Part 4: Rates and Ratios | Beginner’s Guide to Maths Std 2

Understanding rates and ratios are crucial real-world skills. To help you get across them for you HSC, we've put together this introductory guide with concept check questions!

Worried about the ratio of your marks to your effort? Well in this article, we’re going to take a load off you and give you an introduction to Maths Standard 2 rates and ratios.

 

In this article, we’ll discuss:

 

Practise your rates and ratios skills!

 

Download your free Rates and Ratios worksheet

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Year 12 Standard Mathematics: Rates & Ratios

Measurement is the application of numbers and geometry, and applying it in practical situations to solve problems.

Measurement includes ratios, where interpretation of maps and plans are needed to effectively know their size.

Rates are a ratio which the two quantities are measured in different units. Rates you may have seen before include speed (km/h or m/s) when driving, or hourly wage ($/hour).

 

 

NESA Syllabus Outcomes

NESA requires students to be proficient in the following outcomes:

  • Solve practical problems involving ratio, for example capture-recapture, mixtures for building materials or cost per item
    • Work with ratio to express a ratio in simplest form, to find the ratio of two quantities and to divide a quantity in a given ratio
    • Use ratio to describe map scales
  • Use, simplify and convert between units of rates, for example \( km/h \ \text{and} \ m/s, mL/min \ \text{and} L/h \)
  • Use rates to solve practical problems

 

Assumed Knowledge

Students should be familiar with the ratio form and be able to simplify the ratio form.

Students should also be able to calculate quantities from a given ratio.

A ratio is defined as: “a quotient or proportion of two numbers, magnitudes or algebraic expression.” It is assumed that students know the standard SI prefixes (milli, kilo, giga etc.) and how to convert to between these different units of measurements.

 

 

Simplifying Ratios

A ratio is used to compare amounts of the same unit in a set proportion.

It is essentially a fraction, and can be simplified in the same way. For example, \(12:16 \ = \ 3:4\), by dividing each number by \(4\).

Ratios can be manipulated by multiplying or dividing each side by the same number, eg. \(12:16 \ = \ 6:8 \ = \ 24:32\).

 

Example 1

Write the following ratios in the simplest form.

a) \(27:81\)

b) \(28:56:14\)

c)  \(\frac{3}{8}:\frac{9}{25}\)

d) \(7.15:15.80 \)

 

Solution 1

a)

\(27:81 \ = \ 1:3\)

b)

\(28:56:14 \ = \ 2:4:1\)

c)

\(\frac{3}{8}:\frac{9}{25} \ = \ \frac{1}{8}:\frac{3}{25} \ = \ 25:24\)

d)

\(7.15:15.80 \ = \ 715:1580 \ = \ 143:316 \)

 

Converting Ratios Between Units

Ratios can be expressed as between units, for example \( 36 \text{min} : 2 \text{hrs} \).

Ratios can be simplified by changing the units of one side to match the other.

It is generally easier to match the smaller unit.

\( 36 min : 2 hrs \ = \ 36 min :120 min \ = \ 3 min : 10 min \ = \ 3:10 \)

 

 

Example 2

Write the following ratios in the simplest form by making the units the same on both sides.

a) \( 4.57 \ L : 210 \ mL \)

b) \( 1.5 \ ha : 200 \ m^{2} \)

c) \( 6 \ mm : 0.87 \ km \)

d) \( 2 \ hr : 4 \ days \)

 

Solution 2 

a)

\( 4.57L:210mL \ = \ 4570mL:210mL \ = \ 457:21 \)

b)

\( 1.5ha:200m^{2} \ = \ 15000m^{2}:200m^{2} \ = \ 150:2 \ = \ 75:1 \)

c)

\( 6mm:0.87km \ = \ 6mm:870m \ = \ 6mm:87000cm \ =  \ 6mm:870000mm \ = \ 1:145000 \)

d)

\( 2hr:4days \ = \ 2hr:96hr \ = \ 1:48 \)

 

 

Dividing a Quantity in a Given Ratio

Ratio quantities can be solved by dividing the quantity in a given ratio.

Steps in dividing a quantity in a given ratio

  1. Find the total number of parts
  2. Divide the total quantity by the number of parts to determine how much 1 part is worth
  3. Multiply each amount of the ratio by the amount of 1 part
  4. Check answers. The total sum of (3) should be equal to the total quantity

 

Example 3

A \( 450g \) bag of lollies was shared between 3 people in the ratio of \( 2:3:4 \). How many grams does each person get?

 

Solution 3

\begin{align*}
\text{Total number of parts} &:9 \\
1 \ \text{part} &= \frac{450}{9} \\
&=50g
\end{align*}

Therefore it is split into \(100g,150g\ \text{and} \ 200g \)

 

 

Example 4

The four sides of a quadrilateral have the ratio \( 1:2:3:6 \). The longest side has the length \( 18mm \). What is the perimeter of the quadrilateral?

 

Solution 4

\begin{align*}
\text{Total number of parts} &:12 \\
6 \ parts=18mm \\
1 \ part=3mm
\end{align*}

Therefore perimeter \( = 12 \times 3 = 36mm \)

 

 

Map Scale Problems with Ratios

A scale drawing is a drawing which represents the actual object. The scale factor is the ratio of the drawing to the actual size of the object.

Maps are a common example of a scaled drawing.

For example, \(1cm:1km \) means that \(1cm \) on the map represents \(1km \) in real life.

Some scales may not have any units, for example just \(1:100 \).

 

Example 5

A scale drawing has the scale \( 1:200 \). What is the actual length of the object given its drawing lengths? Express answers in metres.

a) \( 4mm \)

b) \( 63mm \)

c) \( 130cm \)

d) \( 9.6cm \)

 

Solution 5

a)

\begin{align*}
&1:200 \\
&= \ 4mm:800mm \\
&800mm \ = \ 80cm \ = \ 0.8m
\end{align*}

b)

\begin{align*}
&1:200 \\
&= \ 63mm:12600mm \\
&12600mm \ = \ 1260cm \ = \ 12.6m
\end{align*}

c)

\begin{align*}
&1:200 \\
&=  130cm:26000cm \\
&26000cm = 260m
\end{align*}

d)

\begin{align*}
& 1:200 \\
&= \ 9.6cm:1920cm \\
&1920cm \ = \ 19.2m
\end{align*}

 

Example 6

A scale drawing has the scale \(1:200 \). What is the drawing length of the object given its actual lengths? Express answers in millimetres.

a) \( 4000 cm \)

b) \( 4 km \)

c) \( 1292m \)

d) \( 3.02km \)

 

Solution 6

a)

\( 20cm \ = \ 200mm \)

b)

\( 20m \ = \ 2000cm \ = \ 20000mm \)

c)

\( 6.46m \ = \ 646cm \ = \ 6460mm \)

d)

\( 15.1m \ = \ 1510cm \ = \ 15100mm \)

 

 

Converting Between Units of Rates

It is important to be able to convert between different units of rates (e.g. m/s to km/h). In order to do this, we must know how to convert between different units of measurement. Let’s look at the following example to see how this can be done.

 

Example 7:

Convert \( 2km/h \) to \( m/s \) .

 

Solution 7:

\begin{align*}
\text{Since} \ km \ \xrightarrow{ \times 1000} m \ &and \ h \ \xrightarrow{ \times 3600} s \\
∴ \frac{1}{h} & \xrightarrow{ \div 3600} \frac{1}{s}
\end{align*}

Hence to convert from \( km/h \) to \( m/s \) , we must \( \times 1000 \div 3600 \ = \)

\begin{align*}
∴ 2km/h \ = \ 2 \times \frac{5}{18}  m/s \\
= \frac{5}{9}  m/s
\end{align*}

 

 

Using Rates to Make Comparisons

Rates can be used to make comparisons between different quantities of the same product. To compare different quantities, we must first convert it to the same quantity.

We do this by converting to unit rate, which is where the rates are expressed with quantity one.

For example, when given a price, the unit price is calculated

Unit Price: Price per one unit of measurement (e.g. per kg, m)

 

Example 8:

Which is cheaper, a \( 2L \) and \( 3L \) bottle of milk, priced at \( $3.25  \) and \( $5 \) respectively?

 

Solution 8:

For 2L bottle,

\begin{align*}
2L \ = \ $3.25 \\
∴ 1L \ = \ $1.625
\end{align*}

For 3L bottle,

\begin{align*}
3L \ = \ $5 \\
∴ 1L \ = \ $1.67
\end{align*}

\( ∴ 2L \)  bottle is cheaper out of the two

 

Energy Problems with Rates

Power is the rate at which energy is generate or consumed. Watt \( (W) \)  is the derived unit of power and is equal to one joule per second.

The standard \(  SI \) prefixes (milli, kilo, giga etc.) can be used to change the unit of measurement.

When comparing how efficiency of energy of different appliances, we measure the energy consumption (amount of energy consumed per unit of time). A common energy consumption unit is kilowatt-hour \( (kWh) \) which measures the electrical energy used in an hour.

The lower the \( kWh \) , the less an appliance uses in a year. By multiplying this by the price per \( kW \), the running cost of appliances can be calculated and compared.

 

Example 9:

Determine the cost of running appliances in a year if an electricity company charges \( $0.23 \ \text{per} \ kWh \) :

a) A \( 3kW  \) heater that has been used for \( 15  \) hours per year.

 

b) A \( 15W \) bulb that is used \( 3 \) hours per day, \( 7 \) days a week.

 

Solution 9

a. \( \text{Cost} \ = 3 \times 15 \times 0.23 \ = \$10.35 \)

 

b. The bulb has been used for \( 3 \times 7 \times 52 \ = \ 1092 \) hours in a year, since there are 52 weeks in a year.

\( \text{Cost} \ = \ 0.015 \times 1092 \times 0.23 \ = \ $37.67 \)

 

 

Concept Check Questions

1. Jam is made by adding 1-part gelatin to \( 50 \) parts fruit and \( 49 \) parts sugar. If \( 1.25kg \) sugar was added, then how much fruit should be added to the mixture?

 

2. Which is slower/cheaper? Support your answer with mathematical reasoning.

a. A \( 200g \) packet of chips for \( $2.20  \) or \( 500g  \) packet of chips for \( $5.00  \)

b. An Olympic swimmer swimming at \( 0.5m/s  \) or a cycler riding at \( 2km/h  \)?

 

3. In a \( 5m \times 5m  \)section of land, a farmer counted \( 7  \) cows. How many cows are expected to be in a \( 3.5km \times 3.5km  \) block of land?

 

4. The following appliances are used by a family in a usual week. \( X \) company charges \( $0.35 \) this family per kWh.

  • A \( 3kW \) air-conditioning unit used for \( 2 \) hours every Monday
  • A \( 500W \) fridge that is used for \( 8 \) hours everyday
  • \( 20 \times 10W \) light bulbs that each are used \( 8 \) hours everyday
  • A \( 100W \) television that has been used for \( 2 \) hour each day

Using the above information, answer the following questions.

a) Which appliance has the lowest running cost in a week?

b) Which appliance has the highest running cost in a week?

c) What is the total running cost of all the appliances in a week?

 

Concept Check Solutions

1.

\begin{align*}
2.94 \ kg \ &= \ 49 \ parts \\
1 \ part &=0.06 \ kg \\
50 \ parts & =3 \ kg \ fruit
\end{align*}

 

2. The solution for question 2 is as follows:

a) Let’s calculate the price per \( 100g \) :

For the \( 200g \) packet of chips :

\begin{align*}
\text{Price per 100g} \ = \ \frac{$2.20}{2} = $1.10
\end{align*}

For the \( 500g \) packet of chips:

\begin{align*}
\text{Price per 100g} = \frac{$5}{5} =$1.00
\end{align*}

Hence the \( 500g \) packet of chips is cheaper.

 

b) Olympic swimmer \( = 0.5m/s = \ 1 \div \ 1000 \ \times \ 3600 \ = \ 1.8km/h \)

Hence, the Olympic is slower.

 

3. In a \( 5m \times 5m \) section of land, a farmer counted \( 7 \) cows. How many cows are expected to be in a \( 3.5km \times 3.5km \) block of land?

\begin{align*}
5m:3500m \ = \ 1:700 \\
7 \times 700 \ = \ 4900 \ cows
\end{align*}

 

4.

Cost of air conditioning unit:

\( 2 \times 3 \times 0.35 \ = \ $2.1 \)

Cost of fridge:

\( 0.5 \times 8 \times 7 \times 0.35 \ = \ $9.8 \)

Cost of light bulbs:

\( 20 \times 0.01 \times 8 \times 7\times 0.35 \ = \ $3.92 \)

Cost of television:

\( 0.1\times 2 \times 7 \times 0.35 \ = \ $.49 \)

 

a) The television has the lowest running cost of \(  49c \).

b) The fridge has the highest running cost of \( $9.80 \)

c) Total running cost \( = $2.10 + $9.80 + $3.92 + $.49 = $16.31 \)

 

 

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