Chapter 8 — Finance: Percentages, Compound Interest, Depreciation
Part 1: Percentage Change
Percentage increase:
$\displaystyle \text{New value} = Original \times (1 + \frac{r}{100})$
Percentage decrease:
$\displaystyle \text{New value} = Original \times (1 - \frac{r}{100})$
Percentage change formula:
Percentage change = (new − original) / original × 100
Example 1: A population was 12500 in 2022 and 13900 in 2023. Find the percentage increase.
Percentage increase = (13900 − 12500) / 12500 × 100 = 1400/12500 × 100 = 11.2%
Example 2: A shop buys a jacket for €80 and sells it with a 45% markup.
Selling price = 80 × 1.45 = €116
During a sale, the price is reduced by 20%: $\text{sale price} = 116 \times 0.80 = €92.80$
Overall profit = (92.80 − 80)/80 × 100 = 16% (not 45% − 20% = 25%!)
Part 2: Compound Interest
$\displaystyle A = P(1 + \frac{r}{100})^{n}$
- P = principal (initial amount)
- r = annual interest rate (%)
- n = number of years (or compounding periods)
- $A = \text{final amount }$
Example 3 (ISMA Prep Q11): P = 200000 yen, 3 years. Year 1: 1.8%, Years 2–3: x%.
After 3 years: 209754 yen.
$\displaystyle 200000 \times 1.018 \times (1 + \frac{x}{100})^{2} = 209754$
$\displaystyle 203600 \times (1 + \frac{x}{100})^{2} = 209754$
$\displaystyle (1 + \frac{x}{100})^{2} = \frac{209754}{203600} = 1.03023$
$\displaystyle 1 + \frac{x}{100} = \sqrt{1.03023} = 1.01502$
$x = 1.5$
Example 4: An investment of £5000 grows to £5800 in 3 years at compound interest. Find the annual rate.
$\displaystyle 5000(1+r)^{3} = 5800 \rightarrow (1+r)^{3} = 1.16 \rightarrow 1+r = (1.16)^{\frac{1}{3}} = 1.0507 \rightarrow r \approx 5.1%$
Part 3: Simple Interest
A = P(1 + rn/100) or equivalently I = Prn/100
- $I = \text{interest earned }$
- Unlike compound, interest is NOT added to principal each year
Example 5: Marco invests €4000 at 3.5% simple interest.
Interest per year = 4000 × 0.035 = €140
In 5 years: $I = €140 \times 5 = €700$
Time to reach €5400: need €1400 interest → 1400/140 = 10 years
Part 4: Depreciation
$\displaystyle V = P(1 - \frac{r}{100})^{n}$
Same formula as compound interest but with MINUS (value goes down).
Example 6: A car depreciates by 15% per year. It cost £20000 new.
After 3 years: $20000 \times 0.85^{3} = 20000 \times 0.6141 = £12282.50$
Example 7: A painting appreciates (increases) by 8% per year. Currently worth £12000.
In 4 years: $12000 \times 1.08^{4} = 12000 \times 1.3605 = £16326$
$3 \text{years ago }: 12000 \div 1.08^{3} = 12000 \div 1.2597 = £9526$
Example 8: Phone costs USD 800. Depreciates 25% in year 1, then 15% per year.
Year 1: $800 \times 0.75 = 600. Year 2: 600 \times 0.85 = 510. Year 3: 510 \times 0.85 = USD 433.50$
After year 1, each subsequent year: $600 \times 0.85^{n-1}. \text{First below USD } 200:$
Year 7: 600 × 0.85⁶ = USD 226.30. Year 8: 600 × 0.85⁷ = USD 192.36 → 8 complete years
Exam Traps / Ловушки на экзамене
❌ Trap 1: Compound interest vs simple interest
200000 at 3% for 5 years: 200000 × 0.03 × 5 = 30000 — SIMPLE interest!
✅ Compound: A = 200000 × (1.03)⁵ ≈ 231855. On exams, use compound unless told otherwise.
Ловушка: на экзамене почти всегда СЛОЖНЫЕ проценты: A = P(1 + r/100)ⁿ. Простые: A = P(1 + rn/100). Формула разная!
❌ Trap 2: Successive percentage changes are not additive
45% markup then 20% discount ≠ 25% profit! Multiply the factors: 1.45 × 0.80 = 1.16 → 16% profit.
Ловушка: последовательные проценты перемножаются, а не складываются!
❌ Trap 3: Finding original value — divide, don't subtract
"After 20% increase, price is €120. What was the original?"
$\text{WRONG:}\ 120 - 20% of 120 = 120 - 24 = 96.$
✅ Original × 1.20 = 120 → Original = 120/1.20 = €100
Ловушка: чтобы найти исходное значение, дели на множитель, а не вычитай процент от нового значения.
❌ Trap 4: (1 + r) instead of (1 + r/100)
Interest rate 3% means r = 3, so multiplier is (1 + 3/100) = 1.03, NOT (1 + 3) = 4.
Ловушка: (1 + r/100), НЕ (1 + r) — процент нужно перевести в десятичную дробь.