Interest (SI & CI)

Why CI and SI differ

For the first year, CI and SI on the same sum at the same rate are identical. They part ways from the second year: CI earns interest on the earlier interest, SI never does. So the gap is exactly the interest earned on the accumulated interest.

Two years — the one-line formula

The difference is the first year's interest, re-earned once in the second year: $CI - SI = P\left(\frac{R}{100}\right)^2$ ₹12,500 at 12%: $12500 \times 0.12^2 = ₹180$. Reverse: difference ₹50 at 10% → $P = \frac{50 \times 10000}{100} = ₹5{,}000$. Notice the shape: P × rate-fraction × rate-fraction. The rate fraction $\frac{R}{100}$ appears once per extra year.

Three years

The second year's interest also re-earns in the third, and the first year's interest re-earns twice: $CI - SI = P\left(\frac{R}{100}\right)^2\left(3 + \frac{R}{100}\right)$ ₹10,000 at 10%: $10000 \times 0.01 \times 3.1 = ₹310$. At small rates the factor is roughly 3 (for 3 years) or 2 (for 2 years) times $\left(\frac{R}{100}\right)^2 P$.

CI and SI both given

Given both figures for the same sum, the difference isolates the rate:

• 2 years: difference $= P\left(\frac{R}{100}\right)^2$ and SI $= \frac{2PR}{100}$ → divide: $\frac{R}{100} = \frac{\text{difference} \times 2}{SI}$ → $R = \frac{200 \times \text{difference}}{SI}$. SI ₹800, CI ₹820 → $R = \frac{200 \times 20}{800} = 5\%$, then $P = \frac{800 \times 100}{2 \times 5} = ₹8{,}000$.
• Then the CI itself is just SI + difference.

Doubling chains at CI

At CI a sum multiplies by the same factor every fixed period. If it doubles in T years:

• 4 times in 2T, 8 times in 3T — powers of 2.
• "16 times" = $2^4$ → 4T. ₹P doubling in 6 years → 8 times in 18 years. At SI this does not work (linear growth: 3 times takes 2T) — check which interest the question uses.

Multiplier → rate

Amounts given as multiples: "becomes 1.44 times in 2 years at CI" → chip² $= 1.44$ → chip $= 1.2$ → 20%. Common squares: 1.21 → 10%, 1.44 → 20%, 2.25 → 50%; cubes: 1.728 → 20%, 1.331 → 10%.

A quick worked case

₹8,000 at 5% for 2 years: SI $= 800$, CI $= 8000(1.05^2 - 1) = 820$ — difference ₹20, which is exactly $8000 \times 0.05^2$. Every 2-year question is this single multiplication in disguise; train yourself to see the rate fraction squared.

Quick revision

• 2 years: $CI - SI = P\left(\frac{R}{100}\right)^2$.
• 3 years: $CI - SI = P\left(\frac{R}{100}\right)^2\left(3 + \frac{R}{100}\right)$.
• Difference and SI given: $R = \frac{200 \times \text{diff}}{SI}$ (2 years), then $P = \frac{100 \times SI}{2R}$.
• CI doubling chain: ×2 in T → ×$2^k$ in kT.
• Multiplier m in 2 years → rate $= 100(\sqrt{m} - 1)$.