 straight into (1 + R/100)^2.5 is wrong unless the question explicitly says interest compounds every month at 1/12 of the rate.

Compounding frequency matters when the question states “compounded half-yearly” or “quarterly.” Divide R by n and multiply T by n: A = P(1 + R/(100·2))^(2T) for half-yearly. The effective annual rate is the single rate that, compounded once a year, would produce the same return.

Equal annual installments: If X is paid at the end of each year for T years to settle a loan of P, then P = X[1 − (1 + R/100)^(−T)] / (R/100). Solve for X or P by substitution. This is a common NAT-I twist on the basic formula.

Common Mistakes

• Dividing by 100 twice (once for the rate, once for the percent) and halving the answer.
• Treating Amount as the interest — CI is always A − P.
• Ignoring the instruction “compounded half-yearly” and using annual compounding.
• For T = 1 year, students sometimes subtract SI from CI assuming a difference exists — they are equal.

Practice Prompts

1. A sum of Rs 8,000 at 12% per annum compounded annually for 18 months yields what amount? (Compute 1 year compounded → apply SI for 6 months.)
2. A loan of Rs 50,000 is repaid in 5 equal annual installments at 10% compound interest. Find the installment value using the PV annuity formula.

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