Proportion and the Relationship Between Ratios and Functions

Simple Proportion

Proportion is used to determine how quantities scale relative to one another. If we know the cost or value of a specific number of items, we can calculate the value for any other number of those items. The guide establishes that if $x$ bars of chocolate cost $£y$, then the cost of 1 bar of the same chocolate is $£\frac{y}{x}$. Using this as a base, we find that $a$ identical bars of the same chocolate will cost $£\frac{ay}{x}$.

Methods for Solving Proportion Problems

There are two primary numerical techniques for solving these problems: the unitary method and the method using the Highest Common Factor (HCF).

Worked Example: Chocky Biscuits

If 12 boxes of Chocky Biscuits cost $£42$, how much do 20 boxes of Chocky Biscuits cost?

Method 1: Unitary Method

1. Find the cost of a single unit (one box). 1 box costs $£42 \div 12$. It is best to leave this as a fraction at this stage rather than a decimal to maintain exactness: $£\frac{42}{12}$.
2. Multiply the unit cost by the target number of units (20 boxes). The total is $£\frac{42}{12} \times 20$.
3. Simplify the expression: $\frac{42 \times 20}{12}$.
4. Divide the numerator and denominator by 6: $\frac{7 \times 20}{2}$.
5. Divide the numerator and denominator by 2: $7 \times 10 = 70$.

The cost for 20 boxes is $£70$.

Method 2: Using the HCF

1. Identify the HCF of the starting quantity (12) and the target quantity (20). The HCF is 4.
2. Find the cost of this HCF quantity. Since 12 boxes cost $£42$, then 4 boxes (which is $12 \div 3$) must cost $£42 \div 3 = £14$.
3. Scale this to the final target. Since 20 boxes is $4 \times 5$, the total cost is $£14 \times 5 = £70$.

Relating Ratios to Fractions

In any mixture consisting only of two components $x$ and $y$, if the ratio of $x$ to $y$ is $a:b$, then the fraction of $x$ in the entire mixture is $\frac{a}{a+b}$. This principle allows us to solve problems involving multiple overlapping ratios.

Worked Example: Multi-Part Ratios

A packet of sweets contains only yellow, green, and red sweets. If the ratio of yellow to green is $3:4$ and the ratio of green to red is $6:9$, what fraction of the sweets in the pack is yellow?

1. To find the fraction, we must first determine the unified ratio yellow:green:red. We need a common value for the 'green' part across both ratios.
2. The green values are 4 and 6. The lowest common multiple is 12, but we can also scale the first ratio so its green value matches the second.
3. Multiply the first ratio ($3:4$) by $1.5$ to get $4.5:6$.
4. Now we can combine them: yellow:green:red is $4.5:6:9$.
5. Multiply by 2 to work with integers: $9:12:18$.
6. The fraction of yellow sweets is the yellow part divided by the sum of all parts: $\frac{9}{9+12+18} = \frac{9}{39}$.
7. Simplify by dividing by 3: $\frac{3}{13}$.

Relating Ratios to Linear Functions

Ratios can be expressed as linear algebraic functions. This is a common requirement in the TMUA where variables must be manipulated within larger equations. If the ratio of $x$ to $y$ in a mixture is $a:b$, the relationship is described by the equation $bx = ay$ or $y = \frac{bx}{a}$.

Worked Example: Linear Functions

The ratio $x:y$ is $2:3$. Write $y$ as a linear function of $x$.

1. Start by converting the ratio into a fraction: $\frac{x}{y} = \frac{2}{3}$.
2. Multiply both sides of the equation by $y$ to clear the denominator: $x = \frac{2}{3}y$.
3. Multiply both sides by 3: $3x = 2y$.
4. Isolate $y$ to express it as a function of $x$: $y = \frac{3x}{2}$.

This shows that $y$ is a linear function of $x$ with a gradient of $1.5$.