🧮 The SI Formula

The interest earned is $SI=\dfrac{P\times R\times T}{100}$. The$100$ is there because the rate is a percent (“out of 100”).

Because the base never changes, the interest for one year is a fixed amount — $\dfrac{P\times R}{100}$ — and $T$ years is simply that same amount added $T$ times. So the interest grows in equal yearly steps, like a staircase, never a curve.

The total money you end up with is the Amount: $A=P+SI$. So if a question gives you the Amount, peel the interest back off first: $SI=A-P$. Forgetting this and plugging the Amount in as the principal is the single most common mistake.

Finding a missing letter. Rearrange the same formula:$\;R=\dfrac{SI\times100}{P\times T},\quad T=\dfrac{SI\times100}{P\times R},\quad P=\dfrac{SI\times100}{R\times T}$. Put in the three values you know and read off the fourth.

The exam shortcut. Multiply rate by time into one effective percent: at $6\%$ for $3$ years the interest is just $18\%$ of $P$. And mind the units — the rate is per annum, so months must become years first ($18\text{ months}=\dfrac{18}{12}=1.5\text{ years}$).

SI as a fraction of the principal. Some questions say the interest is a slice of the sum — e.g. $SI=\dfrac{9}{25}P$ — and add that the number of years equals the rate ($T=R$). Put $T=R$ into the formula:$\;SI=\dfrac{P\times R^2}{100}$. Divide by $P$ so the principal cancels (you never need its value): $\dfrac{SI}{P}=\dfrac{R^2}{100}$, which gives the clean rule $R=10\sqrt{\dfrac{SI}{P}}$. For $\dfrac{9}{25}$ that is $10\times\dfrac{3}{5}=6\%$. A percent like $36\%$ becomes the fraction $\dfrac{36}{100}$ first, and remember to take the square root at the very end.