Direct and Inverse Proportionality Models
- Direct Proportion ($y \propto x$): As variable $x$ scales by factor $c$, variable $y$ scales by the exact same factor:
$$y = kx \iff \frac{y}{x} = k \quad (\text{constant } k > 0)$$
• The graph of $y$ against $x$ is a straight line passing through the origin $(0, 0)$ with gradient $k$.
• Powers: $y \propto x^2 \iff y = kx^2$; $y \propto \sqrt{x} \iff y = k\sqrt{x}$. - Inverse Proportion ($y \propto \frac{1}{x}$): As variable $x$ multiplies, variable $y$ divides proportionally:
$$y = \frac{k}{x} \iff xy = k \quad (\text{constant } k > 0)$$
• The graph of $y$ against $x$ is a rectangular hyperbola with asymptotes at $x = 0$ and $y = 0$.
• Powers: $y \propto \frac{1}{x^2} \iff y = \frac{k}{x^2}$ (inverse-square laws e.g. gravitation, light intensity).
Example 1 · Criterion A
Solving Direct and Inverse Proportion Problems
(a) The electrical resistance $R$ of a wire is directly proportional to its length $L$. A wire of length $12\text{ m}$ has resistance $4.8\text{ }\Omega$. Find the resistance of a $35\text{ m}$ wire of the same material.
(b) The illumination intensity $I$ of a light source is inversely proportional to the square of distance $d$. At $d = 2\text{ m}$, $I = 150\text{ lux}$. Find $I$ when $d = 5\text{ m}$.
- (a) Direct proportion: $$R = kL \implies 4.8 = k(12) \implies k = \frac{4.8}{12} = 0.4\text{ }\Omega\text{ m}^{-1}$$ For $L = 35\text{ m}$: $$R = 0.4 \times 35 = 14\text{ }\Omega$$
- (b) Inverse-square proportion: $$I = \frac{k}{d^2} \implies 150 = \frac{k}{2^2} \implies k = 150 \times 4 = 600\text{ lux m}^2$$ For $d = 5\text{ m}$: $$I = \frac{600}{5^2} = \frac{600}{25} = 24\text{ lux}$$
Answers: (a) $14\text{ }\Omega$; (b) $24\text{ lux}$.
Percentage Change & Reverse Percentages
- Decimal Multipliers:
• An increase of $r\%$ corresponds to multiplier $M = 1 + \frac{r}{100}$ (e.g., $+15\% \to \times 1.15$).
• A decrease of $r\%$ corresponds to multiplier $M = 1 - \frac{r}{100}$ (e.g., $-20\% \to \times 0.80$). - Percentage Change Formula: $$\% \text{ Change} = \frac{\text{New Value} - \text{Original Value}}{\text{Original Value}} \times 100\%$$
- Reverse percentages: When given the final value after a percentage change and asked for the original value: $$\text{Original Value} = \frac{\text{Final Value}}{\text{Multiplier}} = \frac{\text{Final Value}}{1 \pm \frac{r}{100}}$$ Never calculate $r\%$ of the final value and add/subtract it!
Example 2 · Criterion A
Forward Multipliers and Reverse Percentages
(a) A retailer purchases a coat for $\$80$ and marks it up by $35\%$. Find the retail price.
(b) In a end-of-season sale, a laptop is discounted by $20\%$ to a sale price of $\$640$. Calculate the original price before the discount.
- (a) Markup: $$\text{Multiplier} = 1 + 0.35 = 1.35$$ $$\text{Retail Price} = \$80 \times 1.35 = \$108$$
- (b) Reverse percentage: $$\text{Multiplier for } 20\% \text{ reduction} = 1 - 0.20 = 0.80$$ $$\text{Original Price} = \frac{\$640}{0.80} = \$800$$ (Check: $20\%$ of $\$800 = \$160$; $\$800 - \$160 = \$640$ ✓).
Answers: (a) $\$108$; (b) $\$800$.
Financial Mathematics: Compound Interest & Depreciation
- Compound Interest (with compounding frequency $k$ times/year):
$$FV = PV \left(1 + \frac{r}{100k}\right)^{kn}$$
where:
• $PV$ = Present Value (initial principal deposited/borrowed)
• $FV$ = Future Value (total balance after interest)
• $r$ = annual nominal interest rate (%)
• $k$ = compounding periods per year (annual $k=1$, semi-annual $k=2$, quarterly $k=4$, monthly $k=12$, daily $k=365$)
• $n$ = time in years
• $\text{Total Interest Earned} = FV - PV$. - Depreciation (Reducing Balance Decay): For an asset losing $r\%$ of its value each year: $$V(n) = V_0 \left(1 - \frac{r}{100}\right)^n$$
Example 3 · Criterion A & D
Comparing Compounding Frequencies and Asset Depreciation
(a) $\$6000$ is invested for $4\text{ years}$ at $5.4\%$ p.a. compounded monthly ($k = 12$). Calculate the total interest earned to the nearest cent.
(b) A delivery van purchased for $\$32\,000$ depreciates at $15\%$ per annum. Calculate its book value after $3\text{ years}$.
- (a) Monthly compound interest: $$PV = 6000, \quad r = 5.4, \quad k = 12, \quad n = 4 \implies kn = 12 \times 4 = 48$$ $$\text{Periodic rate} = \frac{5.4}{12} = 0.45\% = 0.0045$$ $$FV = 6000(1 + 0.0045)^{48} = 6000(1.0045)^{48} \approx 6000 \times 1.240509 = \$7443.05$$ $$\text{Interest} = FV - PV = 7443.05 - 6000 = \mathbf{\$1443.05}$$
- (b) Vehicle depreciation: $$V_3 = 32000(1 - 0.15)^3 = 32000(0.85)^3 = 32000 \times 0.614125 = \mathbf{\$19\,652}$$
Answers: (a) $\$1443.05$; (b) $\$19\,652$.
Divide $\$840$ in the ratio $3 : 4 : 7$. Find each share.
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Step 1: Total parts [1 mark]
$\text{Total parts} = 3 + 4 + 7 = 14$ [A1]
Step 2: Value per part [1 mark]
$\text{Value per part} = \frac{\$840}{14} = \$60$ [A1]
Step 3: Calculate individual shares [1 mark]
• Share 1: $3 \times \$60 = \mathbf{\$180}$
• Share 2: $4 \times \$60 = \mathbf{\$240}$
• Share 3: $7 \times \$60 = \mathbf{\$420}$ [A1]
A house is purchased for $\$450\,000$ and sold 5 years later for $\$558\,000$.
Calculate the percentage profit made on the sale.
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Step 1: Profit amount [1 mark]
$\text{Profit} = 558\,000 - 450\,000 = \$108\,000$ [A1]
Step 2: Percentage calculation [2 marks]
$\% \text{ Profit} = \frac{108\,000}{450\,000} \times 100\% = 0.24 \times 100\% = \mathbf{24\%}$ [A2]
The price of a smartphone including $12\%$ sales tax is $\$784$.
Find the price of the smartphone before tax was added.
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Step 1: Identify multiplier [1 mark]
$\text{Multiplier} = 1 + 0.12 = 1.12$ [M1]
Step 2: Reverse percentage [2 marks]
$\text{Pre-tax Price} = \frac{\$784}{1.12} = \mathbf{\$700}$ [A2]
Variable $y$ is inversely proportional to $x$. When $x = 8$, $y = 15$.
(a) Find the equation linking $y$ and $x$. [2 marks]
(b) Find $y$ when $x = 24$. [1 mark]
(c) Find $x$ when $y = 5$. [1 mark]
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(a) Formula: [2 marks]
$y = \frac{k}{x} \implies 15 = \frac{k}{8} \implies k = 15 \times 8 = 120$
$\mathbf{y = \frac{120}{x}}$ [A2]
(b) $y$ when $x = 24$: [1 mark]
$y = \frac{120}{24} = \mathbf{5}$ [A1]
(c) $x$ when $y = 5$: [1 mark]
$5 = \frac{120}{x} \implies x = \frac{120}{5} = \mathbf{24}$ [A1]
The braking distance $d$ (metres) of a vehicle is directly proportional to the square of its speed $v$ ($\text{km/h}$).
At a speed of $v = 40\text{ km/h}$, the braking distance is $d = 12\text{ m}$.
(a) Express $d$ in terms of $v$. [2 marks]
(b) Calculate the braking distance at $v = 100\text{ km/h}$. [2 marks]
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(a) Constant of proportionality: [2 marks]
$d = k v^2 \implies 12 = k(40)^2 = 1600k$
$k = \frac{12}{1600} = \frac{3}{400} = 0.0075$
$\mathbf{d = 0.0075 v^2}$ [A2]
(b) Distance at 100 km/h: [2 marks]
$d = 0.0075 \times (100)^2 = 0.0075 \times 10\,000 = \mathbf{75\text{ metres}}$ [A2]
An investor has $\$10\,000$ to invest for $3\text{ years}$. Two banks offer different terms:
• Bank Alpha: $5.0\%$ simple interest per annum.
• Bank Beta: $4.8\%$ compound interest per annum compounded annually.
Calculate the interest earned from each bank and state which bank yields the higher return.
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Bank Alpha (Simple Interest): [2 marks]
$I = \frac{P \cdot r \cdot t}{100} = \frac{10000 \times 5 \times 3}{100} = \mathbf{\$1500}$ [A2]
Bank Beta (Compound Interest): [2 marks]
$FV = 10000(1 + 0.048)^3 = 10000(1.048)^3 = 10000(1.15101) = \$11\,510.10$
$I = 11510.10 - 10000 = \mathbf{\$1510.10}$ [A1]
Bank Beta yields a higher return by $\$10.10$. [A1]
A sum of $\$8000$ is deposited into a fixed deposit account paying $6\%$ annual nominal interest for $5\text{ years}$.
(a) Calculate the future value if interest is compounded quarterly ($k = 4$). [2 marks]
(b) Calculate the future value if interest is compounded monthly ($k = 12$). [2 marks]
(c) Find the additional interest earned by compounding monthly rather than quarterly. [1 mark]
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(a) Quarterly compounding: [2 marks]
$kn = 4 \times 5 = 20$; periodic rate $= \frac{6}{4} = 1.5\% = 0.015$
$FV = 8000(1.015)^{20} \approx 8000 \times 1.346855 = \mathbf{\$10\,774.84}$ [A2]
(b) Monthly compounding: [2 marks]
$kn = 12 \times 5 = 60$; periodic rate $= \frac{6}{12} = 0.5\% = 0.005$
$FV = 8000(1.005)^{60} \approx 8000 \times 1.348850 = \mathbf{\$10\,790.80}$ [A2]
(c) Difference: [1 mark]
$\text{Difference} = 10\,790.80 - 10\,774.84 = \mathbf{\$15.96}$ [A1]
A laser CNC cutting machine is purchased for $\$75\,000$. Its salvage value depreciates by $18\%$ per annum.
(a) Write a formula for its book value $V(t)$ after $t$ years. [1 mark]
(b) Calculate the value of the machine after $4\text{ years}$. [1 mark]
(c) Determine the minimum number of years after which the machine’s value will drop below $\$20\,000$. [2 marks]
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(a) Formula: [1 mark]
$V(t) = 75000(1 - 0.18)^t = \mathbf{75000(0.82)^t}$ [A1]
(b) Value after 4 years: [1 mark]
$V(4) = 75000(0.82)^4 = 75000(0.45212) \approx \mathbf{\$33\,909.13}$ [A1]
(c) Time below $20\,000: [2 marks]
$75000(0.82)^t < 20000 \implies (0.82)^t < \frac{20000}{75000} = \frac{4}{15} \approx 0.2667$
$t \ln(0.82) < \ln(4/15) \implies t > \frac{\ln(4/15)}{\ln(0.82)} = \frac{-1.32176}{-0.19845} \approx 6.66$
It drops below $\$20\,000$ after $\mathbf{7\text{ years}}$. [A2]
A traveler exchanges $\$1200\text{ USD}$ to Euros ($\text{EUR}$).
• Bank rate: $1\text{ USD} = 0.92\text{ EUR}$.
• The bank charges a flat commission fee of $2.5\%$ deducted from the USD amount before exchange.
(a) Calculate the amount of USD remaining after the commission fee is deducted. [2 marks]
(b) Calculate the total Euros received by the traveler. [2 marks]
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(a) Fee deduction: [2 marks]
$\text{Commission fee} = 2.5\% \text{ of } 1200 = 0.025 \times 1200 = \$30$
$\text{Remaining USD} = 1200 - 30 = \mathbf{\$1170\text{ USD}}$ [A2]
(b) Currency conversion: [2 marks]
$\text{EUR} = 1170 \times 0.92 = \mathbf{1076.40\text{ EUR}}$ [A2]
A department store advertises a "Double Discount Sale": an item is discounted by $20\%$, and customers with a loyalty card receive an additional $10\%$ off the sale price.
(a) Show that the combined overall discount is $28\%$ and NOT $30\%$. [2 marks]
(b) A loyalty customer pays $\$144$ for a jacket under this promotion. Find the original price of the jacket. [2 marks]
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(a) Combined multiplier: [2 marks]
Multiplier $1 = 1 - 0.20 = 0.80$
Multiplier $2 = 1 - 0.10 = 0.90$
$\text{Overall Multiplier} = 0.80 \times 0.90 = 0.72$
$\text{Overall discount} = (1 - 0.72) \times 100\% = \mathbf{28\%} \ne 30\%$ $\blacksquare$ [R2]
(b) Original price: [2 marks]
$\text{Original} = \frac{\$144}{0.72} = \mathbf{\$200}$ [A2]
Ratio & Percentage Quick Reference
Proportionality Models
Percentage Rules
Compound Interest
Depreciation (Decay)
Top 4 Percentage Pitfalls
1. The Reverse Percentage Subtraction Error
If a price increases by $20\%$ to $\$120$, the original is NOT $\$120 - 20\%(\$120) = \$96$. The original is $\frac{120}{1.20} = \$100$. Percentages apply to the original base, not the final result.
2. Adding Successive Percentages
A $10\%$ rise followed by another $10\%$ rise is NOT a $20\%$ rise! Multipliers multiply: $1.10 \times 1.10 = 1.21$, which is a $21\%$ overall rise.
3. Forgetting Compounding Multiplier in Exponent
For monthly compounding over $3$ years, the exponent is $12 \times 3 = 36$ periods, and rate is divided by $12$. Writing $(1 + 0.06)^3$ calculates annual compounding, missing monthly earnings.
4. Confusing Direct vs Inverse Graphs
Direct proportion graphs are straight lines through $(0,0)$. Inverse proportion graphs are curves that never touch the axes.
Ratio & Percentage Check
Answer these 5 questions to test ratio division, proportion, percentage change, and financial interest.
A jacket on sale for $\$180$ has been discounted by $25\%$. What was the original price of the jacket?
If $y$ is inversely proportional to $x$ and $y = 6$ when $x = 10$, what is the value of $y$ when $x = 15$?
Divide $\$450$ in the ratio $2 : 3 : 4$. What is the largest share?
An investment of $\$4000$ earns $5\%$ compound interest compounded annually. What is its total value after $2\text{ years}$?
A car bought for $\$25\,000$ depreciates by $10\%$ in year 1 and by $10\%$ in year 2. What is its value at the end of year 2?