📈 Different Rates & Finding Time

When the rate changes annually you cannot raise one factor to a power. Instead you multiply a separate growth factor for each year: $A=P\left(1+\dfrac{R_1}{100}\right)\left(1+\dfrac{R_2}{100}\right)\left(1+\dfrac{R_3}{100}\right)$, then $CI=A-P$. Easiest by hand: take each year's percent of the running balance and add it on.

• Two rates? Use the net percent. For just two years at a% then b% the whole thing collapses to a single rate $a+b+\dfrac{ab}{100}$. So 5% then 8% is $5+8+\dfrac{5\times8}{100}=13.4\%$ — apply it once.

• Never just add the rates. Adding $4+5+6=15\%$ gives simple interest; CI is always a touch more because each year's interest itself earns interest the following year.

Finding the time — match A/P to a power. Start from $A=P\left(1+\dfrac{R}{100}\right)^{T}$ and divide by P, so $\dfrac{A}{P}=\left(1+\dfrac{R}{100}\right)^{T}$. Reduce the fraction $\dfrac{A}{P}$ to lowest terms and write the yearly multiplier as a small fraction (10% $=\dfrac{11}{10}$, 12.5% $=\dfrac{9}{8}$). The power you need is the number of years. For example $\dfrac{1331}{1000}=\dfrac{11^{3}}{10^{3}}=\left(\dfrac{11}{10}\right)^{3}$, so $T=3$ years.

• “Becomes k times” shortcut: if a sum becomes k times in n years, it becomes $k^{m}$ times in $m\times n$ years. Doubles in 5 yr ⇒ $8=2^{3}$ times in $3\times5=15$ yr. Never reason linearly.