Ratios and Rates
A ratio compares quantities in the same units; a rate compares quantities in different units, such as kilometres per hour. This NSW Year 9 Core maths subtopic revises both from Stage 4 (NESA outcome MA4‑RAT‑C‑01), with mixed units, ratios of four terms and problems that work backwards.
You will write ratios from worded comparisons, simplify ratios of whole numbers and decimals, convert to a common unit before simplifying, such as \(500\) g to \(2.2\) kg, find a missing term, and split an amount into shares that follow a ratio. You will also find the whole from one share, write rates in simplest form and use them, from a car's speed to a city's birth rate, the price of lamb chops per kilogram and an hourly rate of pay. It is part of Year 9 Core Working with Numbers.
Theory
A ratio compares quantities measured in the same unit, and a rate compares quantities measured in different units. This Year 9 maths page revises the skills from earlier years: simplifying ratios, dividing a quantity in a ratio, working back from one share, and writing and using rates such as dollars per kilogram or km/h.
A ratio compares two or more quantities of the same kind, such as \(3 : 4\). The quantities must be in the same unit, and the ratio itself has no units. The order matters: \(3 : 4\) is not \(4 : 3\).
A ratio is in simplest form when its terms are whole numbers with no common factor except \(1\). Divide every term by the highest common factor (HCF).
In the ratio \(a : b\) there are \(a + b\) parts altogether. The first share is \(\dfrac{a}{a+b}\) of the whole.
A rate compares quantities in different units, such as \(60\) km/h or \(\$4.60\) per kg. A rate is simplified by making the second quantity \(1\) unit. Finding the value of one part, or one unit, first is called the unitary method.
Dividing a quantity in the ratio \(a : b\):
Using a rate:
For speed, \(\text{distance} = \text{speed} \times \text{time}\). At \(70\) km/h for \(4\) hours, a car travels \(280\) km.
How to solve a ratio or rate problem
- Write the quantities in the order the question names them, in the same unit for a ratio.
- Simplify by dividing every term by the HCF. Clear decimals first by multiplying by \(10\) or \(100\).
- Count the parts. Match the total, one share or the difference to the right number of parts.
- Find one part (or one unit of a rate), then multiply up to what is asked.
- Check that the shares add to the total and simplify back to the ratio.
(i) \(54 : 72 : 90\),
(ii) \(0.45 : 1.2\) and
(iii) \(750\) g \(: 2.5\) kg.
(i) The HCF of \(54\), \(72\) and \(90\) is \(18\). Divide every term by it:
| \(54 : 72 : 90\) | \(=\) | \(\mathbf{3 : 4 : 5}\) |
(ii) Make whole numbers first by multiplying both terms by \(100\), then divide by the HCF, \(15\):
| \(0.45 : 1.2\) | \(=\) | \(45 : 120\) |
| \(\) | \(=\) | \(\mathbf{3 : 8}\) |
(iii) Change to the same unit, then drop the units and divide by \(250\):
| \(750 \text{ g} : 2.5 \text{ kg}\) | \(=\) | \(750 \text{ g} : 2500 \text{ g}\) |
| \(\) | \(=\) | \(750 : 2500\) |
| \(\) | \(=\) | \(\mathbf{3 : 10}\) |
Check each answer: the terms have no common factor other than \(1\).
Count the parts, find one part, then multiply up:
| \(\text{parts}\) | \(=\) | \(3 + 4 + 5 = 12\) |
| \(\text{one part}\) | \(=\) | \(168 \div 12 = \$14\) |
| \(\text{Ivy}\) | \(=\) | \(3 \times 14 = \mathbf{\$42}\) |
| \(\text{Jai}\) | \(=\) | \(4 \times 14 = \mathbf{\$56}\) |
| \(\text{Kim}\) | \(=\) | \(5 \times 14 = \mathbf{\$70}\) |
Check: \(42 + 56 + 70 = 168\).
(i) A recipe uses sugar and flour in the ratio \(3 : 7\). It uses \(350\) g of flour. How much sugar?
(ii) Mia and Tom share money in the ratio \(4 : 7\). Tom gets \(\$45\) more than Mia. How much is shared altogether?
(i) Flour is \(7\) parts, so find one part from the flour:
| \(7 \text{ parts}\) | \(=\) | \(350 \text{ g}\) |
| \(1 \text{ part}\) | \(=\) | \(50 \text{ g}\) |
| \(\text{sugar}\) | \(=\) | \(3 \times 50 = \mathbf{150} \text{ g}\) |
(ii) The \(\$45\) is the DIFFERENCE, so match it to the difference in parts, \(7 - 4 = 3\):
| \(3 \text{ parts}\) | \(=\) | \(\$45\) |
| \(1 \text{ part}\) | \(=\) | \(\$15\) |
| \(\text{total}\) | \(=\) | \(11 \times 15 = \mathbf{\$165}\) |
Check: Mia \(\$60\), Tom \(\$105\), and \(105 - 60 = 45\).
(i) Write this as a rate in dollars per kilogram.
(ii) What do \(10\) kg cost?
(iii) How many kilograms can be bought for \(\$69\)?
(i) Make the second quantity ONE kilogram:
| \(\text{rate}\) | \(=\) | \(27.60 \div 6\) |
| \(\) | \(=\) | \(\mathbf{\$4.60} \text{ per kg}\) |
(ii) Multiply the rate by the mass:
| \(\text{cost}\) | \(=\) | \(4.60 \times 10 = \mathbf{\$46.00}\) |
(iii) Divide the money by the rate:
| \(\text{mass}\) | \(=\) | \(69 \div 4.60 = \mathbf{15} \text{ kg}\) |
A rate keeps its units: dollars per kilogram, km/h, litres per minute.
Common pitfalls
Frequently asked questions
How do you simplify a ratio?
Divide every term by the highest common factor. If there are decimals, multiply every term by \(10\) or \(100\) first. If the units differ, change them to the same unit first.
How do you divide an amount in a ratio?
Add the terms to get the number of parts, divide the amount by that to get one part, then multiply one part by each term.
What is the difference between a ratio and a rate?
A ratio compares quantities in the same unit and has no units. A rate compares quantities in different units, such as kilometres per hour, and keeps both units.
How do you find the total from one share of a ratio?
Divide the share by its number of parts to get one part. Then multiply one part by the total number of parts.
What is the unitary method?
It means finding the value of one part or one unit first, then multiplying to find any other amount.