Ratio and Proportion Applications in Real Contexts

Introduction to Ratio Applications

Ratio and proportion are powerful tools for solving problems in real world contexts. In the Test of Mathematics for University Admission, you are expected to apply these concepts to conversion, comparison, scaling, mixing, and concentrations. A fundamental skill is expressing a multiplicative relationship between two quantities as either a ratio or a fraction.

Multiplicative Relationships

To express a multiplicative relationship, you should first create an equation. If there is $a$ times as much of $X$ in a mixture as there is of $Y$, then the relationship is defined by $aY = X$. This can be written as the ratio (quantity of $X$) : (quantity of $Y$) = $a : 1$. In fractional form, this relationship is expressed as $\frac{\text{quantity of } X}{\text{quantity of } Y} = \frac{a}{1} = a$.

Application of Ratio to Conversion

Conversion problems involve changing a value from one unit to another using a fixed ratio. For example, consider the conversion of dinars to dollars with a ratio of $11 : 2$.

Example: Converting Dinars to Dollars

The conversion ratio of dinars to dollars is $11 : 2$. How many dollars can you get for $2530$ dinars?

1. Recognise that every $11$ dinars are worth $2$ dollars.
2. Divide the total dinars by the dinar part of the ratio: $2530 \div 11 = 230$.
3. Multiply this result by the dollar part of the ratio: $230 \times 2 = 460$ dollars.

Application of Ratio to Comparison

Comparison problems often require linking two different ratios that share a common third quantity. To solve these, you must ensure the 'linking' quantity has the same number of parts in both ratios.

Example: Comparing Orange and Mango Pur��ee

In a mixture of orange pur��ee, mango pur��ee and water, the ratio by volume of orange pur��ee to water is $3 : 5$. The ratio of mango pur��ee to water is $11 : 15$. What is the ratio of orange pur��ee to mango pur��ee by volume?

1. Identify the linking quantity, which is water. The first ratio is Orange : Water ($3 : 5$) and the second is Water : Mango ($15 : 11$).
2. To compare them, the water parts must match. In the first ratio, multiply both sides by $3$ so that the water parts become $15$. Thus, $3 : 5$ becomes $9 : 15$.
3. Now that water is $15$ in both, you can compare orange to mango directly: $9 : 11$.

Application of Ratio to Scaling

Scaling involves using a ratio to relate measurements on a model or map to real life dimensions. A scale of $1 : 10,000$ means every measurement on the drawing is multiplied by $10,000$ to find the actual measurement.

Example: Park Scale Drawing

A scale drawing is made of a park using a ratio of $1 : 10,000$. The length of the park on the drawing is $30$ cm. What is the actual length of the park in km?

1. Multiply the drawing length by the scale factor: $30 \times 10,000 = 300,000$ cm.
2. Convert cm to metres: $300,000 \div 100 = 3000$ m.
3. Convert metres to kilometres: $3000 \div 1000 = 3$ km.

Application of Ratio to Mixing

Mixing problems require calculating the total amounts of specific components when two different mixtures are combined. There are two main methods for this.

Example: Combining Mixtures X and Y

Mixture $X$ contains orange concentrate and water in the ratio $3 : 7$. Mixture $Y$ contains orange concentrate and water in the ratio $1 : 4$. If $1$ litre of $X$ and $2$ litres of $Y$ are mixed to form Mixture $Z$, what is the ratio of orange to water in $Z$?

Method 1: Volumetric Calculation

1. For Mixture $X$ ($1000$ ml total): $3 + 7 = 10$ parts. One part is $100$ ml. Orange = $300$ ml, Water = $700$ ml.
2. For Mixture $Y$ ($2000$ ml total): $1 + 4 = 5$ parts. One part is $400$ ml. Orange = $400$ ml, Water = $1600$ ml.
3. Total Orange in $Z$: $300 + 400 = 700$ ml. Total Water in $Z$: $700 + 1600 = 2300$ ml.
4. Ratio Orange : Water in $Z$ = $700 : 2300 = 7 : 23$.

Method 2: Scaling Parts

1. Mixture $X$ ($3 : 7$) has $10$ parts. To compare, we write Mixture $Y$ ($1 : 4$) with the same number of parts, which is $2 : 8$.
2. Since we have $2$ litres of $Y$ for every $1$ litre of $X$, we multiply the components of $Y$ by $2$.
3. The new ratio is $(3 + (2 \times 2)) : (7 + (2 \times 8)) = 7 : 23$.

Application of Ratio to Concentrations

Concentrations relate the volume of one state or substance to another during a process like compression.

Example: Gas Compression

When a gas is compressed to a liquid, the ratio of gas volume to final liquid volume is $100 : 3$. How many litres of gas are needed for $1$ litre of liquid?

1. The ratio means $100$ litres of gas produce $3$ litres of liquid.
2. Divide the gas volume by the liquid parts: $100 \div 3 = 33 \frac{1}{3}$ litres.

Multiplicative Relationships in Smoothies

Consider a smoothie containing apple juice and mango juice where there is twice as much apple as mango.

1. Let $a$ be apple and $m$ be mango. The equation is $a = 2m$.
2. The ratio $a : m$ is $2m : m$, which simplifies to $2 : 1$.
3. The fraction of the smoothie that is apple is $\frac{2}{2+1} = \frac{2}{3}$.