1. No category

Apply numeric reasoning in solving problems 17
Direct linear proportions
Various practical problems involving rates can also
be solved using proportional reasoning.
If two quantities are directly proportional, then
the ratio of the two quantities is constant. The
ratio can be scaled up (or down) by multiplying (or
dividing) both quantities in the ratio by the same
number.
Example
Which mixture has a greater proportion of eggs:
3 eggs in a 825 g cake or 2 eggs in a 560 g cake?
Solution
The first mixture has eggs:cake = 3:825.
Example
The second mixture has eggs:cake = 2:560.
13 cups of fruit drink are made using 3 cups of
cordial and 10 cups of water.
Scaling up both ratios, compare 6:1 650 with
6:1 680. [double first, triple second]
The ratio of cordial to water is 3:10 (using any unit
of measurement).
The first cake has the greater proportion of eggs.
[6 eggs to smaller mixture]
The table shows other quantities of fruit drink, with
cordial and water in this proportion.
Note: This problem could also be solved using
3
2
fractions: compare
with
.
825
560
1
1
with
.
Simplifying fractions: compare
275
280
The first fraction is larger.
Cordial
Water
Explanation
9
30
multiplying 3:10 by 3
tablespoons tablespoons
1
1
cups
2
5 cups
10
litres
3
50
litres
3
1 litre
5 litres
dividing 3:10 by 2
dividing 3:10 by 3
10
multiplying 1:
by 5
3
3
Note: Each of these quantities of fruit drink is
13
10
cordial and
water.
13
[each quantity has 3 parts cordial and 10 parts water
making 13 parts altogether]
[same numerator, larger denominator]
Exercise K: Direct linear
proportions
1.
Complete the tables by scaling up/down the
mixtures made up of two quantities in the
given ratios. Remember to include units.
a.
cordial : water = 1:7
Cordial
Fractions are useful for comparing proportions.
Water
2 cups
Example
10.5 tablespoons
Each working day, Dave spends 3 hours in his car
and 5 hours at his desk, while Kim spends 3.5 hours
in her car and 5.5 hours at her desk. Who spends a
greater proportion of their working day in the car?
3 litres
What fraction of the mixture is water?
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Solution
3
Dave has car:desk ratio 3:5 so he spends of his
8
day in his car.
Kim’s ratio is 3.5:5.5 or 7:11 (doubling) so she
7
of her day in her car.
spends
18
28
3
7
27
and
[changing to equivalent
=
=
8 72
18 72
fractions]
So Kim spends the greater proportion of time in her
car.
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b.
milk : cream = 8:3
Milk
Cream
20 mL
5 tablespoons
1 litre
What fraction of the mixture is cream?
Apply algebraic procedures in solving problems 69
Solving problems with quadratic
equations
2.
The product of two numbers is 96. The smaller
number is 4 less than the larger number. Solve
the equation:
96 = x(x + 4)
to find the smaller number. (There are two
possible answers.)
3.
The formula for the sum, S, of the first n
counting numbers is:
When solving quadratic equations in a practical
context, answers must be meaningful. For example,
physical quantities such as length and area cannot
be negative.
Example
Rod checks the price of a book in two different
bookshops. He finds that the book costs $3 more in
one bookshop than in the other.
If the product of the prices in dollars is 180, find the
cheaper price by solving the equation:
x(x + 3) = 180
Solution:
x represents the cheaper price (since (x + 3) is a
higher price)
x2 + 3x = 180
[expanding]
[rearranging]
x2 + 3x – 180 = 0
(x + 15)(x – 12) = 0
[factorising]
x = –15 or x = 12
n (n + 1)
.
2
How many numbers would need to be added
to make a sum of 300?
S=
The cheaper price is $12 (since a price of –15 dollars
is impossible).
Exercise V: Solving problems with
quadratic equations
1.
A square room is made 2 m longer and 1 m
wider. The area of the new room is 72 m2.
Solve the equation:
(x + 2)(x + 1) = 72
to find the area of the original square room.
x
2
x
4.
A box of jelly beans is
5 cm high. Its length is
2 cm longer than twice
its width. Its volume is
120 cm3. Solve the
equation:
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to find the width of
the cuboid.
72 Achievement Standard 91027 (Mathematics and Statistics 1.2)
Exercise W: Solving straightforward
exponential equations
Solving straightforward
exponential equations
Exponential equations may involve finding the
value of an unknown exponent (index or power).
It is useful to be familiar with the powers of the
first few whole numbers. For example:
2n
1
2
4
8
16
32
n
0
1
2
3
4
5
1.
3n
4n
5n
1
1
1
3
4
5
9
16
25
27
64 125
81
256 625
243 1 024 3 125
If x n = x m then n = m
Note: This rule does not hold if x = 0 or ±1.
Example
Solve:
[since 34 = 81]
2. (–4)x = –64
(–4)x = (–4)3
x=3
[since (–4)3 = –64]
3. 7x + 1
7x + 1
x+1
x
= 49
= 72
=2
=1
2.
6a = 36
b.
10b = 10 000
c.
2c = 128
d.
5d = 3 125
e.
1
 1
  = 243
3
f.
(–2)f = 4
g.
7g = 1
h.
(0.4)h = 0.0256
Solve for x.
a.
3x + 1 = 10
b.
5x – 5 = 120
[since 49 = 72]
In some equations involving exponents, there is an
unknown base number.
Example
a.
e
The following general rule can be used to solve
exponential equations.
1. 3x = 81
3x = 34
x=4
Find the unknown exponents.
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1. x3 = 125 has solution x = 5
[since 53 = 125]
343
3
5
has solution x = [since 35 = 243
2. x =
1024
4
and 45 = 1 024]
c.
64 – 8x = 0
It is important to note that even powers of negative
numbers are positive, eg (–2)4 = 16. This means that
there are two solutions to an exponential equation
such as x4 = 16, namely x = 2 or –2.
d.
9x
= 243
3
Note: Using the ‘plus or minus’ sign (±), this
solution is written x = ±2.
e.
2x + 2 = 8
f.
32x = 81
Example
1. a 4 = 625 has solution a = ±5 (ie, a = –5 or 5)
2. y 2 – 4 = 140 rearranges to y 2 = 144, which has
solution y = ±12
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Achievement Standard 91028
Investigate relationships between
tables, equations and graphs
Mathematics
and Statistics
1.3
Externally assessed
4 credits
Chapter 3
Sequences of numbers
2. By adding 1 to each term of the sequence
3, 6, 9, 12, …
A sequence is an ordered list of numbers (these
numbers are called terms of the sequence).
this linear sequence is formed: 4, 7, 10, 13, ….
The general rule for this sequence is y = 3x + 1.
For example, for the sequence 1, 4, 9, 16, 25, …
the 1st term is 1, the 2nd term is 4, the 3rd term is
9, and so on.
By observing patterns in a sequence, a rule for the
values of the terms may be found. This rule is often
expressed in terms of the position of the term in the
sequence.
y = mx + c
Where x is the position of the term in the sequence,
m is the steady increase, and c is a constant.
Example
The rule for the sequence
Example
Position of term, x
Value of term, y
For the sequence 1, 4, 9, 16, 25, …
The differences between terms are 3, 5, 7, 9, …
[4 – 1 = 3, 9 – 4 = 5, …]
So the next difference is 9 + 2 = 11 and the 5th term
is 25 + 11 = 36.
Similarly the 6th term is 36 + 13 = 49, and so on.
Linear sequences
2 3 4 5 ...
10 14 18 22 ...
By substituting x = 1 into the rule to get
6 = 4 × 1 + c, it can easily be seen that c = 2.
The rule for the sequence is therefore
y = 4x + 2.
Linear sequences arise in many practical situations
which may involve spatial patterns of shapes.
Letters relating to the objects (eg M for the number
of matchsticks) may also be used in the rules.
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A linear sequence increases (or decreases) steadily
by the same amount from one term to the next.
Example
1
6
is y = 4x + c [since the terms increase steadily by 4]
These differences increase by 2 each time.
Example
1. The linear sequence 3, 6, 9, 12, 15, … is formed
by adding 3 to any term to get the next term. A
general relationship between the position of the
term in the sequence, and its value can be found
using a table.
Position of term, x
Value of term, y
The general rule for a linear sequence is always of
the form
1
3
2
6
3
9
4 5
12 15
...
...
Joined hexagons are made with matchsticks.
The table summarises the information.
Number of hexagons, n
Number of matchsticks, M
Clearly, the value of each term is equal to
3 × the position of the term.
The rule is M = 5n + c
This can be written in the form y = 3x.
The rule is M = 5n + 1.
1
6
2 3 ...
11 16 ...
[M increases steadily by 5]
Substituting n = 1 gives 6 = 5 + c, so c = 1.
Investigate relationships between tables, equations and graphs 115
Finding the equation of a parabola
Example
By examining the features of a parabola, its
equation can be found.
The graph below is the same shape as y = –x2 and its
x-intercepts are –2 and 4.
Transformation method
10
If the coordinates of the vertex can be seen, then
the transformation method can be used.
8
6
If the basic parabola y = x2 has been:
4
• Translated c units horizontally, its equation is
y = (x – c)2. The parabola moves right if c > 0
and left if c < 0.
2
–4
• Translated d units vertically, its equation is
y = x2 + d. The parabola moves up if d > 0 and
down if d < 0.
• Translated c units horizontally and d units
vertically, its equation is y = (x – c)2 + d.
x2
If the parabola being translated is y = – (an
‘upside down’ parabola) then the corresponding
equations are y = –(x – c)2 and y = –x2 + d and
y = –(x – c)2 + d.
Example
In the graph below the vertex of the parabola is at
(–2, 3).
10
y
y
–2
2
The equation of the parabola is y = –(x – –2)(x – 4)
which simplifies to y = –(x + 2)(x – 4) or
y = (x + 2)(4 – x).
To write down the equation of a parabola which
has had a change of scale, see the following section
on applications for examples of how to do these.
Exercise Q: Finding the equation of
a parabola
1.
Write down the equations of the following
parabolas in factored form. All have the same
shape as y = x2 (or y = –x2)
a.
4
8
y
3
6
2
4
1
2
–4
4
–2
2
4
–5 –4 –3 –2 –1
x
x2
which simplifies to y = (x + 2)2 + 3
Factored form method
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If the x-intercepts of the parabola can be seen then
the factored form of the equation is useful.
If the parabola has the same shape as y = x 2 and its
x-intercepts are c and d then the equation of the
parabola is y = (x – c)(x – d ).
If the parabola is ‘upside down’ then the
corresponding equation is y = –(x – a)(x – b).
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2 x
–2
The basic parabola y = has been translated
–2 units horizontally and 3 units vertically.
So the equation is y = (x – –2)2 + 3
–1
1
–3
–4
b.
The parabola above is moved 2 units right
and 3 units up. Give the equation of the
resulting parabola.
126 Achievement Standard 91028 (Mathematics and Statistics 1.3)
b.
A table is calculated of x-values between 4.5 and
4.7, with increments of 0.01.
Complete the table below.
Width
x
5
Length
y =100 – 2x
100 – 2 × 5 = 90
Area
xy
450
10
15
20
25
30
35
c.
Complete the statement.
The enclosure with largest area has width
between
d.
and
Use five more pairs of values for the width
and length to explore this problem further.
Width
x
Length
y =100 – 2x
Area
xy
Minimum surface area occurs between 4.63 and
4.65. A table can then be calculated with values
between 4.63 and 4.65 in increments of 0.001. This
shows that the minimum surface area occurs when
the side length of the square is 4.64 cm to 3 sf.
e.
Exercise V: Optimal solutions using
numerical methods
1.
A farmer uses a 100 metre length of fencing
to make a rectangular enclosure against an
existing hedge. He wants to fence off as large
an area as possible.
hedge
x
x
y
a.
Complete the statements.
Width of enclosure with largest area is
m.
Largest possible area of enclosure is
m2.
2.
A square-based cuboid has surface area
240 cm2.
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Using x for the width and y for the length
of the enclosure, show that y = 100 – 2x.
y
x
x
What is the largest volume the cuboid can
have?
a.
Using x for the side length of the square
base, and y for the length of the cuboid,
60 x
show that y =
– .
x
2
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152 Achievement Standard 91031 (Mathematics and Statistics 1.6)
Exercise H: Angles in a circle
Angles in a circle
1.
Parts of a circle are labelled in the diagrams below.
VˆÀVՓviÀi˜Vi
ˆÕÃ
À>`
Vi˜ÌÀi
>ÀV
`ˆ
>“
ÃiV̜À
Find the angle marked x in each diagram. Give
reasons for each answer. O is the centre of a
circle.
a.
A
iÌ
V…œÀ`
iÀ
46°
O
B
x
Ãi}“i˜Ì
Ì>˜}i˜Ì
C
Rules for angles in a circle
The following rules (and their abbreviations) need
to be known.
Rule
Angles at the
circumference
standing on the
same arc are
equal.
Diagram
Example
A
b.
a
b
30°
x
B
50°
28°
a=b
(∠s on same arc)
x = 50°
(∠s on same arc)
36°
x
D
C
The angle at the
centre is twice
the angle at the
circumference.
80°
x
x = 160°
( at centre)
(∠ at centre)
Four different
cases are
illustrated.
120°
c.
A
O
x
The angle in a
semi-circle is a
right angle.
x
a
a = 90°
x
B
50°
x = 60°
( at centre)
(∠ in a semi)
E
C
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35°
x = 55°
( in a semi)
d.
A
46°
When finding unknown angles in circle problems,
you may need to use any of the geometry rules you
have previously learned.
For example:
• Parallel line rules.
• Isosceles triangle rules.
x
O
D
B
C
• Vertically opposite angles, and so on.
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Apply geometric reasoning in solving problems 177
Achievement Standard 91031
Practice test questions
c.
A triangular plastic track marker DEF has
length DF = 21.5 cm and angle DFE = 36˚.
F
Question 1
a.
21.5 cm
A cross-country running course is in the shape
of a right-angled triangle, ABC as shown.
A
655 m
E
36°
D
Find the length EF.
B
438 m
C
The distance AB is 655 m and the distance BC
is 438 m.
i.
What is the length of AC?
d.
A post of height 3.1 m is held up by a guy
rope, which makes an angle of 58˚ with the
ground.
guy rope
3.1 m
58°
ii.
b.
What is the size of the angle BAC?
What is the length of the guy rope?
The cross-section of a spectator stand is also in
the shape of a right-angled triangle, PQR, as
shown.
e.
The side view of the sports pavillion is shaped
as shown below,
C
P
5.36 m
19.25 m
Q
67°
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A
P
B
R
The length PQ is 5.36 m and the length PR is
19.25 m.
Find the length QR.
9.2 m
2.9 m
A
D
where AB = 2.9 m, CD = 9.2 m and
∠ BCD = 67˚.
Find the length of the roof section, BC.
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182 Achievement Standard 91035 (Mathematics and Statistics 1.10)
Comparison questions
This Achievement Standard requires students to
investigate a given multivariate data set using
the statistical enquiry cycle to make comparisons
between or among groups in the population.
This means that the data set will be supplied in
the assessment, with the values of several variables
given for each individual, eg gender, height,
weight, length of foot, resting heart rate, etc.
h.
What is the circumference of your wrist?
i.
What is your popliteal* length?
j.
What is the length of your index finger?
k.
Are you left-handed, right-handed or
ambidextrous?
l.
What is your main method of transport to
school?
m. How long does it usually take you to get to
school?
The assessment involves comparisons in which the
same variable is used for two different groups (eg
heights of boys are compared with heights of girls).
n.
What is the weight of your school bag today?
o.
What is your favourite learning area?
p.
How fast is your reaction time?
Note: In this assessment, it would be incorrect to
ask a relationship question using bivariate data,
in which the relationship between two different
variables for the same individual is investigated (eg
how the height of a boy is related to the weight of
a boy).
q.
What sport or activity do you most enjoy?
r.
How physically fit do you think you are?
s.
What is your resting pulse rate?
t.
How long have you had your current
cellphone for?
It is important to remember that the data set you
are working with is usually a random sample from
a larger population. The information from the
sample is used to make a statement or an inference
about the population.
Exercise A : Statistical variables
and questions
*For a seated person, the popliteal length is
the measurement from the underside of the
leg right behind the knee to the floor.
1.
For the above list, write down the variable
involved in each question, and its unit or
possible values (where appropriate).
a.
b.
c.
A nationwide online survey for Year 5–13 students
which provides real, relevant data and classroom
activities to enhance statistical enquiry across the
curriculum.
d.
e.
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CensusAtSchool www.censusatschool.org.nz is
funded by Statistics New Zealand and the Ministry
of Education.
f.
g.
h.
The following are some of the questions that are
used by CensusAtSchool in its survey.
i.
a.
Are you male or female?
j.
b.
How old are you?
c.
Which country were you born in?
d.
Which ethnic group do you belong to?
e.
How tall are you?
f.
What is the length of your right foot?
g.
What is your arm span?
k.
l.
m.
n.
o.
p.
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Demonstrate understanding of chance and data 239
The probability of the first sequence is
Pascal’s Triangle
P(MMFFF) = 0.4 × 0.4 × 0.6 × 0.6 × 0.6
Pascal’s Triangle is an array of numbers as shown
below:
1
1
2
3
4
1
6
5
Each of the 10 sequences above has the same
probability, so the chance of 2 males and 3 females
out of the next 5 people is
1
3
1
= 0.42 × 0.63
1
1
1
[since p(male) = 0.4, p(female) = 0.6]
10 × 0.42 × 0.63 = 0.3456.
1
4
10
The number of ways of placing 2 males in a
sequence of 5 people can also be found using
Pascal’s Triangle, using the diagram above.
1
10
5
1
The numbers in each row are found by adding two
adjacent numbers in the row above. For example,
the left hand ‘10’ in row 5 is found by adding the
left hand ‘4’ and the ‘6’ in row 4.
This array of numbers can be used to show the
number of pathways (without going backwards ie
‘up’) from the top point to any other point below it
in the array.
D
1
G
1
H
3
K
1
P
1
L
4
Q
5
R
10
Alternatively, start at A and move diagonally
right for 2 places (2 males) to F, then diagonally
left for 3 places (3 females) to R (10 ways).
Peter is late to school one in every five school days.
What is the probability that he will be late 3 times in
the next 10 school days?
F
1
I
3
M
6
•
Example
C
1
E
2
Start at A and move diagonally left for 2 places
(2 males) to D, then diagonally right for 3 places
(3 females) to S (which gives 10 ways).
Note: This method will work only when there are
only TWO possible outcomes.
A
1
B
1
•
Solution:
P(late) = 0.2,
J
1
N
4
O
1
S
10
T
5
P(not late) = 0.8
U
1
1
= 0.2]
5
[1 – 0.2 = 0.8]
[
The number of combinations of Peter being late
to school 3 times in 10 days is 120 (using Pascal’s
Triangle – see Appendix, page 277).
The diagram shows that there are
The probability that Peter will be late on 3 of the
next 10 school days is:
• 2 pathways from A to E: ABE, ACE
120 × (P(late))3 × (P(not late))7
• 3 pathways from A to H: ABDH, ABEH and ACEH.
= 120 × (0.2)3 × (0.8)7
• 6 pathways from A to M: ABDHM, ABEHM,
ACEHM, ABEIM, ACEIM and ACFIM.
= 0.2013
and so on.
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Pascal’s Triangle can be used in solving probability
problems.
Example
In a doctor’s practice, 40% of the patients are male.
What is the probability that out of the next 5 people
to enter the waiting room at this doctor’s surgery, 2
are male and 3 are female?
Solution
There are 10 ways that there can be 2 male (M)
and 3 female (F) in a sequence of 5 people, as listed
below.
MMFFF
FMFMF
MFMFF
FMFFM
MFFMF
FFMMF
MFFFM
FFMFM
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FMMFF
FFFMM
Note: The first eleven rows of Pascal’s triangle are
given as an appendix at the back of this workbook.
Exercise P: Probabilities using
Pascal’s Triangle
1.
What percentage of families with 5 children
will have 3 boys and 2 girls? (Assume that the
birth of a boy or a girl is equally likely.)
Year 11
Mathematics
Index
A
adjacent 167
adjacent angles on a straight line 133
algebraic expression 41
algebraic fractions 45, 51
alternate angles 134
analysis 181
angle of depression 175
angle of elevation 175
angle sum of a triangle 138
angles at a point 133
arrowhead 141
asymptote 123
average 203
axis of symmetry 105, 111
B
back-to-back stem-and-leaf plot 187
bar 207
base 43
base angles 139
bearings 173
BEDMAS 4, 42
BEMA 42
biased 199
bivariate data 182
C
census 199
certain 223
changing the subject of a formula 59
chords 154
classes 201
clinometer 175
clusters 186
coefficient 41
co-interior angles 134
comparison questions 181
composite 1
composite bar graph 207
compound interest 36
conclusion 193
concyclic 157
conditional probabilities 236
continuous 200, 201
corresponding angles 134
cosine 167
cube root 7
cyclic quadrilateral 157
D
data 181
data source 199
decay curve 123
decimal 22
decimal place 24
denominator 9
difference of two squares 48
directly proportional 17
discrete 200, 201
distribution 186, 190
distributive law 47, 49
divide 45
dot plots 185
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SA
S
E
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A
P
E
elimination method 61
equally likely outcomes 227
equation of the line 86, 91
equilateral 138
equivalent fractions 9
estimates 25
event 223
experimental probability 223
exponent 43
exponential 5
exponential curve 122
exponential equations 72
exterior angle 138