Because each change applies to the new value, not the original — the base shifts every step. A 20% decrease followed by a 30% decrease is not a 50% decrease.

Convert every percentage change to a decimal multiplier: an increase of $R\%$ is $\times\left(1+\frac{R}{100}\right)$, a decrease is $\times\left(1-\frac{R}{100}\right)$. Chain the multipliers together — either step by step, or combine them first.

£80 increased by 10%, then that result discounted by 15%: multiplier $=1.10 \times 0.85 = 0.935$, so final price $= £80\times 0.935 = £74.80$ (an overall 6.5% decrease).

Both are just the multiplier applied $n$ times:

£18,000 car depreciating 30% in year 1, then 14% in each of years 2 and 3: $18000 \times 0.70 \times 0.86^2 = £9318.96$. Rates don't have to be equal each year — just multiply each year's multiplier in sequence.

Rearrange $\text{Final} = \text{Original} \times (\text{multiplier})^n$ to $\text{Original} = \dfrac{\text{Final}}{(\text{multiplier})^n}$.

A car depreciates 35% a year; it's worth £5460 after 1 year. Find its value a year earlier: multiplier $=0.65$, so original $=\dfrac{5460}{0.65}=£8400$.

Unknown rate: isolate the multiplier, then take the $n$th root. A car's value halves in 5 years: $M^5 = 0.5 \Rightarrow M = \sqrt[5]{0.5} \approx 0.8706$, so the annual decrease is $(1-0.8706)\times100 \approx 12.9\%$.

Unknown number of years: isolate $(\text{multiplier})^n$ and use trial and improvement, since solving for $n$ directly needs logarithms (not on this syllabus). £4500 grows to £5469.78 at 5% p.a.: $(1.05)^n = 1.2155$; try $n=3$ (too low, $1.158$), try $n=4$ (matches, $1.2155$) — so $n=4$ years.

If every linear dimension is scaled by multiplier $M$, area scales by $M^2$ and volume scales by $M^3$ — this links percentage change directly to geometry.

A circle's radius increases by 8%: find the % increase in area. Linear multiplier $1.08$; area multiplier $=1.08^2=1.1664$, so area increases by $16.64\%$.

Reverse direction: if volume increases by 25%, the linear multiplier is $\sqrt[3]{1.25}\approx1.0772$ — a $7.72\%$ increase in side length (3 s.f.).

Mia invests £4000 for 3 years: 3% p.a. in year 1, then $x\%$ p.a. for years 2–3. She has £4413.45 at the end. Find $x$.

1. After year 1: $4000 \times 1.03 = £4120$.
2. This grows at $x\%$ for 2 years: $4120 \times M^2 = 4413.45$, so $M^2 = \dfrac{4413.45}{4120} = 1.071225$.
3. $M = \sqrt{1.071225} = 1.035$, so $x = 3.5$.

Multi-stage rate problems like this always split into "known stage → forward calculation" then "unknown stage → reverse/root calculation."