Real-Life Ratio Applications and Best-Buys

You must memorise the following AQA-specific metric and imperial approximations:

• $5\text{ miles} \approx 8\text{ km}$
• $1\text{ kg} \approx 2.2\text{ lbs}$ (pounds)
• $1\text{ litre} \approx 1.75\text{ pints}$
• $1\text{ gallon} \approx 4.5\text{ litres}$
• $1\text{ inch} \approx 2.5\text{ cm}$

Worked Example: Map Scale Conversion

A map scale is $1 : 500,000$. Two towns are $8\text{ cm}$ apart on the map. Find the actual distance in km.

Step 1: Multiply by the scale factor to find the real distance in cm.

• $8\text{ cm} \times 500,000 = 4,000,000\text{ cm}$

Step 2: Convert cm to m (since $1\text{ m} = 100\text{ cm}$).

• $4,000,000\text{ cm} \div 100 = 40,000\text{ m}$

Step 3: Convert m to km (since $1\text{ km} = 1000\text{ m}$).

• $40,000\text{ m} \div 1000 = 40\text{ km}$

Comparing Ratios and Concentration

If you have ever poured too much squash into your glass, you already intuitively understand concentration. Concentration is the ratio of a specific ingredient (like syrup) to the total mixture or solvent.

To compare two different mixtures, you can use the unitary method. This involves calculating the value of a single unit first (e.g., finding the amount of water per $1$ unit of syrup) to make a fair comparison. Alternatively, you can use a common total comparison by finding the Least Common Multiple (LCM) of the total number of parts.

Worked Example: Comparing Concentration

Which is stronger: Mixture A ($2$ parts syrup to $5$ parts water) or Mixture B ($3$ parts syrup to $7$ parts water)?

Step 1: Calculate the amount of water per $1$ part syrup for Mixture A.

• $5 \div 2 = 2.5\text{ parts water per }1\text{ part syrup}$

Step 2: Calculate the amount of water per $1$ part syrup for Mixture B.

• $7 \div 3 \approx 2.33\text{ parts water per }1\text{ part syrup}$

Step 3: Compare and write a concluding statement.

• Mixture B is stronger because it contains less water ($2.33\text{ parts}$) per $1$ unit of syrup compared to Mixture A ($2.5\text{ parts}$).

Mixing Quantities and Limiting Factors

Builders cannot just guess the amount of sand and cement they need; they must follow strict mathematical mixing ratios so the concrete does not crumble under pressure. Mixing problems typically use a part-to-part ratio (comparing individual components, like $1$ part cement to $2$ parts sand) or a part-to-whole ratio (comparing one component to the entire mixture).

To calculate specific ingredient quantities for a required total mixture, use the unitary method to find the volume or weight of a single part, then multiply by the ratio shares.

Worked Example: Calculating Ingredient Quantities and Cost

$18\text{ L}$ of paint is mixed in the ratio $3 : 2$ (White : Blue). White paint costs $£2.80$ per litre and Blue paint costs $£3.50$ per litre. Calculate the total cost of the mixture.

Step 1: Find the total number of parts and calculate the volume of $1$ part (the unitary value).

• Total parts = $3 + 2 = 5$ parts
• $1\text{ part} = 18\text{ L} \div 5 = 3.6\text{ L}$

Step 2: Multiply by the ratio to find the volume of each specific ingredient.

• White paint: $3 \times 3.6\text{ L} = 10.8\text{ L}$
• Blue paint: $2 \times 3.6\text{ L} = 7.2\text{ L}$

Step 3: Calculate the cost for each ingredient and sum them for the total cost.

• Cost of White: $10.8 \times £2.80 = £30.24$
• Cost of Blue: $7.2 \times £3.50 = £25.20$
• Total Cost: $£30.24 + £25.20 = £55.44$

When you have fixed amounts of available ingredients, you must apply the limiting ingredient rule. The maximum total mixture you can make is always determined by the ingredient with the smallest scale factor.

Worked Example: Limiting Factor in Mixing

Concrete is mixed in the ratio Cement : Sand : Gravel = $1 : 2 : 4$. You have $37\text{ kg}$ Cement, $120\text{ kg}$ Sand, and $200\text{ kg}$ Gravel. Calculate the maximum amount of concrete you can make.

Step 1: Find the scale factor multiplier for each ingredient (Available Amount $\div$ Ratio Part).

• Cement: $37 \div 1 = 37$
• Sand: $120 \div 2 = 60$
• Gravel: $200 \div 4 = 50$

Step 2: Identify the limiting ingredient.

• The lowest multiplier is $37$, meaning Cement limits the total amount you can make.

Step 3: Multiply the total parts by the limiting multiplier.

• Total ratio parts = $1 + 2 + 4 = 7$
• Maximum concrete = $7 \times 37 = 259\text{ kg}$

Solving Best-Buy Problems

Supermarkets deliberately package items in different sizes to make it tricky to spot the best deal. To solve these problems, we must determine true value for money by comparing products on an equivalent, fair basis to find the best buy.

There are two primary methods to do this. You can calculate the Price per Unit (where the lowest value wins) or you can calculate the Amount per £1 (where the highest value wins). Whichever method you choose, you must end with a clear statement comparing your results.

Worked Example: Price per Unit Method

Brand A is $400\text{ g}$ for $£2.56$. Brand B is $750\text{ g}$ for $£5.10$. Which is the best buy?

Step 1: Calculate the cost per $100\text{ g}$ for Brand A.

• $£2.56 \div 4 = £0.64\text{ per }100\text{ g}$

Step 2: Calculate the cost per $100\text{ g}$ for Brand B.

• $£5.10 \div 7.5 = £0.68\text{ per }100\text{ g}$

Step 3: Compare and conclude.

• Brand A is the best buy because $£0.64 < £0.68$, meaning it is cheaper per unit of weight.

Worked Example: Amount per £1 Method

Which offer is better: Offer X ($1.2\text{ L}$ for $£1.80$) or Offer Y ($2\text{ L}$ for $£3.20$)?

Step 1: Divide the quantity by the cost for Offer X.

• $1.2 \div 1.80 = 0.667\text{ litres per }£1$

Step 2: Divide the quantity by the cost for Offer Y.

• $2 \div 3.20 = 0.625\text{ litres per }£1$

Step 3: Compare and conclude.

• Offer X is the best buy because $0.667 > 0.625$, meaning you get a greater amount of product per $£1$ spent.