Curriculum for Excellence - National 5 resources from Pegasys
MA T H EMA T I C S
CFE Mathematics
National 5
Practice Unit Tests
Contents & Information

9 Practice Unit Tests...... (3 for each unit)

Answers & marking schemes
1
Pegasys Educational Publishing
Detailed marking schemes
Practice Unit Assessment (1) for National 5 Expressions and Formulae
1.
Simplify, giving your answer in surd form:
2.
(a)
Simplify
(b)
The number of people attending a football match was 3·12 × 104. If each person paid £27,
how much was collected? Give you answer in Scientific Notation.
3.
(i)
32
x4  x6
x3
(ii)
5x 4  4 x
 52
Expand and simplify where appropriate:
(a)
d(4d – e)
y² – 6y
(g + 4)(g + 9)
t² – 49
4.
Factorise:
5.
Express x² + 6x + 7 in the form (x + p)² + q.
6.
Write
7.
Write each of the following as a single fraction:
(a)
(a)
(b)
(b)
(c)
x² + 7x + 12
(4 x  3)( x  4)
( x  4) in its simplest form.
( x  4) 2
3 5

a b
( a , b  0)
(b)
f e

5 g
( g  0)
8.
Points P and Q have coordinates (–5, –4) and (6, 3) respectively. Calculate the gradient of PQ.
9.
Calculate the volume of a sphere with radius 2·3 cm, giving your answer correct to 2 significant
figures.
2·3 cm
2
10.
The logo for Cyril's Cars is shown below. The logo is a sector of a circle of radius 6∙2 cm. The
reflex angle at the centre is 240o.
A
240o
B
11.
(a)
Calculate the length of the arc AB.
(b)
Cyril wants to jazz up the logo by outlining it with coloured rope. He buys 20 metres of
rope. How many logos would he be able to makeup? (#2.1 and #2.2)
Sherbet in a sweet shop is stored in a cylindrical container like the one shown in diagram 1.
20cm
32cm
Diagram 1
The sherbet is sold in conical containers with diameter 5 cm and height 6 cm as shown in
diagram 2.
5 cm
6 cm
Diagram 2
The shop owner thinks he can fill 260 cones from the cylinder. Is he correct? (#2.1 and #2.2)
End of Question Paper
3
Practice Unit Assessment (1) for Expressions and Formulae:
Marking Scheme
Points of reasoning are marked # in the table.
Question Main points of expected responses
1
2
1
2
3
4
5
start of process
simplified surd
simplify numerator
correct answer
correct coefficient
simplify indices
calculation of amount
1
2
1
2
3
6
express in standard form
4 (a)
(b)
1
2
3
1
2
(c)
3
4
multiply out brackets
multiply out the brackets
collect like terms
factorise expression
factorise difference of two
squares
start to factorise trinomial
expression
complete factorisation
6
1
2
3
1
2
5
1
2
6
1
2 (a) (i)
(ii)
(b)
3 (a)
(b)
4
5
3
√16√2 (or equivalent)
4√2
x10
x7
20
3
3
x 2 in answer 20 x 2
27 × 3·12 × 104
=84·24 × 104
£8·424 × 105
4d2 – de
g2 + 4g + 9g + 36
g2 + 13g + 36
y(y – 6)
(t + 7)(t – 7)
4
(x 3)(x 4) ie evidence of
brackets, x, 3 and 4
(x + 3)(x + 4)
start of process
complete process
1
2
(x + 3)2
(x + 3)2 – 2
1
reduce to simplest form
1
4x  3
x4
7 (a)
1
denominator correct
1
(b)
2
numerator correct
3
multiply by inversion of
fraction
correct answer
3
1
evidence of gradient
calculation
1
Uses
2
correct gradient
2
7
11
4
8
2
4
///
ab
3b  5a
ab
g

e
fg
5e
y 2  y1
or equivalent
x 2  x1
4
9
10 (a)
(b)
11
1
substitute and start
calculation
1
2
complete calculation
2
4
   2  33
3
4
   12  167 or
3
equivalent
50·939 cm³ or equivalent
3
round calculation to 2
significant figures
3
51 cm3
1
correct ratio and substitution
1
2
240
   12  4
360
calculate arc length
2
25·957 cm or equivalent
#2.1
#2.2
valid strategy
interpretation of answer
#2.1
uses valid strategy to find
volumes of cone and
cylinder
1
2
calculate volume of cylinder
calculate volume of cone
# 2.2 states conclusion
#2.1 eg 2 000 ÷ 38
#2.2 (for 52∙63) 52 logos can
be made.
# 2.1 Substitutes relevant values
into correct formulae
1
2
10 048 cm3 or equivalent
39∙25 cm3 or equivalent
# 2.2 Shop owner is wrong
because only 256 cones
can be filled
5
Practice Unit Assessment (2) for National 5 Expressions and Formulae
1.
Simplify, giving your answer in surd form:
2.
(a)
Simplify
(b)
The number of people attending a musical was 2·64 × 103. If each person paid £34, how
much was collected. Give you answer in Scientific Notation.
3.
(i)
54
x 7  x 3
x2
1
(ii)
2 x 2  3 x 3
Expand and simplify where appropriate:
(a)
g(6g – h)
k² – 7k
(d + 3)(d – 7)
x² – 81
4.
Factorise:
5.
Express x² – 8x + 1 in the form (x + p)² + q.
6.
Write
7.
Write each of the following as a single fraction:
(a)
(a)
(b)
(b)
(c)
z² + 10z + 21
(3 x  1)( x  3)
( x  3) in its simplest form.
( x  3) 2
5 7

c d
(c, d  0)
(b)
k k

7 h
( h  0)
8.
Points R and S have coordinates (3, –2) and (–6, –3) respectively. Calculate the gradient of RS.
9.
Calculate the volume of a sphere with radius 3·7 cm, giving your answer correct to 2 significant
figures.
3·7 cm
6
10.
The diagram shows a sector of a circle with radius 5·6 cm and angle at the centre 230o.
A
230o
B
11.
(a)
Calculate the length of the arc AB.
(b)
The sector has to be made up into a cone with a fur trim round its base. How many cones
could be trimmed from 40 metres of fur? (#2.1 and #2.2)
During a cross country race, juice is distributed to the runners in conical containers with diameter
6 cm and height 8 cm as shown in diagram 1.
6 cm
8 cm
Diagram 1
At the end of the race juice from 60 cones is poured into a cylinderical container with dimensions
as shown in Diagram 2.
15cm
25cm
Diagram 2
Will this container be large enough to hold the juice? (#2.1 and #2.2)
End of Question Paper
7
Practice Unit Assessment (2) for Expressions and Formulae:
Marking Scheme
Points of reasoning are marked # in the table.
Question Main points of expected responses
1
2
1
2
3
4
5
start of process
simplified surd
simplify numerator
correct answer
correct coefficient
simplify indices
calculation of amount
1
2
1
2
3
4
5
6
express in standard form
4 (a)
(b)
1
2
3
1
2
(c)
3
4
multiply out brackets
multiply out the brackets
collect like terms
factorise expression
factorise difference of two
squares
start to factorise trinomial
expression
complete factorisation
6
1
2
3
1
2
5
1
2
6
1
2 (a) (i)
(ii)
(b)
3 (a)
(b)
3
√9√6
3√6
x4
x2
6
5
5
x 2 in answer 6 x 2
34 × 2·64 × 103
=89·76 × 103
8·976 × 104
6g2 – gh
d2 – 7d + 3d – 21
d2 – 4d – 21
k(k – 7)
(x + 9)(x – 9)
4
(z 3)(z 7) ie evidence of
brackets, z, 3 and 7
(z + 3)(z + 7)
start of process
complete process
1
2
(x – 4)2
(x – 4)2 – 15
1
reduce to simplest form
1
3x  1
x3
7 (a)
1
denominator correct
1
(b)
2
numerator correct
3
multiply by inversion of
fraction
correct answer
3
1
evidence of gradient
calculation
1
Uses
2
correct gradient
2
1
9
4
8
2
4
///
cd
5d  7 c
cd
h

k
h
7
y 2  y1
or equivalent
x 2  x1
8
9
10 (a)
1
substitute and start
calculation
1
2
complete calculation
2
4
   3  73
3
4
   50  653 or
3
equivalent
212·067 cm³ or equivalent
3
round calculation to 2
significant figures
3
210 cm3
1
correct ratio and substitution
1
230
   11  2
360
2
calculate arc length
2
22·468 cm or equivalent
(b)
11
#2.1
#2.2
valid strategy
interpretation of answer
#2.1 eg 4 000 ÷ 22·468
#2.2 (for 178∙02) 178 cones can
be trimmed.
#2.1
uses valid strategy to find
volumes of cone and
cylinder
# 2.1 Substitutes relevant values
into correct formulae
1
2
calculate volume of cone
calculate volume of cylinder
# 2.2 states conclusion
1
2
75·36 cm3 or equivalent
4415∙625 cm3 or equivalent
# 2.2 cylinder is not big enough
since 75·36 × 60 >
volume of cylinder
9
Practice Unit Assessment (3) for National 5 Expressions and Formulae
1.
Simplify, giving your answer in surd form:
2.
(a)
Simplify
(b)
A factory produces 2·4 × 104 cakes every day. How many cakes will it produce in the
month of April? Give you answer in Scientific Notation.
3.
(i)
147
x 2  x8
x 3
1
(ii)
6 x 3  3x 2
Expand and simplify where appropriate:
(a)
m(3m – n)
(p + 5)(p + 8)
h² – 11h
(b)
q² – 144
4.
Factorise:
5.
Express x² + 7x + 9 in the form (x + p)² + q.
6.
Write
7.
Write each of the following as a single fraction:
(a)
(a)
(b)
(c)
a² – 12a + 32
(2 x  5)( x  7)
( x  2  5) in its simplest form.
(2 x  5) 2
4 9

m n
(m, n  0)
(b)
4 k

k l
( h  0)
8.
Points C and D have coordinates (–8, –2) and (6, –4) respectively. Calculate the gradient of CD.
9.
Calculate the volume of a cone with diameter 4·6 cm and height 7 cm giving your answer correct
to 2 significant figures.
4·6 cm
7 cm
10
10.
(a)
Calculate the area of the sector of a circle in the diagram which has radius 6∙8cm.
o
O135
42o
(b)
These sectors have to be cut from a piece of card with an area of 6500 cm².
Assuming there is not waste, how many sectors can be cut from the card?
(#2.1 and #2.2)
11.
A candle is in the shape of a sphere with a diameter of 10 cm.
(a)
Calculate the volume of the candle.
The candle was melted down and poured into a conical container like the one shown in this
diagram.
11 cm
18 cm
(b)
Will the cone be big enough to hold the wax? [assume there is no wax lost during the
melting process] (#2.1 and #2.2)
End of Question Paper
11
Practice Unit Assessment (3) for Expressions and Formulae:
Marking Scheme
Points of reasoning are marked # in the table.
Question Main points of expected responses
1
2
1
2
3
4
5
start of process
simplified surd
simplify numerator
correct answer
correct coefficient
simplify indices
calculation of distance
1
2
1
2
3
√49√3
7√3
x10
x13
18
4
5
6
express in standard form
4 (a)
(b)
1
2
3
1
2
(c)
3
4
multiply out brackets
multiply out the brackets
collect like terms
factorise expression
factorise difference of two
squares
start to factorise trinomial
expression
complete factorisation
6
1
2
3
1
2
x 3 in answer 18x
30 × 2·4 × 104
=72 × 104
7·2 × 105
3m2 – mn
p2 + 8p + 5p + 40
p2 + 13p + 40
h(h – 11)
(q + 12)(q – 12)
5
1
2
6
1
2 (a) (i)
(ii)
(b)
3 (a)
(b)
3
(or equivalent)
5
 53
4
(a 4)(a 8) ie evidence of
brackets, a, 4 and 8
(a – 4)(a – 8)
start of process
complete process
1
2
(x + 3·5)2
(x + 3·5)2 – 3·25
1
reduce to simplest form
1
x7
2x  5
7 (a)
1
denominator correct
1
(b)
2
numerator correct
3
multiply by inversion of
fraction
correct answer
3
1
evidence of gradient
calculation
1
Uses
2
correct gradient
2

4
8
2
4
///
mn
4n  9m
mn
l

k
4l
k2
y 2  y1
or equivalent
x 2  x1
1
7
12
9
10 (a)
1
substitute and start
calculation
1
2
complete calculation
2
1
   2  32  7
3
1
   37  03 or
3
equivalent
38·75806 cm³ or equivalent
3
round calculation to 2
significant figures
3
39 cm3
1
correct ratio and substitution
1
135
   6  82
360
2
calculate sector area
2
54·4476 cm or equivalent
(b)
11
#2.1
#2.2
valid strategy
interpretation of answer
#2.1 eg 6 500 ÷ 54·4476
#2.2 (for 119∙38) 119 sectors
can be cut.
#2.1
uses valid strategy to find
volumes of cone and
sphere
# 2.1 Substitutes relevant values
into correct formulae
1
2
calculate volume of sphere
calculate volume of cone
# 2.2 states conclusion
1
2
523·33 cm3 or equivalent
569∙91 cm3 or equivalent
# 2.2 cone is big enough
since 523·33 < 569∙91
13
Practice Unit Assessment (1) for National 5 Relationships
1.
A straight line with gradient – 3 passes through the point (– 2, 5).
Determine the equation of this straight line.
4p – 12 < p + 6.
2.
Solve the inequation
3.
The Stuart family visit a new attraction in Edinburgh. They paid £32.25 for 3 adult tickets and 2
child tickets.
Write an equation to represent this information.
4.
Solve the following system of equations algebraically:
3a + 5b = 39
a–b= –3
5.
Here is a formula
S
2x
6
3
Change the subject of the formula to x.
y
6.
The diagram shows the parabola with
equation y  kx 2 .
40
What is the value of k?
35
30
25
20
15
10
5
–3 –2
–1
0
1
2
3x
14
7.
The equation of the quadratic function whose graph is shown below is of the form y = (x + a)2 + b,
where a and b are integers.
Write down the values of a and b.
y
8
6
4
2
–2 –1
8.
0
1
2
3
4 x
Sketch the graph y = (x – 1)(x + 3) on plain paper.
Mark clearly where the graph crosses the axes and state the coordinates of the turning point.
9.
A parabola has equation y = (x – 3)2 + 4.
(a)
Write down the equation of its axis of symmetry.
(b)
Write down the coordinates of the turning point on the parabola and state whether it is a
maximum or minimum.
10.
Solve the equation (x – 3)(x + 7) = 0
11.
Solve the equation x2 + 2x – 7 = 0 using the quadratic formula.
12.
Determine the nature of the roots of the equation 3x2 + 2x – 1 = 0 using the discriminant.
15
13.
To check that a room has perfect right angles, a builder measures two sides of the room and its
diagonal. The measurements are shown in this diagram.
6·3m
3·3m
5·4m
Are the corners of the room right – angled?
14.
The diagram shows kite ABCD and a circle with centre B.
AD is the tangent to the circle at A and CD is the tangent
to the circle at C.
Given that angle ABC is 126°, calculate angle ADC.
B
126o
A
C
D
15.
A water container is in the shape of a cylinder which is 150 cm long. The volume of water in the
container is 12 000 cm3.
A similar miniature version is 15cm long.
Calculate how much water the miniature version would hold.
16
16.
Here is a regular, 5 – sided polygon.
Calculate the size of the shaded angle.
17.
Sketch the graph of y = 4sin xo for 0° ≤ x ≤ 360°.
18.
Write down the period of the graph of the equation y = cos 3xo.
19.
Solve the equation 4sin x° – 1 = 0, 0° ≤ x ≤ 360°.
End of Question Paper
17
Practice Unit Assessment (1) for Relationships:
Marking Scheme
Points of reasoning are marked # in the table.
Question
Main points of expected responses
1
1
correct substitution
1
y – 5 = –3(x –(–2))
(or equivalent)
2
1
2
3
simplify for p
simplify numbers
solve
1
2
3
3p
18
p<6
3
#2.1 uses correct strategy and sets
up equation
1

multiply by appropriate
Factor
#2.1 3a + 2c = 32·25
2
3
solve for a
solve for b
2
3
1
2
subtract 6
multiply by 3
1
2
3
divide by 2
3
6
1
correct value of k
1
k=5
7
1
2
find value of ‘a’
find value of ‘b’
1
2
a = –1
b=2
8
1
1
–3, 1 and (0, –3)
2
(–1, –4)
3
identify and annotate roots
and y-intercept
identify and annotate turning
point
draw correct shape of graph
3
correctly annotated graph
9 (a)
1
axis of symmetry
1
x=3
(b)
2
3
turning point
nature
2
3
(3, 4)
minimum turning point
10
1
solve equation
1
x = –7, x = 3
11
1
correct substitution
1
4
5
2
1
3a + 5b = 39
5a – 5b = –15
or equivalent
a=3
b=6
S–6
(S – 6) × 3
(or equivalent)
3( S  6)
2
(or equivalent)
 2  2 2  4  1  7
2
18

3
4
evaluation discriminant
solve for 1 root
complete solution
1
2
correct substitution
evaluate discriminant
2
12
13
14
15
16
2
3
4
32
x = 1·8
x = –3·8
(rounding not required)
1
2
(2)2 – 4 × 3 × –1
16
#2.2 interpret result
#2.2
real and unequal roots
Since b2 – 4ac > 0
1
calculates and adds squares
of two short sides
1
3·32 +5·42 = 40·05
2
squares longest side
2
6·32 = 39·69
#2.2 interprets result
#2.2 so 3·32 + 5·42 ≠ 6·32
and hence triangle is not rightangled using converse of
Pythagoras.
The corners of the room are not
right angled.
1
radius and tangent
1
2
3
subtract
correct answer
2
3
either angle BAD or
angle BCD = 90°
360 – (90 + 90 + 126)
54°
1
use volume scale factor
1
(15/150)3 × 12000
2
correct answer
2
12 cm3
#2.1 use a valid strategy
1
correct answer
#2.1 eg
1
108°
centre angles
360/5 = 72° each
17
1
2
correct amplitude and period
correctly annotated graph
complete with roots and
amplitude.
1
2
4 / – 4 and 360°
Correct graph
18
1
correct period
1
120°
19
1
2
3
solve for sin x°
solve for x
complete solution
1
2
3
sin x° = 0·25
14·5°
165·5°
19
Practice Unit Assessment (2) for National 5 Relationships
1.
A straight line with gradient 4 passes through the point (2, –4).
Determine the equation of this straight line.
2.
Solve the inequation
7m + 5 < 2m + 30.
3.
The Clelland family visit a new attraction in Inverness. They paid £29.40 for 2 adult tickets and 4
child tickets.
Write an equation to represent this information.
4.
Solve the following system of equations algebraically:
7x + 2y = 32
2x – y = 6
5.
Here is a formula
A
4B
2
5
Change the subject of the formula to B.
y
6.
The diagram shows the parabola with
equation y  kx 2 .
16
What is the value of k?
14
12
10
8
6
4
2
–3 –2
–1
0
1
2
3x
20
7.
The equation of the quadratic function whose graph is shown below is of the form y = (x + a)2 + b,
where a and b are integers.
Write down the values of a and b.
y
8
6
4
2
–2 –1
8.
0
1
2
3
4 x
Sketch the graph y = (x – 5)(x – 7) on plain paper.
Mark clearly where the graph crosses the axes and state the coordinates of the turning point.
9.
A parabola has equation y = (x + 4)2 – 3.
(a)
Write down the equation of its axis of symmetry.
(b)
Write down the coordinates of the turning point on the parabola and state whether it is a
maximum or minimum.
10.
Solve the equation (x – 10)(x + 5) = 0
11.
Solve the equation x2 – 3x – 2 = 0 using the quadratic formula.
12.
Determine the nature of the roots of the equation 4x2 + 3x + 5 = 0 using the discriminant.
21
13.
A shape has dimensions as shown in the diagram.
12·5m
10m
7·5m
Kalen thinks it is a rectangle. Is he correct?
14.
The diagram shows kite PNML and a circle with centre M.
PL is the tangent to the circle at L and PN is the tangent
to the circle at N.
L
Given that angle LMN is 142°, calculate angle LPN.
142o
P
M
N
15.
A cuboid has length 30 cm and a volume of 1500 cm³
A similar miniature version is 10 cm long.
Calculate the volume of the miniature cuboid.
.
22
16.
Here is a regular, 12 – sided polygon.
Calculate the size of the shaded angle.
17.
Sketch the graph of y = 7cos xo for 0° ≤ x ≤ 360°.
18.
Write down the period of the graph of the equation y = sin 5xo.
19.
Solve the equation 7cos x° – 2 = 0, 0° ≤ x ≤ 360°.
End of Question Paper
23
Practice Unit Assessment (2) for Relationships:
Marking Scheme
Points of reasoning are marked # in the table.
Question
Main points of expected responses
1
1
correct substitution
1
y + 4 = 4(x ––2)
(or equivalent)
2
1
2
3
simplify for m
simplify numbers
solve
1
2
3
5m
25
m <5
3
#2.1 uses correct strategy and sets
up equation
1

multiply by appropriate
factor
#2.1 2a + 4c = 29·4
2
3
solve for x
solve for y
2
3
1
2
add 2
multiply by 5
1
2
3
divide by 4
3
6
1
correct value of k
1
k=2
7
1
2
find value of ‘a’
find value of ‘b’
1
2
a = –1
b=4
8
1
1
5, 7 and (0, 35)
2
(6, –1)
3
identify and annotate roots
and y-intercept
identify and annotate turning
point
draw correct shape of graph
3
correctly annotated graph
9 (a)
1
axis of symmetry
1
x = –4
(b)
2
3
turning point
nature
2
3
(–4, –3)
minimum turning point
10
1
solve equation
1
x = –5, x = 10
11
1
correct substitution
1
4
5
2
1
7x + 2y = 32
4x – 2y = 12
or equivalent
x=4
y=2
A+2
(A + 2) × 5
(or equivalent)
5( A  2)
4
(or equivalent)
3  3 2  4  1  2
2
24

3
4
evaluation discriminant
solve for 1 root
complete solution
1
2
correct substitution
evaluate discriminant
2
12
13
14
15
2
3
4
17
x = 3·6
x = –0·6
(rounding not required)
1
2
(3)2 – 4 × 4 × 5
–71
#2.2 interpret result
#2.2
roots are not real
since b2 – 4ac < 0
1
calculates and adds squares
of two short sides
1
7·52 +102 = 156·25
2
squares longest side
2
12·52 = 156·25
#2.2 interprets result
#2.2 so 7·52 +102 = 12·52
and hence triangle is rightangled using converse of
Pythagoras.
The shape is a rectangle
1
radius and tangent
1
2
3
subtract
correct answer
2
3
either angle PLM or
angle MNP = 90°
360 – (90 + 90 + 142)
38°
1
use volume scale factor
1
(10/30)3 × 15000
2
correct answer
2
55·6 cm3
#2.1 use a valid strategy
#2.1 eg
1
correct answer
1
150°
17
1
2
correct amplitude and period
correctly annotated graph
complete with roots and
amplitude.
1
2
7 / – 7 and 360°
Correct graph
18
1
correct period
1
72°
19
1
2
3
solve for cos x°
solve for x
complete solution
1
2
3
cos x° = 2/7
73·4°
286·6°
16
centre angles
360/12 = 30° each
25
Practice Unit Assessment (3) for National 5 Relationships
1.
A straight line with gradient ½ passes through the point (1, 5).
Determine the equation of this straight line.
5k – 3 < 2k + 9.
2.
Solve the inequation
3.
A group of friends met in a coffee bar. They paid £9.40 for 4 cappuccinos and 2 lattes.
Write an equation to represent this information.
4.
Solve the following system of equations algebraically:
5c – 2d = 36
c + d = 17
5.
Here is a formula
k 7
5m
4
Change the subject of the formula to m.
y
6.
The diagram shows the parabola with
2
equation y  kx
24
What is the value of k?
21
18
15
12
9
6
3
–3 –2
–1
0
1
2
3x
26
7.
The equation of the quadratic function whose graph is shown below is of the form y = (x + a)2 + b,
where a and b are integers.
Write down the values of a and b.
y
8
6
4
2
–4 –3
8.
–2
–1
0
1
2 x
Sketch the graph y = (x – 4)(x + 2) on plain paper.
Mark clearly where the graph crosses the axes and state the coordinates of the turning point.
9.
A parabola has equation y = 5 – (x + 3)2 .
(a)
Write down the equation of its axis of symmetry.
(b)
Write down the coordinates of the turning point on the parabola and state whether it is a
maximum or minimum.
10.
Solve the equation (x – 7)(x + 1) = 0
11.
Solve the equation x2 + 5x – 7 = 0 using the quadratic formula.
12.
Determine the nature of the roots of the equation 9x2 + 6x + 1 = 0 using the discriminant.
27
13.
A shape has dimensions as shown.
A
6·7m
8m
D
B
4·6m
C
Is angle DAB = 90o in this shape?
14.
The diagram shows kite WXYZ and a circle with centre X.
WZ is the tangent to the circle at W and YZ is the tangent
to the circle at Y.
Given that angle WXY is 139°, calculate angle WZY.
X
139o
W
Y
Z
15.
A tube of toothpaste is 21 cm long and has a volume of 50cm³
A similar miniature version is 9cm long.
Calculate how much toothpaste the miniature version would hold.
28
16.
Here is a regular, 10 – sided polygon.
Calculate the size of the shaded angle.
17.
Sketch the graph of y = –3sin xo for 0° ≤ x ≤ 360°.
18.
Write down the period of the graph of the equation y = sin ½ xo.
19.
Solve the equation 5tan x° – 7 = 0, 0° ≤ x ≤ 360°.
End of Question Paper
29
Practice Unit Assessment (3) for Relationships:
Marking Scheme
Points of reasoning are marked # in the table.
Question
Main points of expected responses
1
1
correct substitution
1
y – 5 = ½ (x –1)
(or equivalent)
2
1
2
3
simplify for k
simplify numbers
solve
1
2
3
3k
12
k<4
3
#2.1 uses correct strategy and sets
up equation
1

multiply by appropriate
Factor
#2.1 4c + 2l = 9·4
2
3
solve for c
solve for d
2
3
1
2
subtract 7
multiply by 4
1
2
3
divide by 5
3
6
1
correct value of k
1
k=3
7
1
2
find value of ‘a’
find value of ‘b’
1
2
a=1
b=3
8
1
1
–2, 4 and (0, –8)
2
(1, –9)
3
identify and annotate roots
and y-intercept
identify and annotate turning
point
draw correct shape of graph
3
correctly annotated graph
9 (a)
1
axis of symmetry
1
x = –3
(b)
2
3
turning point
nature
2
3
(–3, 5)
maximum turning point
10
1
solve equation
1
x = –1, x = 7
11
1
correct substitution
1
4
5
2
1
5c – 2d = 36
5c + 2d = 34
or equivalent
c = 10
d=7
k–7
(k – 7) × 4
(or equivalent)
4( k  7 )
5
(or equivalent)
 5  5 2  4  1  7
2
30

3
4
evaluation discriminant
solve for 1 root
complete solution
1
2
correct substitution
evaluate discriminant
2
12
13
14
15
2
3
4
53
x = 1·1
x = –6·1
(rounding not required)
1
2
(6)2 – 4 × 9 × 1
#2.2 interpret result
#2.2
equal roots
since b2 – 4ac = 0
1
calculates and adds squares
of two short sides
1
4·62 +6·72 = 66·05
2
squares longest side
2
82 = 64
#2.2 interprets result
#2.2 so 4·62 +6·72 ≠ 82
and hence triangle is not rightangled using converse of
Pythagoras.
Angle DAB is not a right
angle.
1
radius and tangent
1
2
3
subtract
correct answer
2
3
either angle ZWX or
angle ZYX = 90°
360 – (90 + 90 + 139)
41°
1
use volume scale factor
1
(9/21)3 × 50
2
correct answer
2
4 cm3
#2.1 use a valid strategy
#2.1 eg
1
correct answer
1
144°
17
1
2
correct amplitude and period
correctly annotated graph
complete with roots and
amplitude.
1
2
– 3 / 3 and 360°
Correct graph
18
1
correct period
1
720°
19
1
2
3
solve for tan x°
solve for x
complete solution
1
2
3
tan x° = 1·4
54·5°
234·5°
16
centre angles
360/10 = 36° each
31
Practice Unit Assessment (1) for National 5 Applications
1.
A farmer wishes to spread fertiliser on a triangular plot of ground.
The diagram gives the dimensions of the plot.
35 m
58o
44 m
Calculate the area of this plot to the nearest square metre.
2.
The diagram shows the paths taken by two runners, Barry and Charlie. Barry runs 350 metres from
point S to position R. Charlie runs 300 metres to position T.
T
300 m
12o
S
`
350 m
R
What is the shortest distance between the two runners? [i.e. the distance TR on the diagram]
32
3.
On an orienteering course there are three checkpoints at points U, V and W as shown in the diagram
below.
N
U
400 km
o
V 125
220 km
W
W is 220 kilometres from V and 400 kilometres from U.
W is on a bearing of 125° from V.
Calculate the bearing of W from U. i.e. the size of angle NUW in the diagram.
Give your answer to the nearest degree.
33
4.
The diagrams below show 2 directed line segments u and v.
v
u
Draw the resultant of 3u+ v.
5.
The diagram below shows a square based model of a glass pyramid of height 8 cm. Square OPQR
has a side length of 6 cm.
The coordinates of Q are (6, 6, 0). R lies on the y-axis.
z
S
y
R
Q (6, 6, 0)
x
O
P
Write down the coordinates of S.
6.
The forces acting on a body are represented by three vectors a, b and c as given below.
 5 


a 2 
 2  5


 3


b 7 
 5  5


1  5 


c 6 
  2


Find the resultant force.
34
7.
 5
 1 
Vector p    and vector q    .
 3
  3
Calculate 2 p  q
8.
Kashef bought a new car for £24 000. Its value decreased by 12% each year. Find the value of the
car after 5 years.
9.
A desk top has measurements as shown in the diagram.
3
2 m
7
2
1 m
3
Calculate the exact area of the desk top (in m2).
10.
A man invested some money in a Building Society last year.
It has increased in value by 15% and is now worth £2760.
Calculate how much the man invested.
11.
The cost of a set menu meal in 7 different café style restaurants were as follows:
£14
£17
£13
£14
£11
£19
£17
(a)
Calculate the mean and standard deviation of these costs.
(b)
In 7 up market restaurants the mean cost of a meal was £22 with a standard deviation of 2·2.
Using these statistics, compare the profits of the two companies and make two valid comparisons.
35
A primary teacher took a note of the results in a spelling test and the number of hours of TV that
some of her pupils watched in a week. She then drew the following graph.
25
20
Spelling Test (S)
Result, s
12.
15
10
5
0
0
10
20
30
40
50
Hours spent watching TV, (h)
(a)
Determine the gradient and the y-intercept of the line of best fit shown.
(b)
Using these values for the gradient and the y-intercept, write down the equation of the line.
(c)
Estimate the mark in the spelling test if the pupil spent 25 hours watching television.
End of Question Paper
36
Practice Unit Assessment (1) for Applications:
Marking Scheme
Points of reasoning are marked # in the table.
Question
1
2
Main points of expected responses
1
substitute into
formula
1
1
 35  44  sin 58
2
2
correct answer
2
653 m2
1
use correct formula
1
selects cosine rule
2
substitute correctly
2
s 2  3002  3502  2  300  350  cos12

3

4
3
process to s
2
3
7 089
4
84·1 metres (rounding not
required)
take square root
#2.1 uses correct strategy
#2.1 sin U 
220sin125
400
then valid
steps below
4
5
1
finds angle U
1
26·8
2
states bearing from
U
2
153·2o (rounding not required)
1
draws 3u
2
applies head-to-tail
method when adding
v
3
draws resultant from
tail of 3u to head of
v.
1
correct point
v
3u
3u + v
1
(3, 3, 8)
37
6
1
2
add to get resultant
correct answer
1
2
7
8
1
correct scalar
multiplication then
addition
2
calculate magnitude
2
112  32
3
correct answer
3
130
1
2
3
start calculation
process calculation
correct answer
1
2
3
1
Note: repeated subtraction
method can be used
9
10
11 (a)
(b)
 5    3  1  5 

 
 

 2  7  6 
 2  5  5  5   2 

 
 

3  5


 15 
 6 


1
area calculation
2
correct answer
10   1  11
       
 6    3  3 
0·88
24 000 × 0·885
£12 665·57
equivalent
#2.1 appropriate strategy
17 5

7 3
85
1
 4 m²
2
21
21
#2.1 eg 1 + 0·15 x = £2760
1
correct answer
1
1
mean for A
1
2
calculates ( x  x ) 2
2
3
substitute into
formula
3
4
correct standard
deviation
4
2·77 (rounding not required)
(Equivalent calculations can be used)
#2.2 Compares mean and
standard deviation in a
valid way for data
1
£2 400
105 ÷ 7 = 15
1, 4, 4, 1, 16, 16, 4
46
6
#2.2 On average up market prices more
expensive
There is less of a spread in up market
restaurants
38
12 (a)
(b)
1
chooses 2 distinct
points and
substitutes into
gradient formula
1
m
22  5  7  5
10  40
2
calculates gradient
1
(or based on gradient
2
line of best fit
3
finds intercept
3
c = 27·5 (approximately or by
calculation or from
graph)
4
S= 
4
writes down
equation
(c)
# 2.2 estimate mark
2
m
1
h + 27·5
2
(or equivalent)
#2.2 Approximately 15 hours
39
Practice Unit Assessment (2) for National 5 Applications
1.
A children’s play park, which is triangular in shape, has to be covered with a protective matting.
The diagram gives the dimensions of the plot.
22 m
24 m
56o
Calculate the area, to the nearest square metre, of protective matting needed.
2.
The diagram shows the courses followed by two ships, the Westminster and the Bogota, after they
leave Port A. The Westminster sails 520 metres to position W and the Bogota 580 metres to
position B.
W
520 m
14o
A
`
580 m
How far apart are the ships?[i.e. the distance WB on the diagram]
 Pegasys 2013
B
3.
On a radar screen, three planes, P, Q and R are at the positions shown in the diagram.
N
P
450 km
o
Q 132
300 km
R
R is 300 kilometres from Q and 450 kilometres from P.
R is on a bearing of 132° from Q.
Calculate the bearing of R from P. i.e. the size of angle NPR in the diagram.
Give your answer to the nearest degree.
 Pegasys 2013
4.
The diagrams below show 2 directed line segments a and b.
b
a
Draw the resultant of 2a + 2b.
5.
The diagram below shows a square based model of a glass pyramid of height 10 cm. Square OPQR
has a side length of 8 cm.
The coordinates of R are (0, 8, 0). P lies on the x-axis.
z
S
y
R (0, 8, 0)
Q
x
O
P
Write down the coordinates of S.
6.
The forces acting on a body are represented by three vectors k, l and m as given below.
 3 


k   2  5
4


Find the resultant force.
 Pegasys 2013
 2 


l  4 
1  5 


  3  5


m  0 
 4 


7.
 3
  2
Vector a    and vector b    .
6
  5
Calculate a  2b
8.
Due to inflation, house prices are expected to rise by 3∙6% each year.
What will the average house price be in 3 years if it is £142,000 today?
9.
A room has dimensions as shown in the diagram.
3
4 m
8
3
2 m
4
Calculate the exact amount of carpet that would have to be bought for the room.
10.
A woman bought an antique painting last year.
It has increased in value by 35% and is now worth £3 510.
Calculate how much the woman paid for the painting.
11.
A quality control examiner on a production line measures the weight, in grams, of cakes coming off
the line. In a sample of eight cakes the weights were
150
147
148
153
149
143
145
149
(a)
Find the mean and standard deviation of the above weights.
(b)
On a second production line, a sample of 8 cakes gives a mean of 148 and a standard
deviation of 6·1.
Using these statistics, compare the profits of the two companies and make two valid
comparisons.
 Pegasys 2013
The diagram below shows the connection between the thickness of insulation in a roof and the
heat lost through the roof. The line of best fit has been drawn.
Thickness of insulation in centimetres (T)
12.
25
20
15
10
5
0
0
1
2
3
4
5
Heat loss from roof in kilowatts (H)
(a)
Determine the gradient and the y-intercept of the line of best fit shown.
(b)
Using these values for the gradient and the y-intercept, write down the equation of the line.
(c)
Estimate the thickness of insulation for a heat loss of 2·5 kilowatts.
End of Question Paper
 Pegasys 2013
Practice Unit Assessment (2) for Applications:
Marking Scheme
Points of reasoning are marked # in the table.
Question
1
2
3
Main points of expected responses
1
substitute into
formula
1
1
 22  24  sin 56
2
2
correct answer
2
219 m2
1
use correct formula
1
selects cosine rule
2
substitute correctly
2
a 2  5202  5802  2  520  580  cos14
3
process to a2
3
21517
4
take square root
4
146·7 metres (rounding not
required)
#2.1 uses correct strategy
#2.1 sin P 
300sin132
450
then valid
steps below
4
5
 Pegasys 2013
v
1
finds angle P
1
29·7
2
states bearing from
P
2
150·3o (rounding not required)
1
draws 2a
2
applies head-to-tail
method when adding
2b
3
draws resultant from
tail of 2a to head of
2b
1
correct point
2b
2a
2a + 2b
1
(4, 4, 10)
6
1
2
7
8
1
 3   2    3 5

   

 2  5   4    0 
  4  1  5    4 

   

2
 1 5 


 65 
  6  5


1
 3   4   1 
   
   
 6    10    4 
add to get resultant
correct answer
1
correct scalar
multiplication then
addition
2
calculate magnitude
2
(1) 2  (4) 2
3
correct answer
3
17
1
2
3
start calculation
process calculation
correct answer
1
2
3
Note: repeated addition
method can be used
9
10
11 (a)
(b)
 Pegasys 2013
1
area calculation
2
correct answer
1·036
142 000 × 1·036³
£157 894
equivalent
#2.1 appropriate strategy
35 11

8 4
385
1
 12 m²
2
32
32
#2.1 eg 1 + 0·35 x = £3510
1
correct answer
1
1
mean for A
1
2
calculates
2
3
substitute into
formula
3
4
correct standard
Deviation
4
3·07 (rounding not required)
(Equivalent calculations can be used)
#2.2 Compares mean and
standard deviation in a
valid way for data
1
£2 600
1184 ÷ 8 = 148
4, 1, 0, 25, 1, 25, 9, 1
66
7
#2.2 On average weights the same
Wider spread on second line.
12 (a)
1
2

(b)
(c)
 Pegasys 2013
3
4
chooses 2 distinct
points and
substitutes into
gradient formula
1
m
20  10
1 5  3  5
2
m  5 (or based on gradient line
of best fit
calculates gradient
3
c = 27·5 (approximately or by
calculation or from
graph)
4
T =  5 H + 27·5
(or equivalent)
finds intercept
writes down
equation
# 2.2 estimate mark
#2.2 Approximately 15 cm
Practice Unit Assessment (3) for National 5 Applications
1.
.
Turf has to be laid on a triangular plot of garden
The diagram gives the dimensions of the plot.
27 m
102o
25 m
Calculate the area, to the nearest square metre, of turf that is required.
2.
Billy and Peter are bowlers. They are playing a game and after they each throw their first bowl they
are in the positions shown in the diagram.
B
24 m
17o
P
`
26 m
How far apart are the bowls after this first throw?[i.e. the distance PB on the diagram]
 Pegasys 2013
T
3.
The positions of three players, K, L and M are shown in this diagram.
N
K
40 m
o
L 125
30 m
M
Player M is 30 metres from player L and 40 metres from player K.
M is on a bearing of 125° from L.
Calculate the bearing of player M from player K. i.e. the size of angle NKM in the diagram.
Give your answer to the nearest degree.
 Pegasys 2013
4.
The diagrams below show 2 directed line segments k and l.
k
l
Draw the resultant of k + 2l.
5.
The diagram below shows a square based model of a glass pyramid of height 5 cm. The base OPQR
is a square.
The coordinates of S are (2, 2, 5). P lies on the x-axis and R lies on the y – axis.
z
S (2, 2, 5)
R
y
Q
x
O
P
Write down the coordinates of Q.
6.
The forces acting on a body are represented by three vectors x, y and z as given below.
 4 


x   2  3
 1 


Find the resultant force.
 Pegasys 2013
 2


y  2  7
 0  5


  2
 
z 1 
  2
 
7.
 3
Vector x    and vector
 6
 2
y    .
5
Calculate 3 x  2 y
8.
Chocolate fountains have become very popular at parties.
At one party 23% of the remaining chocolate was used every 20 minutes.
If 2kg of melted chocolate was added to the fountain at the start of the night,
how much would be left after 1 hour?
9.
Calculate the area of this piece of ground which has dimensions as shown in the diagram.
1
6 m
4
1
10 m
5
10.
I bought a car three years ago.
Since then it has decreased in value by 45% and is now worth £6875.
How much did I pay for the car?
11.
A set of Maths test marks for a group of students are shown below.
35
27
43
18
36
39
(a)
Find the mean and standard deviation.
(b)
Another group had a mean of 37 and a standard deviation of 8∙6.
Compare the test marks of the two classes.
 Pegasys 2013
12.
A selection of the number of games won and the total points gained by teams in the Scottish
Premier League were plotted on this scattergraph and the line of best fit was drawn.
P
80
70
60
Points
50
40
30
20
10
4
8
12
Wins
16
20
W
(a)
Determine the gradient and the y-intercept of the line of best fit shown.
(b)
Using these values for the gradient and the y-intercept, write down the equation of the line.
(c)
Use your equation to estimate the number of points gained by a team who win 27 games.
End of Question Paper
 Pegasys 2013
Practice Unit Assessment (3) for Applications:
Marking Scheme
Points of reasoning are marked # in the table.
Question
1
2
3
Main points of expected responses
1
substitute into
formula
1
1
 27  25  sin102
2
2
correct answer
2
330 m2
1
use correct formula
1
selects cosine rule
2
substitute correctly
2
t 2  24 2  26 2  2  24  26  cos17
3
process to t2
3
58·53
4
take square root
4
7·7 metres (rounding not
required)
#2.1 uses correct strategy
#2.1 sin K 
30 sin125
40
then valid
steps below
4
5
 Pegasys 2013
1
finds angle K
1
38
2
states bearing from
K
2
142o (rounding not required)
1
draws k
2
applies head-to-tail
method when adding
2l
3
draws resultant from
tail of k to head of
2l
1
correct point
2l
k
1
(4, 4, 0)
6
1
2
7
8
1
 4  2   

 
  
 2  3   2  7    1 
 1   0  5   2

 
  
2
 0 


 6 
  2  5


1
 9   4   5
       
18 10   8 
add to get resultant
correct answer
1
correct scalar
multiplication then
addition
2
calculate magnitude
2
52  82
3
correct answer
3
89
1
2
3
start calculation
process calculation
correct answer
1
2
3
Note: repeated addition
method can be used
9
10
11 (a)
(b)
1
area calculation
2
correct answer
0·77
2 000 × 0·77³
913g
equivalent
–3
#2.1 appropriate strategy
25 51

4 5
255
3
 63 m²
2
4
4
#2.1 eg (1 – 0·45) x = £6 875
1
correct answer
1
1
mean
1
2
calculates
2
3
substitute into
formula
3
4
correct standard
deviation
4
9 (rounding not required)
(Equivalent calculations can be used)
#2.2 Compares mean and
standard deviation in a
valid way for data
1
£12 500
198 ÷ 6 = 33
4, 36, 100, 225, 9, 36
410
5
#2.2 On average second group had
higher marks
Second group’s marks less spread out
 Pegasys 2013
12 (a)
1
2

(b)
3
4
chooses 2 distinct
points and
substitutes into
gradient formula
m
2
10
(or based on gradient line
3
of best fit)
c = 0 (approximately or by
calculation or from
graph)
calculates gradient
finds intercept
writes down
equation
40  20
12  6
1
3
4
m
10
W
3
(or equivalent)
P=
#2.2 90 points
(c)
 Pegasys 2013
# 2.2 estimate mark