Interest (SI & CI)

What interest is

When you borrow money, you pay a charge for using it; when you lend or deposit, you earn it. That charge is interest. The money borrowed or invested is the principal (P), the charge is a per-cent per year called the rate (R% per annum), and the period is the time (T, in years).

Simple interest — the formula and its four faces

In simple interest (SI) the interest each year is a fixed slice of the original principal only. The earlier interest earns nothing more. $SI = \frac{P \times R \times T}{100}$ Keep the units straight: T in years (8 months $= \frac{8}{12} = \frac{2}{3}$ year), R per year. Any one of P, R, T, SI can be found from the other three: $P = \frac{100 \times SI}{RT}, \quad R = \frac{100 \times SI}{PT}, \quad T = \frac{100 \times SI}{PR}$ SI on ₹6,500 at 8% for 3 years: $\frac{6500 \times 8 \times 3}{100} = ₹1{,}560$.

Amount

Amount (A) = what you finally get back or owe: $A = P + SI$. Amount questions: first get SI $= A - P$, then use the formula. A sum amounts to ₹1,560 in 2 years at 5%: SI $= 1560 - P$ and SI $= \frac{P \times 5 \times 2}{100} = 0.1P$ → $1.1P = 1560$ → P $= ₹1{,}418.18$ — exam numbers are chosen cleaner: A ₹1,540 → P ₹1,400.

Growth is LINEAR at SI

The amount grows by the same rupee amount every year, so the yearly amounts form an arithmetic progression. Two consequences:

• n times in T years: if a sum becomes n times in T years, the interest earned is $(n-1)P$, so $R = \frac{100(n-1)}{T}$ Doubles in 8 years → $\frac{100}{8} = 12.5\%$. Triples in 16 → $\frac{200}{16} = 12.5\%$.
• T from n and R: $T = \frac{100(n-1)}{R}$. "In what time will a sum become 5 times at 12%?" → $\frac{400}{12} = 33\frac{1}{3}$ years. The word "amounts to n times" includes the principal, so the interest part is only $(n-1)$ times P — the classic trap is using n instead of n−1.

Fraction-of-principal questions

"The SI is $\frac{1}{5}$ of the sum at 4% — find the time." Write $\frac{PRT}{100} = \frac{P}{5}$ → $T = \frac{100}{5 \times 4} = 5$ years. The P cancels; only the fraction, R and T matter. Same shape: "In what time will the SI be $\frac{2}{9}$ of P at 8%?" → $T = \frac{200}{9 \times 8} = 2\frac{7}{9}$ years.

Money scaling and comparison

SI is directly proportional to P, R and T. Doubling the sum doubles the interest; halving the rate halves it; the interest on 2P at half the rate equals the original interest. Difference questions subtract two SI computations, or use ratios: SI on 3P at r% : SI on 2P at 2r% $= 3 : 4$.

Where the marks are lost

• Mixing months and years (R is per annum; convert months to a fraction of a year).
• Using n instead of n−1 in n-times questions.
• Dividing by the amount when the question gives SI, or the reverse.

Quick revision

• $SI = \frac{PRT}{100}$; $A = P + SI$; solve for any missing input.
• n times in T years at SI: $R = \frac{100(n-1)}{T}$, $T = \frac{100(n-1)}{R}$.
• Growth is linear: equal rupee growth every year.
• Fraction of P as SI: $T = \frac{100 \times \text{fraction}}{R}$.
• Months to years: divide by 12.