💰 Amount & Principal

Start from the one formula that runs the whole chapter: $\text{SI}=\dfrac{P\times R\times T}{100}$. The total you walk away with is the amount: $A = P + \text{SI}$. Substituting the interest gives the form you will use most: $A = P\left(1+\dfrac{RT}{100}\right)$ — in words, the amount is $(100+RT)\%$ of the principal.

Splitting an amount back into P. If you are told the amount and need the principal, just undo that multiplier: $P = \dfrac{A\times 100}{100+RT}$. Find $RT$, add 100, divide. For example a sum at $20\%$ for 4 years has $RT=80$, so the amount is $180\%$ of $P$ — divide the amount by $1.8$ to get the principal back.

Why the amounts grow in equal steps. Because the same interest $I=\dfrac{PR}{100}$ is added every year, the amounts after 1, 2, 3 … years are $P+I,\ P+2I,\ P+3I,\dots$ — an arithmetic progression, each value a fixed step $I$ above the last. So if a sum is $A_1$ after $t_1$ years and $A_2$ after $t_2$ years, the gap $A_2-A_1$ is pure interest for the extra $(t_2-t_1)$ years — strip it off to recover $P$.

From a multiple to the rate (or to another multiple). If a sum becomes $n$ times itself, the interest earned is exactly $(n-1)P$ — the lone $P$ in $nP$ is the original principal, which never grows. That gives the shortcut $R=\dfrac{(n-1)\times 100}{T}$. “Doubles” is the tiny case $n=2$, where $R\times T=100$. And to jump from one multiple to another at the same rate, the times sit in the same ratio as the interests: $\dfrac{n_1-1}{T_1}=\dfrac{n_2-1}{T_2}$ — so “doubles in 5 years” means it adds one more principal every 5 years (triples in 10, quadruples in 15, …).

The special case $T=R$. When the question says the number of years equals the rate, $R\times T$ becomes $R^2$, so $\dfrac{\text{SI}}{P}=\dfrac{R^2}{100}$ and $R=10\sqrt{\dfrac{\text{SI}}{P}}$. If the interest is $\dfrac{9}{25}$ of the sum, $R=10\times\dfrac{3}{5}=6\%$ — take the square root, never stop at $R^2$.