Ratio and Proportion for ESAT Mathematics

Expressing Multiplicative Relationships

When two quantities are related multiplicatively, we can express that relationship as an equation, a ratio, or a fraction. If there is $a$ times as much of quantity $X$ in a mixture as there is of quantity $Y$, we can write the equation as $aY = X$.

To express this as a ratio of the quantity of $X$ to the quantity of $Y$, we write $a : 1$. Converting this into a fraction gives:

$\frac{\text{quantity of } X}{\text{quantity of } Y} = \frac{a}{1} = a$

Consider a fruit smoothie containing only apple juice ($a$) and mango juice ($m$). If there is twice as much apple juice as mango juice, we can determine the following:

1. The equation is $a = 2m$.
2. To find the ratio of apple to mango juice ($a : m$), we substitute the equation to get $2m : m$. Dividing both sides by $m$ gives the simplified ratio $2 : 1$.
3. To find the fraction of the smoothie that is apple juice, we look at the parts. Since the ratio is $2 : 1$, there are $2 + 1 = 3$ parts in total. The fraction of apple juice is therefore $\frac{2}{2+1} = \frac{2}{3}$.

Application of Ratio to Conversion

Ratio is a fundamental tool for converting between different units or currencies. If the conversion ratio of dinars to dollars is $11 : 2$, we can calculate how many dollars are equivalent to $2530$ dinars.

The ratio indicates that every $11$ dinars are worth $2$ dollars. First, we find how many sets of $11$ exist in $2530$:

$2530 \div 11 = 230$

Since there are $230$ lots of $11$ dinars, there must be $230$ lots of $2$ dollars:

$230 \times 2 = 460 \text{ dollars}$

Application of Ratio to Comparison

In problems where two different quantities are compared to a common third quantity, we must standardise the third quantity to compare the first two directly.

Suppose a mixture of orange purée, mango purée, and water has a ratio of orange to water of $3 : 5$ and a ratio of mango to water of $11 : 15$. To find the ratio of orange purée to mango purée, we identify water as the linking quantity.

Orange to water is $3 : 5$ and water to mango is $15 : 11$. To compare orange and mango, the number of parts for water must be identical in both ratios. We multiply the orange to water ratio by $3$:

$3 \times 3 : 5 \times 3 = 9 : 15$

Now we have orange to water as $9 : 15$ and water to mango as $15 : 11$. Since the water parts match, we can conclude the ratio of orange to mango is $9 : 11$.

Application of Ratio to Scaling

Scale drawings use ratios to relate measurements on paper to real world distances. A scale of $1 : 10,000$ means every $1$ unit on the drawing represents $10,000$ units in reality.

If a park is $30 \text{ cm}$ long on a scale drawing with a $1 : 10,000$ ratio, the real length is:

$30 \text{ cm} \times 10,000 = 300,000 \text{ cm}$

To convert this to kilometres, we convert through metres:

$300,000 \text{ cm} = 3000 \text{ m} = 3 \text{ km}$

Application of Ratio to Mixing

Mixing problems often involve combining two different mixtures with different ratios.

Example: Mixture $X$ has orange concentrate and water in ratio $3 : 7$. Mixture $Y$ has them in ratio $1 : 4$. If we mix $1$ litre of $X$ and $2$ litres of $Y$ to form mixture $Z$, what is the new ratio?

Method 1: Absolute Volumes

For Mixture $X$ ($1000 \text{ ml}$ total): The total parts are $3 + 7 = 10$. Each part is $1000 \div 10 = 100 \text{ ml}$. Orange is $3 \times 100 = 300 \text{ ml}$ and water is $7 \times 100 = 700 \text{ ml}$.

For Mixture $Y$ ($2000 \text{ ml}$ total): The total parts are $1 + 4 = 5$. Each part is $2000 \div 5 = 400 \text{ ml}$. Orange is $1 \times 400 = 400 \text{ ml}$ and water is $4 \times 400 = 1600 \text{ ml}$.

Combining these in Mixture $Z$: Total orange is $300 + 400 = 700 \text{ ml}$. Total water is $700 + 1600 = 2300 \text{ ml}$. The ratio is $700 : 2300$, which simplifies to $7 : 23$.

Method 2: Scaling Ratios

First, express both ratios with the same total number of parts. Mixture $X$ ($3 : 7$) has $10$ parts. Mixture $Y$ ($1 : 4$) can be rewritten as $2 : 8$, which also has $10$ parts.

Because we are mixing $1$ part of $X$ and $2$ parts of $Y$, the new ratio is:

$(3 + (2 \times 2)) : (7 + (2 \times 8)) = (3 + 4) : (7 + 16) = 7 : 23$

Application of Ratio to Concentrations

Concentration problems often deal with the reduction of volume, such as gas compression. If the ratio of gas volume to liquid volume after compression is $100 : 3$, we can determine the input needed for a specific output.

To find how many litres of gas are needed to produce $1$ litre of liquid, we use the ratio as a scaling factor. If $100$ litres of gas produce $3$ litres of liquid, then for $1$ litre of liquid we need:

$100 \div 3 = 33\frac{1}{3} \text{ litres of gas}$