Mathematics
Clarifications for the syllabus Units 3A and 3B
2012/2567[v2]
Mathematics: Clarifications for the syllabus Units 3A/3B
1
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resources that teachers can use to support their learning programs. Their inclusion does not imply that they are mandatory or
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2
Mathematics: Clarifications for the syllabus Units 3A/3B
Clarifications for the Mathematics syllabus—Units 3A/B
Introduction
The following document has been prepared at the request of the Mathematics Course Advisory
Committee in response to concerns expressed by teachers that the 3A/B MAT course contains too
much content. The committee is of the view that the delivery of the course content at the intended
depth may go part of the way to ameliorating the situation. The examples below are provided to
help clarify the spirit and intent of the syllabus content for 3A/B MAT.
It is important to note that the 3A/B MAT course is being undertaken by two different cohorts of
students (a Year 11 group and a Year 12 group) and the needs of both are slightly different.
Teachers may wish to provide the Year 11 cohort of students with background experiences that
develop concepts to a level beyond those indicated here. However it would be prudent not to
devote too much time to the practise of difficult manipulative skills for their own sake unless they
are required by the syllabus and facilitate essential concept development required for further
learning. Given this possible extension of content teachers should still ensure that assessment
remains within the bounds of the 3A/BMAT syllabus.
The specific syllabus areas for which clarification is provided are:
 The Normal Distribution
 Indices
 Counting and Probability
 Reasoning Geometrically
 Finance
Mathematics: Clarifications for the syllabus Units 3A/3B
3
The Normal Distribution
This content is covered in Unit 3A and it is worth referring to the comment within the unit
description which states that students are to analyse normally distributed data. The calculation and
interpretation of probabilities for normal random variables is covered in unit 3C. Thus the intent in
Unit 3A is to simply give the students a feel for normally distributed data and some of its properties
rather than a heavy focus on problems involving difficult calculations and interpretation of
probabilities associated with the Normal Distribution.
Content associated with Normal distribution from Unit 3A:
3.1.7 calculate probabilities for normal distributions with known mean  and standard deviation 
3.1.8 use the 68%, 95%, 99.7% rule for data one, two and three standard deviations from the
mean
3.1.9 use probability notation for normal random variables such as P(X < x).
3.2.3 calculate quantiles for normally distributed data with known mean and standard deviation
3.2.4 use number of standard deviations from the mean (standard scores) to describe deviations
from the mean in normally distributed data sets.
3.5.10 Compares scores from two or more sets of data using number of standard deviations
from the mean (standard scores)
Background:
In previous studies students come to understand that variation in data occurs naturally. For
example, the heights of people or errors in measurement such as having each student in the class
measuring the length of an object. In Unit 3A students are introduced to scenarios which present
normal distributions and reflect students’ interests and/or the community context, for example, the
volume of a can of soft drink or the lifetime of batteries or reaction times, and in these contexts
calculate proportions or probabilities of occurrences within plus or minus integer multiples of
standard deviations of the mean. These calculations are then extended further by using symmetry
and complementary properties of the normal distribution to calculate other proportions or
probabilities and to find a value on the distribution, given a particular proportion. These problems
should be solved using technology.
Key ideas which need to be emphasised during this unit are:
 Features of a normal distribution
- bell-shaped
- position of the mean
- symmetry about the mean
- the characteristic spread
- the unique positions of one, two and three standard deviations from the mean.
Classroom activities could be of the kind or similar to that found through the following link
http://www.casio.edu.shriro.com.au/cr_one_lesson_wonders.php→Normal Distribution.
4
Mathematics: Clarifications for the syllabus Units 3A/3B
Sample items that are indicative of the depth required are as follows:
Example 1: Question 16 3ABMAT Sample unit package
(4 marks)
Syllabus dot point
use the 68%, 95%, 99.7% rule for data one, two and three standard deviations from the
mean
3.1.8
The scores of 1000 students taking a test are approximately normally distributed, with a mean of
65 and a standard deviation of 15.
(a)
How many students would you expect to score 50 or less?
(2 marks)
(b)
How many students would you expect to score 95 or more?
(2 marks)
Solution
50 is 1 standard deviation below the mean, so we would expect about 16% of 1000
i.e. 160 students to score 50 or less.
(b) 95 is 2 standard deviations above the mean, so we would expect about 2.5% of 1000
i.e. 25 students to score 95 or more.
Specific behaviours
(a)  recognises position of one standard deviation from the mean.
 calculates the number of students.
(b)  recognises position of two standard deviations from the mean
 calculates the number of students.
(a)
Example 2
The heights of plants in a biological experiment are normally distributed with a mean of 54 cm and
standard deviation of 4 cm. Find an interval which will contain the height values of approximately
99.7% of the plants.
Solution
Approximately 99.7% of height values are between the mean plus three standard deviations and
the mean minus three standard deviations i.e.   3  H    3 , which is
54  3  4  H  54  3  4 i.e. 42  H  66 .
So approximately 99.7% of the plants have a height between 42 cm and 66 cm.
Specific behaviours
 relates the proportion of 99.7% to positions three standard deviations from the mean
 calculates the interval.
Mathematics: Clarifications for the syllabus Units 3A/3B
5
Example 3: Question 12 3ABMAT WACE 2010 examination
(7 marks)
Syllabus dot points
3.1.7 calculate probabilities for normal distributions with known mean  and standard deviation 
3.1.8 use the 68%, 95%, 99.7% rule for data one, two and three standard deviations from the
mean
The Western Australian Government collects the prices of petrol every day from all petrol outlets in
Western Australia.
On a particular day, the prices of unleaded petrol from outlets in Perth were close to being normally
distributed, with 95% of outlets charging between 120.2 c/L and 129.8 c/L and 2.5% of outlets
charging more than 129.8 c/L
(a)
Estimate the mean price of unleaded petrol in Perth on that day. Show your reasoning.
(2 marks)
Solution
The middle 95% of prices range between 120.2c and 129.8c, with 2.5% of prices above 129.8c, so
the remaining 2.5% of prices were below 120.2c.
95%
2.5%
A normal distribution is symmetrical about the mean.
therefore the estimated mean price = (120.2 + 129.8)/2 = 125.0 c/L
Specific behaviours
 recognises symmetry (explanation or diagram)
 correctly calculates mean
(b)
Estimate the percentage of outlets charging less than 126 c/L for unleaded petrol that day.
(2 marks)
Solution
In a normal distribution, 95% of
data are approximately two
standard deviations either side
of the mean, therefore
approximate value of standard
deviation = (129.8 – 120.2)/4 =
2.4
or 2 = 129.8 – 125,   = 2.4
P(X<126) = 66% (calculator
solution)
Specific behaviours
 correctly determines the approximate value of 
 correctly determines the percentage of outlets charging less than 126 c/L
6
Mathematics: Clarifications for the syllabus Units 3A/3B
Example 4: Question 9 3ABMAT WACE 2011 examination
(5 marks)
Syllabus dot points
3.1.7 calculate probabilities for normal distributions with known mean  and standard deviation 
3.1.9 use probability notation for normal random variables such as P(X < x).
3.2.3 calculate quantiles for normally distributed data with known mean and standard deviation
The masses of lettuces sold at a fruit and vegetable shop are normally distributed with a mean
mass 600g and standard deviation 20g.
(a)
If a lettuce is chosen at random, determine the probability that its mass lies between 570g
and 610g.
(1 mark)
Solution
P(570  X  610)  0.6247
Specific behaviours
 calculates correct probability
(b)
Determine the mass exceeded by 7% of the lettuces, correct to three (3) significant
figures.
(2 marks)
Solution
P( X  k )  0.07
k  629.52  630 g correct to 3 significant figures
Specific behaviours
 calculates correct probability
states solution correct to three significant figures
Mathematics: Clarifications for the syllabus Units 3A/3B
7
(c)
In one day, 1000 lettuces are sold. Estimate how many weigh less than 545g.
Solution
P( X  545)  0.00298
 The proportion is 0.3% of the 1000 lettuces i.e. 3 lettuces.
Specific behaviours
 calculates the probability
expresses the number of lettuces as a whole number
Example 5
Syllabus dot points
3.5.10 Compares scores from two or more sets of data using number of standard deviations from
the mean (standard scores)
A student has raw scores of 67% in a 3AB Mathematics examination and 73% in a Modern History
examination. Use the following information to compare the student’s performances between the
two examinations.
3AB Mathematics
Modern History
  47.3
  56.9
  15.5
  13.7
Solution
In the Mathematic examination the student’s z-score is
67  47.3
 1.27
15.5
i.e. 1.27 standard deviations above the mean
In the Modern History examination the student’s z-score is
73  56.9
 1.17
13.8
The student has scored better in the Mathematics examination because the higher z-score shows
her score is positioned further above the mean than it is in the Modern History examination.
Specific behaviours
Calculates standardized scores
Compares scores based on standard deviations from the mean.
8
Mathematics: Clarifications for the syllabus Units 3A/3B
Example 6: Question 15 3ABMAT Sample unit package
Syllabus dot point
3.2.3 calculate quantiles for normally distributed data with known mean and standard deviation
The Aussie Snack company produces small packets of fruit and nuts. The label on each packet
states that it contains 150 grams of wholesome goodness. A machine is used to fill the packets.
The amount of fruit and nuts delivered by the machine varies from packet to packet.
Let C denote the weight of the contents of a packet. The weight of the packets can be modelled by
a normal distribution with a mean of 151.9 grams and a standard deviation of 0.44 grams.
(a)
Determine the proportion of packets that will contain less than 152 grams of fruit and nuts.
(1 mark)
(b)
Aussie Snack boasts that less than 1% of its fruit and nut packets ever have less than the
advertised amount of 150 g. Is this a valid statement? Justify your answer
(2 marks)
Question 15
C ~ N(151.9, 0.44)
(a)
P(C < 152) = 0.59
(b)
P(C < 150) = 0.000 008 
Statement is valid, as less than 1%
of packets have less than 150 g in
the packet. 
Mathematics: Clarifications for the syllabus Units 3A/3B
9
Indices
This content is covered in Unit 3A. Students are to develop algebraic skills for solving equations.
The application of the index laws and the level of complexity of the manipulation of numerical and
algebraic expressions that are appropriate are to be based on this underlying need to develop the
skills necessary to solve equations.
The focus of content involving indices for 3AMAT is the development of skills that would be used in
solving problems in either the 3A/3B unit pair or associated with content found in 3C/3DMAT.
Content associated with indices from Unit 3A
1.1.7 use the laws of indices to simplify numerical and algebraic expressions and to solve
equations.
1.3.2 solve algebraically and graphically:
– quadratic equations in factored form
– cubic equations in factored form
kx
 c , b  0 (logarithms not required)
– exponential equations ab
n
- simple power equations x  c , n = 2, 3, ½, ⅓, -1.
Examples of questions involving indices have been split into three types.
Type one
1.1.7
Use the laws of indices to simplify numerical and algebraic expressions and to solve
equations.
Example 1: Question 6 3ABMAT WACE 2011 examination
Syllabus dot point
1.1.7 use the laws of indices to simplify numerical ……. expressions ………….
(a)
Evaluate the following.
3
16 4
(i)
Solution
3
16 4

3
4 4
2
 
 23  8
Specific behaviours
 applies index laws
 evaluates expression
(ii)
 7
2 9 



1
2
Solution
1

7 2

2 9 


1

25  2

 
 9 
1
2
3
 9
  
5
 25 
Specific behaviours
 applies negative index law correctly
 evaluates expression correctly
10
Mathematics: Clarifications for the syllabus Units 3A/3B
Example 2
Syllabus dot point
1.1.7 use the laws of indices to simplify …. algebraic expressions …….
The following type of questions focus on skill development.
Simplify, writing your answers with positive indices
3b6  2b0
8a 4  b3
2a  b
b 3 . b 2 . b
Example 3: Adapted from 3ABMAT Final Sample examination 2010
Syllabus dot point
1.1.7 use the laws of indices to simplify ……… algebraic expressions ……...
Simplify with positive indices and then differentiate with respect to x.
4
 3
1
y   x 2   3
  2x
 
Solution
4
 3
1
y   x 2   3
  2x
 
 x6 
x3
2
so
dy
3x 2
 6 x5 
dx
2
Specific behaviours
applies index laws to simplify expression
differentiates simplified expression
Mathematics: Clarifications for the syllabus Units 3A/3B
11
Type Two
1.1.7 use the laws of indices………and to solve equations
1 1
2 3
1.3.2 solve algebraically and graphically ….simple power equations x = c, n = 2, 3, , , -1
n
Each of the questions below require some rearranging and the use of index laws but they only
1 1
2 3
solve for n = 2, 3, , ,-1
Example 4
Solve each of the following equations:
4 x3  36 x
(a)
Solution
4 x  36 x  4 x  36 x  0
 4 x x2  9  0
 4 x  0 or x 2  9  0
 x  0 or x  3
3
3


Specific behaviours
Rearranges the equation as a polynomial.
Factorises the polynomial on the left side of the equation.
Correctly solves for x.
3x3  1  5 x3
(b)
Solution
3x  1  5 x  8 x  1
1
i.e. x3 
8 3
1

x3   
2
1
 x 
2
3
3
3
Specific behaviours
Rearranges the equation as a polynomial.
Expresses both sides of the equation in index form.
Correctly solves for x.
12
Mathematics: Clarifications for the syllabus Units 3A/3B
8x1  2x
(c)
Solution
8 x 1  2 x  23( x 1)  2 x
i.e 3x - 3  x
i.e. 2 x  3
3
 x 
2
Specific behaviours
Applies index laws to change from base 8 to base 2.
Equates the indices from both sides of the equation.
Correctly solves for x.
2( x  1)3  54
(d)
Solution
2( x  1)  54  ( x  1)  27
i.e. ( x  1)3  33
i.e. x  1  3

x
4
3
3
Specific behaviours
Rearranges the equation as a polynomial.
Expresses both sides of the equation in index form.
Correctly solves for x.
Example 5
1
3
2
3
If r = 3 , what is the value of 2r .
Solution
2
2r 3

2
 1
2 r3 
 
 
 2  32
 18
Specific behaviours
Applies index laws to give the expression in terms of
1
r3
1
3
Substitutes for r = 3
2
Correctly evaluates the expression 2r 3
Mathematics: Clarifications for the syllabus Units 3A/3B
13
Example 6: Question 2 3ABMAT Sample unit Package
Syllabus dot point
1.1.7 use the laws of indices to simplify numerical and algebraic expressions and to solve
equations.
Solve the equations:
7
(a)
3x
9 = 12
x
(3 marks)
(b)
3  27x = 96
(3 marks)
Solution
(a)
3x7
3
 12 and x  0  2  12
9
x
x
1

 x2
4
1
 x 
2
(b) 3  27 x
27 x
i.e. 27 x
so
7x
 96
 32
 25
5
5
 x 
7
Specific behaviours
States the condition x  0
Rearranges the equation as a polynomial.
Correctly solves for x
Example 7: Question 6(b) 3ABMAT WACE 2011 examination
Solve the following for x. Leave your answers as fractions.
(i)
162 x  8
(3 marks)
Solution
162 x  8  28 x  23
 8x  3
3
 x 
8
Specific behaviours
Applies index laws to get a common base.
Equates exponents from both sides of the equation.
Correctly solves for x.
14
Mathematics: Clarifications for the syllabus Units 3A/3B
23 x1  8 2
(ii)
(3 marks)
Solution
1
23 x 1  8 2  23 x 1  23  2 2
3
 23 x 1  2
 3x  1  3
1
2
1
2
 3x

x
9
2
3
2
Specific behaviours
Applies index laws to get a common base.
Equates exponents from both sides of the equation.
Correctly solves for x.
Type 3 Applications
Example 8
Skill required for solving a related rate problem in Question 13 of 3CDMAT sample examination
2010.
Sand is falling onto the top of a pile …………The pile maintains a conical shape in which the radius
of the base is always one half of the height.
Express the volume of the cone in terms of h
Solution
V

3
r 2h


 h
2
  h
32

3

h2
 h3

h

V

12
22
Mathematics: Clarifications for the syllabus Units 3A/3B
15
Example 9: Question 11 3ABMAT WACE 2010 examination
P
18 cm
Q
T
3x cm
S
R
The figure shown in the diagram is obtained by removing the triangle QRT from the rectangle
PQRS. The size of ∠QTR = 90° and QT = TR. PQ = 18 cm and PS = 3x cm.
It can be shown that QT =
3x 2
2
Verify that the area, A , of the figure PQTRS is given by
(a)
A  54 x 
9 x2
cm2
4
Solution
Area of PQTRS = area of rectangle PQRS – area of triangle QTR
1  3x 2 
A  18  3x  

2  2 
2
 2 2
1  9x 22
 54 x  
2  22






 9 x2 
 54 x  
 as required
 4 
16
Mathematics: Clarifications for the syllabus Units 3A/3B
Counting and probability
This content is covered in Unit 3AMAT. The focus for the dot point is the skills and understandings
needed to determine the counts of situations that could be represented or visualised as a sample
space.
3.1.4
use addition and multiplication principles for counting, and use the counts to calculate
probabilities
This content should also be considered in association with the following Unit 3AMAT dot points:
3.1.1
3.1.2
3.1.3
use lists, tree diagrams and two-way tables to determine sample spaces for two- and threestage events
use Venn diagrams to represent sample spaces for two events and to illustrate subset,
intersection, union and complement
use sample spaces to calculate simple probabilities and probabilities for compound events
(event A or B, event A and B, event A given event B, complement of A)
Sample items below are indicative of the depth required. The tree diagrams in Example 1 have
been included to illustrate how counting methods use the multiplication and addition principles and
make the working much simpler. It is not expected that students use both methods when
answering any of these questions.
Example 1: Adapted from Question 3 3ABMAT Final sample examination 2010
Suppose the Prime Minister of Australia wants to visit three of the cities: Sydney, Melbourne,
Brisbane, Adelaide one after another.
(a) How many possible itineraries are there?
By drawing the tree diagram it is clear that the universal sample space has 4 3  2 = 24 possible
itineraries which can be counted.
Using a counting method, the number of possible itineraries which include visits to three cities can
be calculated using the number of choices at each leg of the journey. Hence the four possible
choices of city for the first visit then three choices for the second visit and two for the third visit by
multiplication principle results in 4 3  2 = 24 possible choices.
Mathematics: Clarifications for the syllabus Units 3A/3B
17
(b)
The Prime Minister picks one of the following itineraries at random:
(i)
What is the probability that she will visit Sydney first?
A tree diagram helps illustrate the subset of favourable outcomes for the required event.
2 choices
3 choices
6 itineraries
i.e.1 x 3 x 2
Start
with
Sydney
So P (visit Sydney first)

6 1

24 4
Counting method:
There is only one choice for the first visit (Sydney) leaving three choices for the second visit and
two choices for the third visit. There are 1  3  2  6 possible itineraries which have Sydney as the
first visit.
 P(S first) 
(ii)
1 3  2 1

4  3 2 4
What is the probability that she won’t visit Sydney first?
Looking at the tree diagram, of the 24 possible itineraries 18 do not have Sydney as the starting
point.
 P(do not visit Sydney first) 
18 3

24 4
Counting method:
There are three choices to not visit Sydney first, leaving three choices for the second visit and two
choices for the third visit. There are 3 3 2  18 possible itineraries which do not have Sydney as
the first visit.
Using counting  P(do not visit Sydney first) 
18
3 3 2 3

4  3 2 4
Mathematics: Clarifications for the syllabus Units 3A/3B
(iii)
If she visits Sydney first, what is the probability that she will visit Melbourne second?
Reduced sample space:
A tree diagram helps illustrate the reduced sample space and the subset of favourable outcomes
for the required event.
Melbourne
Reduced
sample space
=1 x 3 x 2
Sydney
So P(S first and M second and other) 
2 1

6 3
Counting method
Reduced sample space:
There is only one choice for the first visit (Sydney) leaving three choices for the second visit and
two choices for the third visit.
i.e. There are 1  3  2  6 possible itineraries which have Sydney as the first visit.
Favourable outcomes:
There is only one choice for the first visit (Sydney), one choice for the second visit (Melbourne) and
two choices for the third visit (Adelaide or Brisbane).
i.e. There are 11 2  2 possible itineraries which have Sydney as the first visit and Melbourne for
the second visit and another for the third visit.
Counting  P((S first and M second and other) given Sydney first) 
1 1 2 2 1
 
1 3  2 6 3
Mathematics: Clarifications for the syllabus Units 3A/3B
19
(iv)
If she visits Sydney first, what is the probability that she will visit Melbourne or Brisbane
second?
Reduced sample space:
Melbourne
Brisbane
Reduced
sample space
=1 x 3 x 2
Sydney
From the tree diagram there are two itineraries which have Melbourne as the second city plus two
Itineraries which have Brisbane as the second city.
This method combines the multiplication and addition principles.
22 2

6
3
1 2 1 2 2
Or P(M given S) or P(B given S) 


6
6
3
So P( (M or B) given S) 
Counting method
Reduced sample space:
There is only one choice for the first visit (Sydney) leaving three choices for the second visit and
two choices for the third visit.
There are 1 2  3  6 possible itineraries which have Sydney as the first visit.
Favourable outcomes:
There is only one choice for the first visit (Sydney), two choices for the second visit (Melbourne or
Brisbane) and two choices for the third visit. There are 1 2  2  4 possible itineraries which have
Sydney as the first visit and Melbourne or Brisbane for the second visit and a choice of two for the
third visit.
Counting  P(S first and M or B second and another) =
20
1 2  2 2

1 3  2 3
Mathematics: Clarifications for the syllabus Units 3A/3B
Example 2: Question 3 3ABMAT Final sample examination 2010
Seven state and territory cities are included in list of destinations. In this case it gets problematic to
start drawing a tree diagram so counting methods are a more efficient approach.
(a) How many possible itineraries involving three cities are there?
Universal sample space:
The number of possible itineraries which include visits to three cities can be calculated using seven
possible choices of city for the first visit then six choices for the second visit and five for the third
visit.
i.e. 7  6  5  210
(b) The Prime Minister picks one of the possible itineraries at random.
(i)
What is the probability that she will visit Sydney first?
Favourable outcomes:
There is only one choice for the first visit (Sydney) leaving six choices for the second visit and five
choices for the third visit. There are 1 6  5  30 possible itineraries which have Sydney as the first
visit.
So P(S first) 
(ii)
1 6  5 30 1


7  6  5 210 7
What is the probability that she won’t visit Sydney first?
P(do not visit Sydney first) = P( visit Sydney first )
1 6
i.e. P( visit Sydney first ) = 1  
7 7
Visualising the tree diagram of the 210 possible itineraries there are 6  6  5  180 itineraries which
do not have Sydney as the starting point.
180 6
 P(do not visit Sydney first) 

210 7
Counting method:
There are six choices for the first visit because Sydney is excluded from being the first visit, six
choices for the second visit and five choices for the third visit.
6  6  5 180 6
i.e. P(do not visit Sydney first) 


7 5 210 7
(iii)
If she visits Sydney first, what is the probability that she will visit Melbourne second?
Reduced sample space:
Given that Sydney is visited first the number of possible itineraries is reduces to 1 6  5  30 .
Favourable outcomes:
There is only one choice for the first visit (Sydney) and one choice for the second visit (Melbourne)
and five remaining cities to choose from for the third visit  11 5  5 .
So P(M given S) 
1 1 5 1
1
  P(M second given S first ) 
1 6  5 6
6
Mathematics: Clarifications for the syllabus Units 3A/3B
21
(iv)
If she visits Sydney first, what is the probability that she will visit Melbourne or
Brisbane second?
Reduced sample space:
Given that Sydney is visited first the number of possible itineraries is reduced to 1 6  5  30.
Favourable outcomes:
There is only one choice for the first visit (Sydney) and two choices for the second visit (Melbourne
or Brisbane) and five remaining cities to choose from the third visit
i.e. P((M or B) given S) 
1 2  5 10 1


1 6  5 30 3
Example 3: Question 3 3ABMAT WACE 2010 examination
(a)
(4 marks)
Write a numerical expression to determine the number of possible eight-digit phone
numbers that
(i)
start with 9 and have no digits the same.
(1 mark)
Solution
1 9  8 7  6  5  4  3
Specific behaviours
 uses multiplication principle correctly
(ii)
start with 9 and have no adjacent digits the same
(1 mark)
Solution
1 9  9  9  9  9  9  9
Specific behaviours
 uses multiplication principle correctly
A child presses ‘9’ on the phone and then presses seven other digits at random. What is
the probability that the number dialled contains no zero, given that no digits are the same?
(2 marks)
(b)
Solution
1 8  7  6  5  4  3  2 2

1 9  8  7  6  5  4  3 9
Specific behaviours
 uses multiplication principle to determine number of possibilities for the event
 uses multiplication principle to determine number of possibilities for the sample space
22
Mathematics: Clarifications for the syllabus Units 3A/3B
Example 4: Question 2 3ABMAT WACE 2011 examination
(4 marks)
Using the letters of the word SPORT (without repetition), what is the total number of four-letter
words that can be made, given:
(a) There are no restrictions.
(1 mark)
Solution
5  4  3  2  120
Specific behaviours
 uses the multiplication principle correctly
(b) The words begin with an S.
(1 mark)
Solution
1  4  3  2  24
Specific behaviours
 makes the correct restriction on the first letter and uses the multiplication principle correctly
(c) The words begin with an S and end with a T.
(1 mark)
Solution
1 3  2  1  6
Specific behaviours
 selects beginning and ending possibilities correctly and hence uses multiplication principle correctly
(d) The words begin with an S or ends with a T.
(1 mark)
Solution
n  A  B   n  A  n  B   n  A  B 
 24  24  6  42
Specific behaviours
 interprets the union of the two sets correctly
Mathematics: Clarifications for the syllabus Units 3A/3B
23
Reasoning geometrically
This content is covered in Unit 3B and it is worth noting that within this unit students are to reason
deductively in algebra and geometry. Further, the ideas here are not new as in the K-10 Syllabus
(Space and Measurement) students are expected to ‘reason geometrically’ by:
 making deductions related to geometric properties of shapes
 exploring demonstrations and informal proofs of general propositions
 testing descriptions by using counter-examples
 using congruent and similar triangles to solve geometric problems.
The specific dot points in Unit 3B referring to reasoning geometrically are:
2.3.1
2.3.2
distinguish general geometric arguments from those based on specific cases
follow geometric deductive arguments and ascertain their validity.
These ideas are further developed in Unit 3D.
A comprehensive treatment of proof and reasoning content for all units is covered in the document;
Mathematics and Mathematics: Specialist
Assessment support package – Proof and mathematical reasoning
This document is available from the Council extranet at:
http://www.scsa.wa.edu.au/internet/Senior_Secondary/Courses/WACE_Courses/Mathematics
It should also be noted that:
 the recommended time allocation for this topic in Unit 3B is approximately 2 hours, which
reflects the view that this topic is meant to be a review of what should have been covered
in year 10.
 the WACE examiners have thus far examined mathematical reasoning under the topic
‘Patterns’ (algebraic reasoning) but no questions have as yet been set specifically on
geometric reasoning. Geometric reasoning has been examined at the 3DMAT level.
Sample assessment items that are indicative of the depth required have been extracted from the
above reference and are provided below.
24
Mathematics: Clarifications for the syllabus Units 3A/3B
Mathematics 3BMAT
Response item and solutions—Geometric proof
2.1
2.3.1
2.3.2
Reason geometrically
distinguish general geometric arguments from those based on specific cases
Follow and ascertain the validity of geometric arguments
1. (8 marks)
(a) ‘The diagonals of a rectangle are equal in length’ is a true
statement, as can be verified using the rectangle pictured
here.
Consider the conjecture:
‘A quadrilateral with diagonals equal in length is a rectangle.’
(i)
Draw a pair of intersecting lines of equal length and join their end-points to create a
quadrilateral.
(1 mark)
(ii)
Repeat part (a) twice more.
(iii)
State whether the conjecture is true or false and justify your answer with reference to
your quadrilaterals in part (a) and (b).
(2 marks)
(2 marks)
(b) ‘A square has all four sides equal in length’ is a true statement.
Consider the conjecture: ‘A quadrilateral with all its sides equal in length is a square.’
State whether the conjecture is true or false and justify your answer.
(3 marks)
2. (3 marks)
An equilateral triangle, a square and a regular hexagon will all tessellate as shown in the
diagram below.
Investigate the conjecture: ‘All regular polygons will tessellate’.
Clearly justify any conclusions.
(3 marks)
Mathematics: Clarifications for the syllabus Units 3A/3B
25
3. (4 marks)
Two triangles are congruent when they are identical in every respect.
Investigate the conjecture: ‘two triangles will be congruent if they each have a side of 3 cm, a
side of 5 cm and an angle of 30’.
Clearly justify any conclusions.
(4 marks)
4. (3 marks)
Investigate the conjecture: ‘Hexagon ABCDEF has all its sides equal in length; therefore it is a
regular hexagon’.
Clearly justify any conclusions.
(3 marks)
26
Mathematics: Clarifications for the syllabus Units 3A/3B
Solutions
1. (8 marks)
Mark
Description
(a)
(i)

(ii)
8
(iii)

The conjecture is false. 
None of the diagrams above is a rectangle but all have two equal
diagonals. 
(b) The conjecture is false. 
A rhombus has all its sides of equal length but it need not have all its angles
equal to 90°. 
2. (4 marks)
Mark
Description
The conjecture is false. 
The regular polygons pictured have angles which are factors of 360° — 60°, 90°
and 120° — and these angles will fit around a vertex with no gaps or overlaps. 
No other regular polygon has an angle which is a factor of 360° and so no other
polygon will tessellate. 
For example, the angle in a pentagon is 108° and 108° is not a factor of 360°. 
(In the diagram below, three regular pentagons have been fitted together at a
vertex but clearly the remaining gap will not accommodate another pentagon.)
4
F
X
D
r
a
w
Mathematics: Clarifications for the syllabus Units 3A/3B
27
3. (4 marks)
Mark
Description
The conjecture is false. 
The two triangles pictured below satisfy the specified measurements but are clearly
not congruent.
In triangle A, the third side is 2.8 cm and the remaining angles are 32° and 118°.
In triangle B, the third side is 6 cm and the remaining angles are 56° and 94°.
B
A
5 cm
5 cm
30
30
4
3 cm
3 cm

Note: This activity can most usefully be done on a CAS calculator where the
triangles can be constructed and the measurements derived directly from the
diagrams.
4. (3 marks)
Mark
Description
The conjecture is false. 
A polygon is regular when all its sides are equal in length AND all its angles are
equal in size. 
Hexagon ABCDEF can be constructed with all its
B
C
sides of the same length but its angles of all
different sizes. See example. 
3
A
D
F
28
Mathematics: Clarifications for the syllabus Units 3A/3B
E
Finance
This content is covered in Unit 3A where students are to apply recursion in practical situations
including for finance.
Content associated with Finance from Unit 3A:
1.5.1 use, construct and interpret spreadsheets for making financial decisions
1.5.2 judge adequacy of spreadsheets and make refinements if necessary
1.5.3 calculate loans with reducible interest, including determining the number of years for the
balance to fall to a specified amount
1.5.4 calculate annuities using a spreadsheet
1.5.5 interpret and make decisions about loan and repayment amounts with reducible interest.
Sample items that are indicative of the depth required are as follows:
Example 1 Question 11 3ABMAT Sample unit package
Syllabus dot points
1.5.1 use, construct and interpret spreadsheets for making financial decisions
1.5.2 judge adequacy of spreadsheets and make refinements if necessary
1.5.3 calculate loans with reducible interest, including determining the number of years
Mike and Mary are planning to buy a house and are exploring loan options.
They have set up a spreadsheet as shown below. Cell R53 is for the interest rate per annum. Cell
R54 is for the number of compounding periods per annum. Cell R55 is for the monthly repayment.
Cell R56 is for the amount that Mike and Mary will borrow.
Row 59 has headings for the: Month, Starting amount at the beginning of the month, Interest for
the month, Repayment for the month, and Amount owing at the end of the month.
P
Q
R
S
T
Interest
Repayment
Amount owing
51 Loan with reducible interest
52
53 % Interest rate p.a.
54 Number of compounding periods p. a.
55 Monthly loan repayment
56 Starting amount
57
58
59
Month
60
1
61
2
62
3
63
4
64
5
Starting amount
Mike and Mary are considering a loan of $250 000 at 9% p.a. compounded monthly with monthly
repayments of $2000.
Mathematics: Clarifications for the syllabus Units 3A/3B
29
(a) Enter the figures in the appropriate cells in the spreadsheet.
Solution
(1 mark)
(a) 9, 12, 2000, 250000 correctly entered in cells R53, R54, R55 and R56 
(b) Write appropriate formulae in cells Q60, R60, S60 and T60 to calculate the amount owing for
any values that are entered in cells R53 to R56.
(7 marks)
Solution
(b) cell Q60 =R56  cell R60 =Q60*$R$53/100/$R$54  cell S60 =$R$55 
cell T60 =Q60+R60-S60 
(same marks can be allocated for other formulae that achieve the same results)
(c)
What other actions would you perform on the spreadsheet to calculate the amount owing
until the loan is paid off?
Solution
(c) cell Q61 =T60  fill down columns Q to T 
Example 2: Question 12 3ABMAT Sample unit package
Syllabus dot points
1.5.3 calculate loans with reducible interest, including determining the number of years for the
balance to fall to a specified amount
1.5.5 interpret and make decisions about loan and repayment amounts with reducible interest.
Susan borrowed $15 000 to purchase a car. She decides to repay $400 at the end of each month.
The table below is incomplete.
Amount Owing
15 000
14 675
Interest
Repayment
400
400
400
400
400
400
Balance
14 675
(a)
What was the annual rate of interest for Susan to borrow her money?
(2 marks)
(b)
How much money will she owe at the end of the first year, if the repayments stay at $400 per
month for the first 6 months, and then increase to $600 for the last 6 months?
(4 marks)
Solution
(b)
After 6 months Susan owes $ 13 025.46 
After 12 months Susan owes $ 9 775.84 
I = Prt
(a)
1
75 = 15000  r 

12
r = 0.06
i.e. Interest Rate was 6% p.a. 
30
Mathematics: Clarifications for the syllabus Units 3A/3B
Example 3: Question 9 3ABMAT WACE 2010 examination
(5 marks)
Syllabus dot points
1.5.3 calculate loans with reducible interest, including determining the number of years for the
balance to fall to a specified amount
1.5.5 interpret and make decisions about loan and repayment amounts with reducible interest.
In order to buy a new hi-fi system, Alex negotiated a personal loan of $4000 with repayments of
$400 to be made at the end of each month. The table below shows the amount owing at the start
of each month (An), the interest payable for that month (I), the repayment (R) and the amount
owing at the end of each month (An+1) for the first eight months.
Month
(n)
1
Amount owing at the
start of the month (An)
$4000.00
Interest
(I)
$30.00
Repayment
(R)
$400.00
Amount owing at the
end of the month (An+1)
$3630.00
2
$3630.00
$27.23
$400.00
$3257.23
3
$3257.23
$24.43
$400.00
$2881.65
4
$2881.65
$21.61
$400.00
$2503.27
5
$2503.27
$18.77
$400.00
$2122.04
6
$2122.04
$15.92
$400.00
$1737.96
7
$1737.96
$13.03
$400.00
$1350.99
8
$1350.99
$10.13
$400.00
$961.12
9
10
11
12
(a)
What is the monthly rate of interest?
(1 mark)
Solution
30  100 = 0.75%
4000
Specific behaviours
 correctly determines monthly rate
(b)
Write a recursive rule to determine the amount owing at the end of each month.
(2 marks)
Solution
An + 1 = 1.0075An – 400 , A1 = 4000 where An + 1 is the amount owing at the end of month n
Specific behaviours
 defines recursive rule correctly
 defines initial term
Mathematics: Clarifications for the syllabus Units 3A/3B
31
(c)
What is the amount of the final payment that Alex will make?
(1 mark)
Solution
A11 = 172.594
= 172.594  1.0075 = 173.888
Final payment = $173.88 or $173.89 or $173.90
plus interest
Specific behaviours
 calculates term 11 plus interest
(d)
What is the total amount paid by Alex for the hi-fi system?
(1 mark)
Solution
400  10 + final payment = $4173.88 (or $4173.89) (or $4173.90)
Specific behaviours
 calculates total amount
Example 4: Question 18 3ABMAT WACE 2011 examination
(5 marks)
Syllabus dot points
1.5.1 use, construct and interpret spreadsheets for making financial decisions
1.5.4 calculate annuities using a spreadsheet
To save money for an overseas holiday, Bruce started an investment account. He placed an initial
deposit of $6000, then deposited an extra $200 at the end of each month for one year.
The table below shows the amount in the account at the beginning of each month ( An ), the interest
added to the account each month ( I n ), the deposit made at the end of each month ( Dn ), and the
amount in the account at the end of each month ( An1 ) for the first six month.
Month (n)
1
2
3
4
5
6
Amount at
beginning of
month ( An )
$6000.00
$6248.00
$6497.98
$6749.96
$7003.96
$7259.99
Interest for
month
( In )
$48.00
$49.98
$51.98
$54.00
$56.03
$58.08
Deposit for
Month
( Dn )
$200.00
$200.00
$200.00
$200.00
$200.00
$200.00
(a) What is the monthly interest rate?
(1 mark)
Solution
48
100  0.8%
6000
Specific behaviours
 Correctly determines the monthly rate.
32
Amount at end of
month
( An1 )
$6248.00
$6497.98
$6749.96
$7003.96
$7259.99
$7518.07
Mathematics: Clarifications for the syllabus Units 3A/3B
(b)
Write a recursive rule to determine the amount in the account at the end of each month.
(2 marks)
Solution
An1  1.008 An  200, A1  6000
Specific behaviours
 Defines the recursive rule correctly
 Defines the initial term correctly
(c) What was the amount in the account at the end of 12 months? Give your answer to the
nearest 10 cents.
(1 mark)
Solution
Amount = $9110.50 [i.e.A13]]
Specific behaviours
 calculates total amount correctly
(d) What was the total amount of interest earned during the year?
(1 mark)
Solution
 $9110.50   $6000  12  $200   $710.50
Specific behaviours
 Calculates interest correctly
Mathematics: Clarifications for the syllabus Units 3A/3B
33