4.3 Direct and Inverse Proportion, Currency Exchange, and Scaling

Direct Proportion ($y = kx$)

Two quantities $x$ and $y$ are in direct proportion when multiplying $x$ by a factor $m$ multiplies $y$ by the exact same factor $m$. Their quotient (the unit rate $k$) remains constant:

$$\frac{y_1}{x_1} = \frac{y_2}{x_2} = k \quad \Longrightarrow \quad y_2 = y_1 \times \frac{x_2}{x_1}$$

School Applications of Direct Proportion

1. Hospitality Recipe and Catering Scaling: If a school food-technology recipe for $16\text{ scones}$ requires $600\text{ g}$ of self-raising flour and $240\text{ mL}$ of milk, scaling the recipe up for $40\text{ scones}$ uses the scale factor $\frac{40}{16} = 2.5$. The required flour is $600\text{ g} \times 2.5 = 1{,}500\text{ g}$ ($1.5\text{ kg}$), and the required milk is $240\text{ mL} \times 2.5 = 600\text{ mL}$.
2. Limiting-Ingredient Pantry Checks: Suppose a hospitality teacher has $3.3\text{ kg}$ ($3{,}300\text{ g}$) of flour in the storeroom and uses the same recipe ($600\text{ g}$ per $16\text{ scones}$). How many scones can be baked? Since $3{,}300 \div 600 = 5.5\text{ batches}$, the teacher can make $5.5 \times 16 = 88\text{ scones}$.
3. Pro-Rata FTE (Full-Time Equivalent) Teacher Salaries: In Australian schools, a full-time teaching load is $1.0\text{ FTE}$ ($5\text{ days per week}$, or $10\text{ days per fortnight}$). A teacher working $3\text{ days per week}$ has a $\frac{3}{5} = 0.6\text{ FTE}$ load. Salary and leave entitlements scale directly with FTE:
   • Finding pro-rata pay from full-time pay: Multiply the $1.0\text{ FTE}$ base salary by the FTE decimal. At a base salary of $92,000, a $0.6\text{ FTE}$ teacher earns $0.6 \times 92{,}000 = 55{,}200$, or $55,200.
   • Finding full-time pay from pro-rata pay: Divide the part-time salary by the FTE decimal. If a $0.8\text{ FTE}$ teacher earns $78,400, the $1.0\text{ FTE}$ base salary is $78{,}400 \div 0.8 = 98{,}000$, or $98,000.
4. Map and Floor-Plan Scales: On a school site plan drawn at a scale of $1 : 250$, every $1\text{ cm}$ on paper represents $250\text{ cm} = 2.5\text{ m}$ on the ground. A gymnasium measuring $14\text{ cm}$ long on the plan has an actual length of $14 \times 2.5\text{ m} = 35\text{ m}$. Conversely, a $45\text{ m}$ oval boundary is drawn with length $45 \div 2.5 = 18\text{ cm}$ on the plan.

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Currency Exchange for Overseas Study Tours

Currency conversion is a direct proportion problem. Exchange rates in Australian questions are normally quoted with 1 Australian dollar ($1\text{ AUD}$) as the base unit—for example, $1\text{ AUD} = 96\text{ JPY}$ (Japanese yen) or $1\text{ AUD} = 0.62\text{ EUR}$ (euros).

$$\text{Foreign Currency} = \text{AUD} \times \text{Rate}, \qquad \text{AUD} = \frac{\text{Foreign Currency}}{\text{Rate}}$$

Worked Example: Converting AUD to Yen and Back

A school organises a language immersion tour to Japan when the exchange rate is $1\text{ AUD} = 96\text{ JPY}$.

• Converting AUD to Foreign Currency (Multiply): A teacher converts $450 AUD into yen for group transport cards: $450 \times 96 = 43{,}200\text{ JPY}$.
• Converting Foreign Currency to AUD (Divide): A museum pass bundle for the group costs $14{,}400\text{ JPY}$. To acquit the receipt in Australian dollars, divide by the exchange rate: $14{,}400 \div 96 = 150$, or $150 AUD.
• Unspent Currency on Return: If the tour group returns to Australia with $11{,}760\text{ JPY}$ unspent and the currency booth buys back yen at $1\text{ AUD} = 98\text{ JPY}$, the school receives $11{,}760 \div 98 = 120$, or $120 AUD.

Always apply a quick magnitude check: since $1\text{ AUD}$ buys $0.62\text{ EUR}$ (less than $1\text{ EUR}$), any euro price must correspond to a larger number of Australian dollars.

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Inverse Proportion ($x_1 y_1 = x_2 y_2$)

Two quantities $x$ and $y$ are in inverse (indirect) proportion when a fixed total workload or capacity is shared: doubling $x$ halves $y$, and multiplying $x$ by $m$ divides $y$ by $m$. Instead of their ratio being constant, their product is constant:

$$x_1 y_1 = x_2 y_2 = \text{Total Work} \quad \Longrightarrow \quad y_2 = \frac{x_1 y_1}{x_2}$$

Think of the constant product $x_1 y_1$ as the total job size measured in teacher-hours, student-days, or printer-minutes.

Worked Example 1: Cohort Essay Marking (Teacher-Hours)

Three English teachers working at the same constant rate take $8\text{ hours}$ to mark all Year 10 examination essays. How long would it take $4$ teachers working at that same rate to mark the same cohort of essays?

• Common Trap: Setting up a direct ratio $\frac{4}{3} \times 8 = 10.67\text{ hours}$, which falsely predicts that adding an extra teacher makes the marking take longer!
• Correct Inverse Proportion Method:
  1. Calculate total marking workload: $3\text{ teachers} \times 8\text{ hours} = 24\text{ teacher-hours}$.
  2. Divide the workload among $4\text{ teachers}$: $24\text{ teacher-hours} \div 4\text{ teachers} = 6\text{ hours}$.

Worked Example 2: Camp Food Provisions (Student-Days)

An outdoor education camp stocks enough food rations to feed $40\text{ students}$ for $6\text{ days}$. If $8\text{ additional students}$ join the camp at the last minute (making $48\text{ students}$ in total) and no extra food is ordered, how many days will the rations last?

1. Total ration supply $= 40\text{ students} \times 6\text{ days} = 240\text{ student-days}$.
2. Duration for $48\text{ students} = 240\text{ student-days} \div 48\text{ students} = 5\text{ days}$.

Joint Rate-Work-Time Problems

When both the number of machines and the number of items change, find the output rate of 1 machine in 1 unit of time first. For example, if $3$ photocopiers print $3{,}600$ exam booklets in $40\text{ minutes}$, then $1$ photocopier prints $3{,}600 \div 3 = 1{,}200$ booklets in $40\text{ minutes}$, which is $1{,}200 \div 40 = 30\text{ booklets per minute}$. Consequently, $5$ photocopiers print $5 \times 30 = 150\text{ booklets per minute}$ and will complete $6{,}000$ booklets in $6{,}000 \div 150 = 40\text{ minutes}$.