Ratio Problems in Context
Use ratios to solve real-life problems for NSW Year 12 Mathematics Standard 1 (NESA outcome MST-12-S1-05). Most everyday ratio problems, like recipes, mixtures, groups of people and shared costs, come down to finding the value of one part, or the amount for one, then multiplying up.
You will scale a recipe or a price with the unitary method, find missing amounts in a mixture when one is known, share a total in a given ratio, and check your answers make sense — skills for cooking for a crowd, mixing paint or concrete, or splitting a bill.
Theory
Most practical ratio problems are solved by finding the value of one part (or the amount for one unit, such as one person or one item) and then multiplying up. This Year 12 Standard 1 (NSW) guide covers recipes, mixtures, groups of people and sharing a profit.
Scaling a recipe or a price uses the unitary method: divide to find the amount for \(1\) (one person, one serve, one item), then multiply by the number you need.
In a mixture with one amount known, one part is the known amount divided by its number of parts. Each other ingredient is its number of parts times one part. When sharing a total, one part is the total divided by the total number of parts.
Always check: the new amounts should still be in the original ratio, and shares should add to the total. Keep units the same and give money to the cent.
Unitary method:
One amount known:
Sharing a total:
How to solve a ratio problem in context
- Read what is given: a total, one share, or an amount for a number of people or items.
- Find one part (or the amount for one unit) by dividing.
- Multiply up to the amount you need.
- Check the answers are still in the ratio and add to any total. Use the same units and give money to the cent.
Use the unitary method: find the mince for one serve, then multiply by \(10\).
| \(\text{One serve}\) | \(=\) | \(\dfrac{450}{6}=75\text{ g}\) |
| \(\text{Ten serves}\) | \(=\) | \(10\times 75=750\text{ g}\) |
Check: \(750\div10=75\text{ g}\) per serve. \(750\text{ g}\) of mince is needed.
The concentrate is \(1\) part, so one part is \(25\text{ mL}\).
| \(\text{Water}\) | \(=\) | \(40\times 25=1000\text{ mL}\) |
| \(\text{Mixture}\) | \(=\) | \(25+1000=1025\text{ mL}\) |
\(1000\text{ mL}\) (\(1\text{ L}\)) of water is needed, making \(1025\text{ mL}\) of mixture.
Girls are \(5\) parts, so divide by \(5\) for one part.
| \(\text{One part}\) | \(=\) | \(\dfrac{40}{5}=8\) |
| \(\text{Boys}\) | \(=\) | \(4\times 8=32\) |
| \(\text{Total parts}\) | \(=\) | \(4+5=9\) |
| \(\text{Total}\) | \(=\) | \(9\times 8=72\) |
There are \(32\) boys and \(72\) students altogether.
Write the investments as a ratio in simplest form (HCF \(5000\)), then share the profit.
| \(15\,000:25\,000\) | \(=\) | \(3:5\) |
| \(\text{Total parts}\) | \(=\) | \(3+5=8\) |
| \(\text{One part}\) | \(=\) | \(\dfrac{\$12\,400}{8}=\$1550\) |
| \(\text{Priya}\) | \(=\) | \(3\times\$1550=\$4650\) |
| \(\text{Sam}\) | \(=\) | \(5\times\$1550=\$7750\) |
Check: \(\$4650+\$7750=\$12\,400\). Priya receives \(\$4650\) and Sam receives \(\$7750\).
Common pitfalls
Frequently asked questions
What is the unitary method in ratios?
The unitary method means finding the amount for one unit first, such as one person or one serve, by dividing. Then multiply by the number of units you need.
How do you scale a recipe up or down?
Divide each ingredient by the number of serves in the recipe to get the amount for one serve, then multiply by the number of serves you want. For example, 450 g for 6 serves is 75 g per serve, so 10 serves need 750 g.
How do you work out a mixture when you know one ingredient?
Divide the known amount by its number of parts to find one part. Then multiply one part by the number of parts for each other ingredient.
How do you share a profit in the ratio of investments?
Write the investments as a ratio and simplify it. Add the parts, divide the profit by the total parts to find one part, then multiply by each partner’s number of parts.
How do you find the total number from one group in a ratio?
Divide the size of the known group by its number of parts to find one part. Then multiply one part by the total number of parts in the ratio.