Comparing simple and compound interest

Learning outcome

Compare simple and compound interest on matching terms, and use a two-year difference shortcut only when its conditions hold.

Concepts and assumptions

A fair numerical comparison fixes the same principal, time and stated rate. Simple interest uses the original principal each year; compound interest reinvests earlier interest. Amount differences equal interest differences when the original principal is the same.

For P > 0, a constant annual rate r%, and n complete years:

SI = P × r × n/100.

Annual-compounding CI = P × (1 + r/100)^n - P.

With the same positive annual rate and annual compounding, SI and CI agree after one year; CI is greater after two or more complete years. At zero rate, both are zero. This first-year equality need not hold with subannual compounding.

For exactly two years of annual compounding, put x = r/100. Compound amount is P × (1 + 2 × x + x^2), whereas simple amount is P × (1 + 2 × x). Therefore:

CI - SI = P × (r/100)^2.

This shortcut requires equal principal, the same constant annual rate, annual compounding and exactly two years. It is not a general formula for three years, varying rates or two half-yearly periods.

Assume no extra deposits, withdrawals, interim payments or fees. Subannual examples use nominal annual rates, divided by periods per year. Use 12 months per year; do not silently assign a compound rule to leftover days or months. Any explicitly simple day extension uses the stated 365-day convention. Keep intermediate values exact and round final money to ₹0.01, rounding half a paise upward.

Worked examples

Example 1 — Compare directly. On ₹6,250 at 8% for 2 years, SI = 6250 × 8 × 2/100 = ₹1,000. Compound balances are 6250 × 1.08 = ₹6,750 and 6750 × 1.08 = ₹7,290. CI = 7290 - 6250 = ₹1,040. The difference is ₹40, also 6250 × 0.08^2.

Example 2 — Recover principal from a difference. The two-year annual CI–SI difference at 7% is ₹147. Thus 147 = P × 0.07^2 = P × 0.0049, giving P = 147 ÷ 0.0049 = ₹30,000. Check: SI = 30000 × 7 × 2/100 = ₹4,200 and CI = 30000 × 1.07^2 - 30000 = ₹4,347; the difference is ₹147.

Example 3 — Three years require recalculation. Compare ₹18,000 at 4% for 3 years. SI = 18000 × 4 × 3/100 = ₹2,160. Compound balances are 18000 × 1.04 = ₹18,720, then 18720 × 1.04 = ₹19,468.80, then 19468.80 × 1.04 = ₹20,247.552. CI = 20247.552 - 18000 = ₹2,247.552 ≈ ₹2,247.55. Difference = 2247.552 - 2160 = ₹87.552 ≈ ₹87.55. The two-year shortcut does not apply.

Common mistakes

Compare interest with interest, not interest with amount. Check compounding frequency before using a shortcut. Subtract exact intermediate results before rounding the final difference.

Practice questions

1. Compare SI and annual CI on ₹9,600 at 5% for 2 years.
2. At 8%, the two-year annual CI–SI difference is ₹96. Find the common principal.
3. Compare SI and annual CI on ₹16,000 at 5% for 3 years.
4. Compare one year’s SI and half-yearly CI on ₹12,000 at a nominal annual 10%.

Worked answers

1. SI = 9600 × 5 × 2/100 = ₹960. CI = 9600 × 1.05^2 - 9600 = ₹984. CI exceeds SI by ₹24.
2. P = 96 ÷ 0.08^2 = 96 ÷ 0.0064 = ₹15,000. Check: SI = ₹2,400, CI = ₹2,496, difference = ₹96.
3. SI = 16000 × 5 × 3/100 = ₹2,400. CI = 16000 × 1.05^3 - 16000 = ₹2,522. Difference = ₹122.
4. SI = 12000 × 10/100 = ₹1,200. Half-yearly rate = 5%, with 2 periods. CI = 12000 × 1.05^2 - 12000 = ₹1,230. Difference = ₹30.