Practice Pack 2
Exam A Paper 1
Practice Paper A: Credit Mathematics
FORMULAE LIST
Standard Deviation: s =
Σ(x − x )2
=
n −1
Σ x2 − ( Σx)2 /n
/n
, where n is the sample size.
n −1
1
ab sinC
2
a
b
c
Sine Rule:
=
=
sin A sinB sinC
Area of a triangle: Area =
Cosine Rule: a = b + c − 2bc cos A or cos A =
2
2
2
The roots of ax2 + bx + c = 0 are x =
b2 + c 2 − a 2
2bc
− b ± (b2 − 4ac)
2a
Page 10
Practice Paper A: Credit Mathematics
1. Evaluate
KU RE
5⋅1 ÷ (8⋅21 − 5⋅21).
2
2. Evaluate
2 23 − 1 51 × 31 .
3
3. Q = R − S where R = 3a2 and S = b2.
Calculate the value of Q when a = −2 and b = 3.
2
4. Express as a single fraction in its simplest form
7
3 .
−
x x +1
3
5. The number of cases of flu treated at 13 health centres on the 1st November was
recorded.
2
23
2
20
12
19
13
11
26
8
27
3
6
Draw a suitable statistical diagram to illustrate the median and the quartiles of
this data.
Page 11
4
Practice Paper A: Credit Mathematics
6. A retailer is offering a 20% discount on the normal price for all 21-inch computer
monitors.
KU RE
This is a 21-inch model and is now being sold at a reduced price of £680.
What is the normal price of this monitor?
3
7. This dice is a regular dodecahedron. Its 12 faces are numbered 1 to 12.
This dice is a regular octahedron. Its 8 faces are numbered 1 to 8.
On each dice each number has an equal chance of being rolled. Haleem has to
roll a multiple of 3 to win the game he is playing. Which dice would give him a
better chance of winning? Show clearly all your working.
Page 12
3
Practice Paper A: Credit Mathematics
8. Number Triangles
A
KU RE
The rule in a Number Triangle is A = B + C
B
(a) Use this rule to copy and complete this Number
Triangle.
C
2
Show that x + 2y = 3
x
(b) The same Number Triangle rule is used for this
Number Triangle. Write down another equation
with x and y.
y
−1
2
2x
2
13
y
2
(c) In both these Number Triangles, x and y have the
same values. Find the values of x and y.
3
9. The graph of y = p sin qqx
x , 0 ≤ x ≤ 120 is shown below.
y
5
x
O
120°
−5
2
Write down the values of p and q.
Page 13
Practice Paper A: Credit Mathematics
10. Two variables x and y are connected by the relationship y = px + q. It is known
that p > 0 and q < 0.
KU RE
3
Sketch a possible graph of y against x to illustrate this relationship.
2
11. (a) Evaluate 83 − 80.
2
(b) Simplify 2 + 8 .
2
12. The circular clock face on Big Ben is set in a
square frame as shown in the diagram.
The tip of the minute hand travels a distance of
27 metres every hour.
Show that the perimeter of the square frame is
exactly 108 metres.
π
[End of Question Paper]
Page 14
4
Practice Pack 2
Exam A Paper 2
Practice Paper A: Credit Mathematics
FORMULAE LIST
Standard Deviation: s =
Σ(x − x )2
=
n −1
Σ x2 − ( Σx)2 /n
/n
, where n is the sample size.
n −1
1
ab sinC
2
a
b
c
Sine Rule:
=
=
sin A sinB sinC
Area of a triangle: Area =
Cosine Rule: a2 = b2 + c2 − 2bc cos A or cos A =
The roots of ax2 + bx + c = 0 are x =
b2 + c 2 − a 2
2bc
− b ± (b2 − 4ac)
2a
Page 16
Practice Paper A: Credit Mathematics
1. On August 27, 2003 Mars was 5⋅6 × 107 km from Earth, its closest approach for
60 000 years.
KU RE
How long would it take to drive this distance in a car (if this were possible!) at an
average speed of 120 kilometres per hour?
4
Give your answer in days in scientific notation.
2. To set up a live video link for a conference a University is charged an initial
connection set-up fee of £120. Thereafter they are charged at a fixed rate
per hour while the conference lasts. They were charged £245 for the 5 hour
conference.
C
245
(£) 120
O
(hours)
5
t
The above graph represents the cost (£C) against the time (t hours).
(a) Find the equation of the line in terms of C and t.
3
(b) The next day the University was charged £282⋅50 for a similar video
conference. How long did this conference last?
3
3. A group of eight gap-year students in Malawi were badly bitten by mosquitoes
and contracted Malaria. Fever normally starts after 10 days.
The time, in days, for the onset of fever in this group were:
7, 10, 8, 9, 13, 7, 10, 8
Calculate the mean and standard deviation of these times.
Page 17
4
Practice Paper A: Credit Mathematics
4. In 2001 the Russian Space Station Mir was destroyed as it burned up in the
upper atmosphere.
KU RE
It had been losing altitude by 6% every month.
At the start of December 2000 its altitude was 340 km.
What was its altitude by the start of March 2001?
3
5. On a snooker table the distance between two pockets A and B is 1⋅78 metres.
1·78 m
A
B
50°
46°
C
From pocket A the ‘line-of-sight’ to the ball at C makes an angle of 46° with the
edge cushion.
Similarly the ‘line-of-sight’ from pocket B to the ball at C makes an angle of 50°
with the edge cushion as shown.
Calculate the distance of the ball at C from pocket A.
Page 18
4
Practice Paper A: Credit Mathematics
6. The diagram shows a triangular flag with a shaded
triangular design.
KU RE
The flag is in the shape of an isosceles triangle
ACD.
In the design the largest side of each shaded triangle A
is at right angles to the lower edge of the flag.
Side AB is 80 cm in length, as shown in the
diagram. The tip of the flag forms a 45° angle.
80 cm
Calculate the length AC of the flag at the
flagpole.
45°
D
5
B
C
7. The diagram shows a gearslider mechanism.
B
Rod AB = 10·5 cm
95°
Rod BC = 18 cm
angle ABC = 95°
A
C
Calculate the length of the slide AC to 1 decimal place.
Page 19
4
Practice Paper A: Credit Mathematics
8. The diagram shows a circular mine shaft with a radius
of 5 metres.
A square steel frame is fitted for a lift. A concrete
cladding fills the gap between the lift frame and the
circular mine shaft.
A
B
KU RE
5m
Calculate the greatest width of the concrete cladding (AB in the diagram) giving
your answer to 3 significant figures.
5
9. The diagram shows a tent in the shape of a triangular prism.
3 metres
1·4 m
1·6 m
This cross-section of the tent shows that it is slightly tilted to the left. The two
sides, one 1⋅4 m and the other 1⋅6 m, meet at an angle of 50° at the top.
50°
1·4 m
1 m
1·6
5
Calculate the volume of the tent.
Page 20
Practice Paper A: Credit Mathematics
10. Solve algebraically the equation
3 cos x° + 4 = 3
KU RE
0 ≤ x < 360
3
11. The diagram shows a swimming pool of width
x metres surrounded by a concrete path.
2m
The length of the swimming pool is 2 metres more
than its width.
x metres
The surrounding concrete path is 1 metre wide
along the length and 2 metres wide along the width
as shown in the diagram.
Pool
Path
1m
2m
1m
The area of the pool is the same as the area of the concrete path.
(a) Show that x2 − 4x − 12 = 0.
4
(b) Hence find the dimensions of the pool.
3
[End of Question Paper]
Page 21
Practice Pack 2
Exam B Paper 1
Practice Paper B: Credit Mathematics
FORMULAE LIST
Standard Deviation: s =
Σ(x − x )2
=
n −1
Σ x2 − ( Σx)2 /n
/n
, where n is the sample size.
n −1
1
ab sinC
2
a
b
c
Sine Rule:
=
=
sin A sinB sinC
Area of a triangle: Area =
Cosine Rule: a = b + c − 2bc cos A or cos A =
2
2
2
The roots of ax2 + bx + c = 0 are x =
b2 + c 2 − a 2
2bc
− b ± (b2 − 4ac)
2a
Page 26
Practice Paper B: Credit Mathematics
1. Evaluate
KU RE
8⋅82 − 2⋅9 × 3.
2
2x2 + 3x − 2.
2
2
3
2
2. Factorise
3. Evaluate
(1 21 − 31).
4. Simplify
x(2x − 1) − x(2 − 3x).
5. f(
f x) =
3
6
, x ≠ 0.
x
(a) Evaluate f(
f −3).
1
(b) Given that f(
f a) = 3, find a.
2
Page 27
Practice Paper B: Credit Mathematics
6. In this diagram the line AB has gradient − 21 and
it cuts the y-axis at the point (0, 3).
A
y
KU RE
(a) Write down the equation of the line AB.
2
0
B
x
(b) Point A has coordinates (−2, a).
Find the value of a.
2
7. Protons and neutrons are subatomic particles. Inside each atom
there is a nucleus made up of protons and neutrons. These
subatomic particles themselves are made up from two types of
quarks: ‘up’ quarks and ‘down’ quarks. The electrical charge of
a proton or neutron is the sum of the electrical charges of its
quarks.
(a) A proton is made from two ‘up’ quarks and one ‘down’ quark
and has a total electrical charge of 1 unit.
Write down an algebraic equation to illustrate this.
PROTON
u
u
d
NEUTRON
u
d
1
d
(b) A neutron is made from one ‘up’ quark and two ‘down’ quarks and has a
total electrical charge of zero.
Write down an algebraic equation to illustrate this.
(c) Find the electrical charge of an ‘up’ quark. Show clearly your reasoning.
Page 28
1
3
Practice Paper B: Credit Mathematics
8. Mia is in class 3M1 for Maths. There are three S3 Maths sets. The table shows
the numbers of boys and girls in each.
3M1
3M2
3M3
Boys
12
10
14
Girls
16
14
6
KU RE
(a) A pupil is chosen at random from Mia’s class. What is the probability that a
boy was chosen?
1
(b) At the end of term the three classes are put together in the school hall to
watch a Maths film. If a pupil is chosen at random from the audience what
is the probability now that a boy is chosen?
1
9. A random sample of cars on Scotland’s roads were tested for fuel efficiency. The
measure used was average miles per gallon or mpg. The results are shown in this
boxplot:
less
efficient 16
15
27
20
25
36 38
30
35
40
more
44 efficient
45
What percentage of sampled cars were less efficient than 38 mpg?
1
10. ‘Clearlawn’ mosskiller/fertiliser treatment contains nitrogen, phosphorus and
potassium mixed in the ratio 3:2:1.
(a) Mr Clevedon’s lawn requires 12 kg of nitrogen. Using ‘Clearlawn’ treatment
how much potassium would his lawn get?
(b) He decides to use ‘Clearlawn’ and finds the treatment is sold in 7 kg packets.
How many should he buy to treat his lawn? Show clearly your reasoning.
Page 29
1
3
Practice Paper B: Credit Mathematics
11. A sequence of numbers has the following first few terms:
KU RE
t1 = 1
t2 = 7
t3 = 19
t4 = 37
(a) Complete the calculation of S3 and S4 in this pattern:
S1 = t1
=1
=1
S2 = t1 + t2
=1+7
=8
S3 = t1 + t2 + t3
=………
=…
S4 = … … … …
=…………
=…
2
(b) Suggest a formula for Sn in terms of n.
1
(c) Hence find a formula for tn+1 the (n + 1)th term of the sequence.
2
12. (a) Evaluate
(b) Simplify
−
1
9 2.
2
t × t2.
2
Page 30
Practice Paper B: Credit Mathematics
13. The diagram shows a ball-sorting device.
Over-sized or under-sized balls are detected
by a sensor which activates the trap door to
remove them.
KU RE
trap door
The proper-sized balls are collected in the
collecting trough.
sensor
reject
trough
collecting
trough
The collecting trough is in the shape of a cuboid with a diagonal shute-plate as
shown in the diagram below with all measurements in centimetres.
5x
shute-plate
blocked-off
volume
B
2x
A
2x
volume available for
ball collection
One quarter of the volume of the cuboid is unavailable for collecting as it is
blocked off by the shute-plate.
Using the dimensions as shown in the diagram find, in terms of x, the
length AB.
[End of Question Paper]
Page 31
4
Practice Pack 2
Exam B Paper 2
Practice Paper B: Credit Mathematics
FORMULAE LIST
Standard Deviation: s =
Σ(x − x )2
=
n −1
Σ x2 − ( Σx)2 /n
/n
, where n is the sample size.
n −1
1
ab sinC
2
a
b
c
Sine Rule:
=
=
sin A sinB sinC
Area of a triangle: Area =
Cosine Rule: a = b + c − 2bc cos A or cos A =
2
2
2
The roots of ax2 + bx + c = 0 are x =
b2 + c 2 − a 2
2bc
− b ± (b2 − 4ac)
2a
Page 34
Practice Paper B: Credit Mathematics
1. The total emissions of Greenhouse Gases by the USA in 2007 amounted to
the equivalent of 7⋅2 million tonnes of carbon dioxide. If the annual increase
in emissions is 1⋅2%, calculate the total amount of emissions of Greenhouse
Gases by the USA expected in 2010. Give your answer in millions of tonnes to 2
significant figures.
KU RE
4
2. The amounts, in £, spent by a sample of six diners at a restaurant one Saturday
evening were:
£44, £47, £38, £97, £40, £52
(a) Find the mean amount spent.
1
(b) Find the standard deviation of the amounts spent.
2
(c) On a weekday evening at the same restaurant the standard deviation of the
amounts spent was £8⋅40.
Make one valid comparison between the amounts spent by diners on a
weekday evening and a Saturday evening.
3. On this map of Fife, Leven lies due
West of Elie.
Cupar is 11·9 km from Leven on a
bearing of 004°.
1
St Andrews
Cupar
Cupar is on a bearing of 320° from
Elie.
Leven
How far is Elie from Leven? Do not use a scale drawing.
Page 35
Elie
4
Practice Paper B: Credit Mathematics
4. An oil drum is in the shape of a cylinder with diameter
58 cm and height 97 cm.
58 cm
KU RE
(a) Calculate the volume of the drum giving your
answer to the nearest litre.
3
97 cm
(b) If 64 litres are poured out of a full drum what will
be the depth of the remaining oil in the drum?
2
5. The ‘hex’ numbers form a sequence of positive integers. The nth ‘hex’ number
is given by 1 − 3n + 3n2.
61 is a ‘hex’ number. Which one is it in the sequence?
4
6. This diagram shows a linkage mechanism
in a machine.
D
A
The lengths and angles between the
various pivot points A, B, C and D are toggle
shown below.
link
connecting
rod
C
B
oscillating
lever
AB = 9 cm
35°
D
Angle CDA = Angle ABC = 90°
A
23°
9 cm
Angle BAC = 23°
Angle CAD = 35°
C
Calculate the length of the toggle link CD.
Page 36
B
4
Practice Paper B: Credit Mathematics
7. The diagram shows the design for a triangular
pendant.
KU RE
The area of the central triangle PQR is 4·2 cm2
2·5 cm
R
P
3·4 cm
Q
P
= 2·5Rcm
P
= 3·4Qcm
Calculate the size of the acute angle QPR at the top of the pendant.
8.
3
The diagram below shows the graph of a quadratic function with the equation
y = k (x − p) (x − q) where p < q.
y
6
−2
2
O
6
x
The graph cuts the x-axis at the points (−2, 0) and (6, 0) and cuts the y-axis at the
point (0, 6).
(a) Write down the values of p and q.
2
(b) Calculate the value of k.
2
(c) Find the coordinates of the maximum turning point of the function.
Page 37
2
Practice Paper B: Credit Mathematics
9. These two organ pipes are mathematically similar in shape.
KU RE
The larger pipe is 240 cm in length and the smaller pipe is 180 cm in length.
The volume of the larger pipe is 43 litres.
Calculate the volume of the smaller pipe to the nearest litre.
3
10. This spanner head is circular with a symmetrical 5-sided gap.
The diagram below shows the dimensions of the circular head. C is the centre of
the circle which has diameter 5 cm. The head measures 4⋅8 cm across as shown.
4·8 cm
A
5 cm
C
B
Calculate the width of the gap AB.
Page 38
4
Practice Paper B: Credit Mathematics
11. Marcus travelled from St Andrews to Thurso in two stages.
KU RE
(a) In the first stage of his journey he covered 180 miles in x hours.
Find, in terms of x, his average speed.
1
(b) He covered the 60 miles of the second stage of his journey in 2 hours less
time than the first stage.
Find an expression for his average speed for the second stage of his
journey.
1
(c) His average speed on both stages of his journey was the same. Calculate the
time taken for the whole of his journey.
3
[End of Question Paper]
Page 39
Practice Pack 2
Exam C Paper 1
Practice Paper C: Credit Mathematics
FORMULAE LIST
Standard Deviation: s =
Σ(x − x )2
=
n −1
Σ x2 − ( Σx)2 /n
/n
, where n is the sample size.
n −1
1
ab sinC
2
a
b
c
Sine Rule:
=
=
sin A sinB sinC
Area of a triangle: Area =
Cosine Rule: a2 = b2 + c2 - 2bc cos A or cos A =
b2 + c 2 − a 2
2bc
− b ± (b2 − 4ac)
The roots of ax + bx + c = 0 are x =
2a
2
Page 44
Practice Paper C: Credit Mathematics
1. Evaluate
KU RE
1⋅3 × (4⋅92 + 0⋅08).
2
2
÷ 1 1.
3
3
2
2. Evaluate
3. Given that f(x) = x(2 - x), evaluate f(-1).
2
4. Solve the inequality 3 + 2x < 4(x + 1).
3
5. (a) Factorise 9y2 - 4.
1
(b) Hence simplify
2
9 y2 - 4 .
15 y - 10
W
4A
Change the subject of this formula to A.
6. P =
Page 45
2
Practice Paper C: Credit Mathematics
7. The triangle below shows the distances between
Cupar (C), Leven (L) and St Andrews (S).
10 miles
KU RE
Cupar
Leven
S
8 miles
C
St Andrews
Elie
CS = 8 miles
12 miles
CL = 10 miles
LS = 12 miles
L
1
8
Show that cos C = .
3
8. A survey was carried out to compare the punctuality of the trains run by two
train companies. 10 trains at random were recorded with the ‘number of minutes
late’ being recorded. The results for the two companies were:
Company A
Company B
11
6
1
1
2
4
1
1
3
1
3
9
7
5
5
5
0
8
0
9
Draw an appropriate statistical diagram to compare the two sets of data.
9. Part of the graphs y = f(x) and y = g(x) are shown
where:
f(x) = 4 + 3x - x2
y = g(x)
3
points of
intersection
y = f(x)
g(x) = 10 - 2x
Find the x-coordinates of the two points of intersection by solving the equation
f(x) = g(x).
Page 46
3
Practice Paper C: Credit Mathematics
10. Simplify fully
KU RE
( 18 )2 + ( 6 )2 .
2
11. Express in its simplest form:
x3 × (x-1)-2.
2
length in inches
12. A baby measured 20 inches in length at birth. The baby’s length was recorded at
regular intervals and a graph plotted.
L
24
20
0
4
12
8
Age in weeks
16
W
The graph is linear and shows that at 16 weeks the baby’s length is 24 inches.
Find the equation of the straight line graph in terms of W and L.
Page 47
4
Practice Paper C: Credit Mathematics
13. Carrie is a tea blender for a large international tea company.
KU RE
She has recently bought quantities of Kenyan and Rwandan tea and has created
two different blends from these teas. They are made in 5 kg packets.
Blend A: 2 kg of Kenyan and 3 kg of Rwandan costing 6⋅10 euros.
Blend B: 3 kg of Kenyan and 2 kg of Rwandan costing 5⋅90 euros.
Let the cost of Kenyan be k euros per kg and the cost of Rwandan be r euros per
kg.
(a) Write down an algebraic equation to illustrate the make up and cost of
Blend A.
1
(b) Write down a similar equation for Blend B.
1
(c) She creates a third blend as follows:
Blend C: 4 kg of Kenyan and 1 kg of Rwandan.
Find the cost of 5 kg of Blend C.
[End of Question Paper]
Page 48
4
Practice Pack 2
Exam C Paper 2
Practice Paper C: Credit Mathematics
FORMULAE LIST
Standard Deviation: s =
Σ(x − x )2
=
n −1
Σ x2 − ( Σx)2 /n
/n
, where n is the sample size.
n −1
1
ab sinC
2
a
b
c
Sine Rule:
=
=
sin A sinB sinC
Area of a triangle: Area =
Cosine Rule: a2 = b2 + c2 - 2bc cos A or cos A =
The roots of ax2 + bx + c = 0 are x =
b2 + c 2 − a 2
2bc
− b ± (b2 − 4ac)
2a
Page 50
Practice Paper C: Credit Mathematics
1. In 2009 scientists at Stanford University broke the record for the smallest writing.
They wrote ‘SU’. Each letter had width 3 × 10-8 cm.
KU RE
In letters this size the entire Bible could be written in a line 1·05 × 10-1 cm long.
Calculate the number of letters in the Bible. Give your answer in scientific
notation.
2
2. Mr Middleton was informed by ‘Petcare’ that his annual premium for his Pet
1
insurance had been increased by 6 %.
2
His annual premium was now £87·82.
Calculate his annual premium before this increase.
3
3. Solve the equation
3x2 - x - 5 = 0.
Give your answers correct to 1 decimal place.
Page 51
4
Practice Paper C: Credit Mathematics
4. From Dublin the bearing of Edinburgh is 035° and
the bearing of London is 118°
From Edinburgh the bearing of London is 158°
The distance from Dublin to Edinburgh is 343 km
N
N
35°
Dublin 118°
KU RE
158°
Edinburgh
Calculate the distance from Edinburgh to London.
5
London
5. A cylindrical plastic water pipe has a uniform
1
thickness of 2 cm as is shown in the cross2
sectional diagram below.
The outside diameter of the pipe is 30 cm and the inside
diameter is 25 cm.
30 cm
(a) Calculate the area of plastic in the cross-section.
(b) 1 cm3 of plastic weighs 0·12 grams.
Calculate the weight of a 1 metre length of pipe.
Page 52
3
25 cm
2
Practice Paper C: Credit Mathematics
6. An aircraft fuselage has a circular cross-section of diameter 4·6 metres.
KU RE
The passenger compartment floor is 1⋅2 metres above the lowest point of the
cargo compartment as shown below:
passenger compartment
C
4·6 m
floor
cargo compartment
1·2 m
x
In the diagram above C is the centre of the
cross-section.
(a) Calculate x, the width of the passenger compartment floor.
(b) The passenger compartment ceiling is the same width as the floor. How high
above the floor is it?
3
1
7. A chocolate syrup manufacturer uses a recipe that requires mixing honey and
cocoa in the ratio 9:2 by weight.
The manufacturer sells the syrup in 1 kg jars.
1
2
At the end of the week they have 49 kg of honey and 12 kg of cocoa left.
What is the maximum number of 1 kg jars of syrup that can be made with these
remaining ingredients?
Page 53
3
Practice Paper C: Credit Mathematics
8. The diagram shows part of the graph y = cos x°.
KU RE
y
0·2
A
B
x
O
The line y = 0⋅2 cuts the graph shown at five points.
Find the x-coordinates of A and B, the 2nd and 5th of these five points.
3
9. A company is hiring an Excavator machine.
They are considering two Plant Hire companies,
‘Earthmove’ and ‘Trenchers’.
The hire rates for these companies are as follows:
‘Earthmove’
‘Trenchers’
Delivery/Collection charge £64
Hourly rate: £30
Delivery/Collection charge £28
Hourly rate: £34
(a) Calculate the cost of a 2 hour hire from each of the companies.
1
(b) For a 20 hour hire which company is cheaper?
1
(c) For each company find a formula for the cost of a ‘n’ hour hire.
2
(d) ‘Trenchers’ claim that they are the cheaper of the two companies.
Find algebraically the greatest number of hours of hire for this claim to be
true.
Page 54
2
Practice Paper C: Credit Mathematics
10. In comparing stars to the sun, astronomers use this information.
KU RE
The diameter, D, of the star varies directly as the square root of its luminosity,
L, and inversely as the square of its temperature, T.
(a) Write down a formula for D in terms of L and T.
1
(b) If the luminosity is multiplied by 4 and the temperature is doubled, what
happens to the diameter?
2
11. Here is a number pattern:
2+1=1×3-2×0
4+1=2×4-3×1
6+1=3×5-4×2
(a) Write down the 4th line of this pattern.
1
(b) Write down the nth line of this pattern.
2
(c) Hence show algebraically that this pattern is always true.
1
Page 55
Practice Paper C: Credit Mathematics
12. This diagram shows the supporting wooden structure for a roof.
The diagram on the right shows the dimensions of one of
the central sections bounded by the two vertical posts AG
and DF.
The ‘web truss’ BC is 2⋅7 m in length and is parallel to the
other ‘web truss’ DE.
Section AC of the ‘King post’ AG is 3 m in length and the
lower section CE is 2 m in length.
EH is a horizontal line parallel to the ‘bottom chord’ GF
with DH 2⋅5 m in length.
Calculate the distance, GF, between the two vertical posts.
[End of Question Paper]
Page 56
KU RE
A
B
3m
C
2m
E
G
2·7m
D
2·5m
H
F
5
Practice Pack 2
Exam D Paper 1
Practice Paper D: Credit Mathematics
FORMULAE LIST
The roots of ax + bx + c = 0 are x =
2
Sine Rule:
−b ±
(b
2
− 4ac
)
2a
a
b
c
=
=
sin A sin B sin C
Cosine Rule: a2 = b2 + c2 − 2bc cos A or cos A =
b2 + c2 − a2
2bc
1
Area of a triangle: Area = ab sinC
2
Standard Deviation: s =
Σ(x − x )2
=
n −1
Σx2 − (Σx)2 /n
, where n is the sample size.
n −1
Page 10
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Practice Paper D: Credit Mathematics
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KU RE
1. Evaluate 35% of £351.
2
2. Evaluate 4.25 ÷ (4·3 + 0·7).
2
2
1 3
3. Evaluate 1 − of 1 .
3
2 5
3
4.
he diagram shows the
T
display of a computer
running a Graph Plotting
software package.
GRAPH PLOTTING DISPLAY
GRAPH
Enter data
x
y
−2
−8
−1
−5
0
−2
1
1
2
4
3
7
y
Lewis has entered his data.
x
When he clicks the ‘Plot’
button a straight line graph
will be shown for the data.
Find the equation of this
graph.
equation
Click to see graph:
3
PLOT
5. Solve the equation 3 +
2
= 1.
k
3
6. Two identical spinners are shown:
1
1
5
5
2
4
3
2
4
3
The two arrows are spun simultaneously.
Find the probability that the total of the resulting two numbers will be less
than 6.
2
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Practice Paper D: Credit Mathematics
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7. Here are cards showing a number pattern:
KU RE
Card 1: 0 × 1 − 1 × (−3)
Card 2: 1 × 2 − 2 × (−2)
Card 3: 2 × 3 − 3 × (−1)
(a) Write down the number pattern that would appear on card 4.
1
(b) Hence or otherwise find the expression that would appear on card n in its
simplest form.
3
8.The times (to the nearest 10 minutes) spent on homework by a group of
pupils are shown below:
Time spent (min)
Frequency
0
1
10
7
20
11
30
22
40
19
50
8
60
2
Construct a cumulative frequency table and hence find the median time spent
on homework for this group of pupils.
3
9. The diagram shows the cross-section of a circular railway tunnel.
Horizontal ceiling and floor segments
have been removed (shaded areas).
ceiling
10 feet
14 feet
The height of the tunnel from floor to
ceiling is 14 feet with the floor width
being 16 feet.
16 feet
floor
The radius of the circular tunnel is
10 feet.
5
Calculate the width of the ceiling.
10. V = 4 3 + 3 A.
(a) Find the value of V when A = 27 expressing your answer as a surd in its
simplest form.
(b) Calculate the value of A when V = 5 3 .
3
3
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Practice Paper D: Credit Mathematics
RE
11. (a) Mark plans to print leaflets advertising his gardening business. He has
a budget of £100. Printing costs are 70p for 25 leaflets. In addition the
printer charges a fixed administration fee of £51 on all orders.
How many leaflets can he afford to print?
3
KU RE
2
(b) A month later with a budget of £B he orders more leaflets. Printing costs are
now c pence for 25 leaflets and the fixed administration fee is now £A.
1
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Find a formula for T, the total number of leaflets he can afford to print.
3
12.The diagram shows the rectangular end of an Mp3 player. The area of the
end is 1 cm2.
(3x + 2) cm
x cm
Calculate the value of x, the thickness of the MP3 player.
5
[End of question paper]
5
3
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Practice Pack 2
Exam D Paper 2
Practice Paper D: Credit Mathematics
FORMULAE LIST
The roots of ax2 + bx + c = 0 are x =
Sine Rule:
−b ±
(b
2
− 4ac
)
2a
a
b
c
=
=
sin A sin B sin C
b2 + c2 − a2
Cosine Rule: a = b + c − 2bc cos A or cos A =
2bc
2
2
2
Area of a triangle: Area =
Standard Deviation: s =
1
ab sinC
2
Σ(x − x )2
=
n −1
Σx2 − (Σx)2 / n
,, where n is the sample size.
n −1
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Practice Paper D: Credit Mathematics
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1.The battery ‘run time’ of 7 laptops was measured. The times, measured to
the nearest hour, were:
KU RE
10, 11, 12, 12, 8, 10, 14.
Calculate the mean and standard deviation of these times.
4
2.The diagram shows an example of a ‘Golden Triangle’. Calculate the area
of Golden Triangle ABC.
A
1·62 m
B
3.The formula a =
72°
4
C
1m
V2
occurs in Physics when investigating motion in a circle.
r
Find the value of a when V = 1·2 × 103 and r = 4 × 108. Give your answer in
scientific notation.
3
4. Pete arranged three sticks to form a triangle as shown.
5·5 cm
14·4 cm
13·2 cm
He thought the triangle was right-angled. Was he correct? Show your reasoning.
of
support
ro
4m
3
ro
of
support
5.The diagram shows the crosssection of the attic space of a
house which is in the shape of
an isosceles triangle.
support
There are three vertical roof
2m
2m
attic floor
supports with the longest
12 m
central support 4 metres in
height. The two smaller supports are each placed 2 metres from the edge of
the floor. The whole floor measures 12 metres across.
Calculate the height of the two small supports.
4
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Practice Paper D: Credit Mathematics
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6. Solve the equation 2x2 + 3 = 6x.
KU RE
Give your answers correct to 1 decimal place.
3
7. (a) Show that the surface area of this cuboid, in cm2, is given by the expression
18x2.
1
4x cm
x cm
x cm
(b) The expression for the volume (in cm3) of this cuboid is equal to the expression
for its surface area (in cm2).
4
Calculate the dimensions of the cuboid.
8. Rock Q lies 7 km from Port P on a bearing of 120°. Rock R lies due west of
rock Q. Port P lies on a bearing of 054° from rock R.
N
P
120º
7 km
R
Q
5
Calculate the distance RQ between the two rocks.
9.Megan rents a van. When she collects the van its petrol tank is full. The
amount of petrol used varies directly as the distance travelled in the van.
P
petrol
(in litres)
56
(500,16)
kilometres travelled
D
The graph shows the amount of petrol, P litres, left in the tank and the
distance travelled, D km.
(a) How much petrol was in the tank initially?
1
(b) Calculate the rate of petrol consumption in litres per 100 km.
3
10. (a) Solve algebraically the equation 4 cos x° – 1 = 0
0 ≤ x < 360
3
(b) Hence write down the solution of the equation
4 cos (–x2)°– 1 = 0
0 ≤ x < 360.
1
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RE
Practice Paper D: Credit Mathematics
11.
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seat
wheel
KU RE
pedal
arm
This diagram shows the various parts of a unicycle.
The diagram on the right represents this
unicycle.
The seat, S, lies vertically above the centre, C,
of the wheel.
The wheel has a radius of 0·75 m.
S
1·1 m
P
x°
The seat, S, is 1.1 m above the top of the
wheel.
The pedal, P, is 1 m above the horizontal road
surface.
(a) Calculate the angle x° which the arm of the pedal, CP, makes with the
vertical.
4
(b) Calculate how far, in metres, the pedal P must travel clockwise to reach
position R vertically below the centre C. Give your answer correct to 3
significant figures.
4
C
1m
R
[End of question paper]
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Practice Pack 2
Exam E Paper 1
Practice Paper E: Credit Mathematics
FORMULAE LIST
The roots of ax2 + bx + c = 0 are x =
Sine Rule:
−b ± (b2 − 4ac)
2a
a
b
c
=
=
sin A sinB sin C
Cosine Rule: a2 = b2 + c2 − 2bc cos A or cos A =
Area of a triangle: Area =
Standard Deviation: s =
b2 + c2 − a2
2bc
1
ab sinC
2
Σ(x − x )2
=
n −1
Σx2 − (Σx)2 / n
,where n is the sample size.
n −1
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Practice Paper E: Credit Mathematics
KU RE
1
1
1. Evaluate ÷ 2 .
6
9
2
2. Evaluate 9.9 – 7.2 ÷ 30.
3. a =
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2
1 + 3b
.
bc
Change the subject of this formula to b.
3
4.Fergus rolls a pair of dice 90 times, hoping to get a total of either 4 or 6.
If the probability of this happening is 2 , how many times out of 90 rolls
9
would you expect one of these totals to appear?
2
5.
B
(10,2)
A
–3
Find the equation of this straight line graph in terms of A and B.
6. Remove brackets and simplify (3 – 2a)(2 + a) – (3 – a2).
4
3
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Practice Paper E: Credit Mathematics
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7. Chloe estimated the height of a lamp post to be more than 6 metres.
mirror
7m
KU RE
1·6 m
2m
She now calculates the height using similar triangles. To do this she places
a small mirror on the ground 7 metres from the base of the lamppost as
shown.
She now walks back from the mirror until she can just see the top of the lamp
post reflected in the mirror.
This happens when she is 2 metres from the mirror. Since the two shaded angles
are equal the two triangles are similar. Her eye is 1.6 metres above the ground.
Was her estimate correct? Show all your working.
3
3
8. Simplify a × a 2 .
2
9. (a) Keiran has saved up 30 coins. He has x twenty pences and y fifty pences.
Write down an equation in x and y to illustrate this information.
1
(b) A twenty pence coin weighs 5 g and a fifty pence coin weighs 8 g. When
Keiran weighs his coins on the kitchen scales the total weight is 210 g.
Write down another equation in x and y to illustrate this information.
(c) How many of each coin does he have? Show your working.
2
3
10. A rectangle has length 2x metres and breadth x metres.
It has a diagonal 100 metres in length.
2x metres
x metres
100 metres
Calculate the value of x, giving your answer as a surd in simplest form.
3
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Practice Paper E: Credit Mathematics
RE
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11. A relationship between A and B is given by the formula
A=
KU RE
1000
.
B
When B is multiplied by 4 what is the effect on A?
2
12. Part of the graph y = a sinbx°is shown below.
y
1
2
0
120°
x
– 12
2
Write down the values of a and b.
13. Archaeologists discovered a prehistoric circular enclosure with diameter 20
metres that was cut through by a roman road.
One edge of the road passes through the centre of the enclosure. The other
edge forms a 16 metre long chord as shown in the diagram.
an
rom ad
ro
16 m
prehistoric
circular
enclosure
w
3
20 m
Calculate w, the width of the road, in metres.
4
3
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Practice Paper E: Credit Mathematics
14. The nth term of a sequence is given by the formula
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KU RE
un = 2n – 1.
(a) Find the 3rd term of this sequence.
(b) There is a theorem that states:
I f 2n – 1 is a prime number then n is prime also. It is known that 127 is a
prime. By calculating a suitable value for n, show that this prime number
is an example of the truth of this theorem.
1
2
[End of question paper]
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Practice Pack 2
Exam E Paper 2
Practice Paper E: Credit Mathematics
FORMULAE LIST
The roots of ax + bx + c = 0 are x =
2
Sine Rule:
−b ±
(b
2
− 4ac
)
2a
a
b
c
=
=
sin A sin B sin C
b2 + c2 − a2
Cosine Rule: a = b + c − 2bc cos A or cos A =
2bc
2
2
2
Area of a triangle: Area =
1
ab sinC
2
Standard Deviation: s =
Σ(x − x )2
=
n −1
Σx2 − (Σx)2 / n
, where n is the sample size.
n −1
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Practice Paper E: Credit Mathematics
1. Solve the equation 5x2 + x – 1 = 0.
DO NOT
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KU RE
Give your answer correct to 2 significant figures.
4
2. (a) Cameron measured the height of six Australian Mountain Ash trees
which are reputedly the tallest of all flowering trees.
The heights, in metres, were 92, 89, 88, 95, 96, 86.
Find the mean and standard deviation for this data.
4
(b) In a second location he again measured the heights of six Mountain
Ash trees. At this second site the heights had a mean of 76 metres and a
standard deviation of 15 metres.
ake two valid comparisons between the heights of the trees at the first
M
and second locations.
3.Kim purchased a plot of land at the start of 2011 for £15 500. She estimated
that its value would appreciate by 6.5% each year. Assuming her estimate is
correct what would the plot be worth at the start of 2014.
2
3
4. Stephen has received a pay rise of 15% and now earns £15.64 per hour.
What was his hourly rate of pay before the rise?
5. Solve the inequality x − 3 > 2
6
3
3
2
6. A lawn sprinkler system waters an area in the shape of a sector of a circle
with radius 20 metres.
The angle at the sprinkler is 48°.
20 m
48°
sprinkler
Calculate the area of lawn covered by the sprinkler.
3
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Practice Paper E: Credit Mathematics
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7. Kirkcaldy is 11 km due East of Cowdenbeath. From Cowdenbeath the
bearing of Leven is 066°. From Kirkcaldy the bearing of Leven is 048°.
KU RE
L
N
66°
C
11 km
K
Calculate the distance between Cowdenbeath and Leven.
4
8. Solve algebraically the equation
3tanx° + 7 = 0, 0 <
− x < 360.
3
9. The diagram shows a flat triangular metal machine component.
Side AB is 3 cm in length.
The angle between sides AB and AC is 110°.
A
110°
3 cm
B
C
If the area of the component is 6.3 cm2 calculate the length of side AC.
3
10. Evie makes a vinaigrette salad dressing. She mixes vinegar and oil in the ratio 4:9.
If she has 248 ml of vinegar and 600 ml of oil what is the maximum volume
of vinaigrette that she can make?
3
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Practice Paper E: Credit Mathematics
RE
11. (a) Aberdeen Art Gallery is planning to restore the rectangular frame of one
of its pictures.
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KU RE
Here is a diagram of the rectangular picture with its surrounding frame:
x metres
x–1
metres
The outside dimensions of the frame are x metres by (x – 1) metres.
The frame has a uniform width of 0.5 metres.
Show that the area, P m2, of the picture is given by
P = x2 – 3 x + 2
2
(b) If the area of the picture is 3 of the total area enclosed (picture plus
5
frame), calculate the value of x.
4
12.
y
f(x) = (x – 5)(x + 2)
O
x
3
A
3
The diagram shows the graph of the function f(x) = (x – 5)(x + 2).
Find the y-coordinate of the minimum turning point A on the graph.
4
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Practice Paper E: Credit Mathematics
13. (a) A garden centre has ‘row covers’ that protect plants from frost.
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KU RE
They are shaped as half-cylinders 8 metres long and stand 2 metres tall
at their highest point.
2m
8m
Calculate the volume of air enclosed by this ‘row cover’.
2
(b) The garden centre plans to replace each ‘row cover’ by a hemispherical
cover that encloses the same volume of air.
height
Calculate the height of this hemisphere.
[The volume of a hemisphere with radius r is given by the formula
2
V = πr3.]
3
3
[End of question paper]
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Practice Pack 2
Exam F Paper 1
Practice Paper F: Credit Mathematics
FORMULAE LIST
The roots of ax + bx + c = 0 are x =
2
Sine Rule:
−b ±
(b
2
− 4ac
)
2a
a
b
c
=
=
sin A sin B sin C
2
2
2
Cosine Rule: a = b + c − 2bc cos A or cos A =
b2 + c2 − a2
2bc
1
Area of a triangle: Area = ab sinC
2
Standard Deviation: s =
Σ(x − x )2
=
n −1
Σx2 − (Σx)2 / n
, where n is the sample size.
n −1
Page 38
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Practice Paper F: Credit Mathematics
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KU RE
4
2
1. Evaluate 2 − 1 .
5
3
2
2. Evaluate 82.8 ÷ (1.8 − 0.9).
2
3. (a) Factorise 2x2 + x − 6.
1
(b) Hence simplify
4. Given that f (x) =
2x + 4
2 x2 + x − 6
2
.
6
, evaluate f(−5).
3+ x
2
5.Find the equation of this straight line graph.
y
10
(8,4)
x
0
3
6.These two hexagonal nuts are mathematically similar in shape.
The thicknesses of the nuts are 10 mm for the larger one and 6 mm for
the smaller one.
large nut
small nut
6 mm
10 mm
If the larger nut weights 15 g calculate the weight of the smaller nut.
(You can assume that, for the two nuts, equal volumes have equal weights,
i.e. the two nuts are made from the same metal).
4
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Practice Paper F: Credit Mathematics
DO NOT
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KU RE
7. Solve the equation 3a – (2 – a) = 5a.
8.
3
A
1·5 m
D
B
2·5 m
2m
C
The diagram shows a kite ABCD with dimensions as shown.
The kite has right-angles at B and D.
1
(a) Calculate the area of the kite.
(b) Use your answer to part (a) to calculate the length of the diagonal BD of
kite ABCD as shown in the diagram below.
3
A
D
B
C
9. The graph of y = x2 has been moved to the position shown in the diagram
below.
y
−4
0
x
−8
Page 40
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Practice Paper F: Credit Mathematics
RE
The equation of the graph in this new position is
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KU RE
y = (x + 4)2 – 8.
Draw a sketch showing the graph with equation
y = (x – 3)2 + 2.
2
10. Mia has 18 coins all of which are 10p-coins or 5p-coins. She has x 10 p-coins
and y 5p-coins.
(a) Write down an equation in x and y to illustrate this information.
(b) In total her coins are worth £1.25.
3
1
2
Write down another equation in x and y to illustrate this information.
3
(c) How many 5p-coins does she have?
11. (a) Mark downloads x tunes at a cost of £0.60 per download from Hometune’s
website. Write down an algebraic expression for the cost, in pence, of the
downloads.
1
(b) Hometune offers a package membership deal:
• Membership fee: only £8.00 per month.
• 10 free downloads to members per month.
• Special download rate for members: £0.45 per tune.
(i) Calculate the total cost to Mark of downloading 25 tunes during the
course of a month if he were to become a member.
(ii) Write down an algebraic expression for the total cost, in pence, for a
member who downloads x tunes during the course of a month where
x is greater than 10.
1
2
(c) Find the minimum number of tunes Mark would have to download
during the course of a month for membership to be a cheaper option
than to download the tunes as a non-member.
3
Show all your working.
[End of question paper]
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Practice Pack 2
Exam F Paper 2
Practice Paper F: Credit Mathematics
FORMULAE LIST
The roots of ax + bx + c = 0 are x =
2
Sine Rule:
−b ±
(b
2
− 4ac
)
2a
a
b
c
=
=
sin A sin B sin C
Cosine Rule: a2 = b2 + c2 − 2bc cos A or cos A =
Area of a triangle: Area =
1
ab sinC
2
Standard Deviation: s =
Σ(x − x )2
=
n −1
b2 + c2 − a2
2bc
Σx2 − (Σx)2 / n
, where n is the sample size.
n −1
Page 44
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Practice Paper F: Credit Mathematics
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1.A school purchased a computer for £273.70. The school later discovered that
this price included V.A.T. at a rate of 15% for which they should not have
been charged.
What should the school have been charged for the computer?
KU RE
3
2.Our sun, along with all the other stars in our galaxy, the Milky Way, orbits
around a black hole that is at the centre of the galaxy.
The orbit is roughly circular with a radius of 2.36 × 1017 km. Calculate the
circumference of the orbit.
Give your answer in scientific notation.
3
3. (a) The average daily circulation figures, in thousands, over the course of
6 months for an Aberdeen local newspaper are:
27, 35, 24, 23, 29, 30.
3
Calculate the mean and standard deviation of this data.
(b) The equivalent data over the same period for a local Dundee newspaper
give a mean of 35 and a standard deviation of 1.2.
ake two valid comparsons between the circulation figures of the
M
Aberdeen and the Dundee newspapers over the 6 months.
4. A block of rubber in the shape of a cuboid
with dimensions 8 cm × 3 cm × 4 cm is
used to make a door wedge.
B
A
C
A slanting rectangular slice, ABCD, is
made in the block through edge AB of
the block.
2
D
4 cm
3 cm
8 cm
This creates a wedge with height 4 cm
at one end and 1 cm at the other end as
shown in the diagram on the right.
B
A
C
The wedge is sitting on a horizontal
surface.
Calculate the angle the wedge makes
with the horizontal.
Give your answer correct to 1 decimal place.
4 cm
D
1 cm
4
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Practice Paper F: Credit Mathematics
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KU RE
5. (a) Expand and simplify (2y – 1)(y – 2).
1
1
(b) Expand x − 2 (x – 3).
(c) Simplify, leaving your answer as a surd
2
108 − 5 3.
2
6. (a) A cattle feeding trough is in the shape of a prism as shown in the diagram.
It has length 4.2 m with the area of each end 1.6 m2.
4·2 m
1·6 m2
1
Calculate the volume of the trough.
(b) An alternative design for the trough is in the shape of half a cylinder. For
this design each end is a semi circle with diameter 2 metres.
2m
If this new trough has the same volume as the trough design in part (a)
above, calculate its length.
7. 4
N
P
N
Q
R
(a) Port Q lies on a bearing of 136° from Port P. From Port Q, the bearing
of Port R is 258°.
Calculate the size of angle PQR.
2
Page 46
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Practice Paper F: Credit Mathematics
RE
DO NOT
WRITE IN
THIS MARGIN
(b) A ferry leaves Port P and travels to Port Q, a distance of 10.2 km, along
the direct route PQ. It then sails from Port Q to Port R, a distance of
12.8 km, along the direct route QR.
Calculate the distance travelled by the ferry when it returns to Port P
from Port R along the direct route RP.
Give your answer to 3 significant figures.
8.
KU RE
4
On this 1 square × 2 squares
grid a total of 3 different
rectangles can be drawn
3 different rectangles
The number of different rectangles, r, that can be drawn on a grid of squares
of size n squares × 2 squares is given by the formula
r=
4
3
n(n + 1)
2
(a) How many different rectangles can be drawn on a grid of squares of size
3 squares × 2 squares?
(b) On a grid of squares of size n squares × 2 squares it is possible to draw
198 different rectangles. Show that for this grid n2 + n – 132 = 0.
2
(c) Hence find the size of the grid of squares.
3
9. The Large Hadron Collider has a circular
tunnel with a radius of 4.3 km. In the tunnel
protons are accelerated to nearly the speed
of light. At this speed they travel 15 km in
51 microseconds (a microsecond is one
millionth of a second). In the tunnel there is
a detector, Alice, at point A and a detector,
LHC ‘B’, at point B and they form an angle
of 96° at the centre of the circle as shown.
2
96°
A
(Alice)
B
(LHC ‘B’)
Calculate the time, in microseconds, a proton takes at this speed to travel
anti-clockwise from Alice to LHC ‘B’.
4
2
Page 47
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Practice Paper F: Credit Mathematics
10.ABCDE is a regular pentagon. Line L is an axis of symmetry. The two
diagonals AC and BD intersect at point F as shown. Each side of the pentagon
has length 1 cm and each diagonal of the pentagon has length x cm.
AFDE is a rhombus.
F
C
x cm
E
KU RE
line L
B
A
DO NOT
WRITE IN
THIS MARGIN
1 cm
D
(a) Write down an expression for the length FC, in cm.
1
(b) Triangle DFC is similar to triangle BDE. Use this information to show that
x2 – x – 1 = 0.
3
11.The depth, d metres, of water in a harbour starting at noon is given by the
formula
d = 7–3 sin (30h)°, where h is the number of hours after noon.
(a) Calculate the depth of the water in the harbour at 3 pm (3 hours after
noon).
(b) How long after noon will the depth of the water in the harbour first be
5 metres?
2
4
[End of question paper]
Page 48
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Worked Answers to Practice Exam A: Credit Mathematics
Practice Exam A: Paper 1 Worked Answers
Q1 5·1 ÷ (8·21 – 5·21)
= 5·1 ÷ 3
✓
= 1·7
✓
2 marks
2
3
1
5
1
3
6
1
Q2 2 – 1 ×
=
8
3
=
8
3
=
40
15
– × 
 5 3
–
6
15
–
✓
✓
6
15
Order of operations
• Calculations in brackets should be
done first. The subtraction is done
before the division in this case.
Division of decimals
• No calculator is allowed in Paper 1.
1⋅ 7
• Working:
3 5 ⋅2 1
)
Order of operations
• Multiplication is done before
subtraction.
Multiplication of fractions
• Change mixed fractions to ‘top-heavy’
fractions 1 51 = 6 .
5
• Multiply ‘top’ numbers, multiply
‘bottom’ numbers
6
5
×1=
3
6 ×1
5×3
=
6
15
=
40 − 6
15
NOTES: 1·2 page 8
=
34
15
Subtraction of fractions
• A common denominator is needed.
3 and 15 both divide exactly into 15.
=2
4
15
✓
3 marks
•
8
3
=
8 ×5
3×5
=
40
15
multiply ‘top’ and ‘bottom’
by 5
• Change back to mixed fractions.
NOTES: 1·2 page 9
Q3 Q = R – S
Substitution
• Replace R by 3a2 and S by b2
= 3a2 − b2
= 3 × (−2)2 − 32
✓
=3×4–9
NOTES: 4·1 page 30
Calculation
• Squares are calculated first:
= 12 – 9
=3
• Replace a by −2 and b by 3.
(−2)2 = −2 × (−2) = 4 and 32 = 3 × 3 = 9
✓
2 marks
• Multiplication is done before
subtraction.
NOTES: 1·3 page 9
Page 59
Worked Answers to Practice Exam A: Credit Mathematics
Q4
7
3
−
x x +1
7(x + 1)
3x
=
–
x(x + 1)
x(x + 1)
7(x + 1) − 3x
=
x(x + 1)
=
=
Subtraction of algebraic fractions
• A common denominator is needed:
multiply x and x + 1 together.
✓
• Always multiply the ‘top’ and the
‘bottom’ by the same expression:
7 × (x + 1)
x × (x + 1)
7 x + 7 − 3x
x(x + 1)
4x + 7
x(x + 1)
and
3 ×x
x+1 × x
Simplify
• The numerators are subtracted and
then simplified.
✓
NOTES: 4·5 pages 38 and 39
2 marks
Q5 Here is the data in order:
The median
• The data as given is unordered. To
calculate the median and quartiles
you must order the data from least to
greatest.
2, 2, 3, 6, 8, 11, 12, 13, 19,
20, 23, 26, 27
✓
Now divide data into equal
groups:
• The middle value or mean of the
middle two values of the ordered data
is the median.
(2, 2, 3, 6, 8, 11) 12 (13, 19, 20, 23, 26, 27)
3+6 4.5
=
2
(Q1)
20+23 21.5
=
2
(Q3)
Median
(Q2)
✓
✓
Quartiles
• If the median is one of the data values
then it is not used to calculate the
lower and upper quartiles.
Here is a boxplot of the data:
2 4 .5
0
5
21.5
12
10
15
20
• The lower quartile (Q1) is the median
of the smaller half of the data.
27
25
• The upper quartile (Q3) is the median
of the larger half of the data.
30
✓
4 marks
• A scale should be used and the five
numbers written on the boxplot as
shown.
Q6 Normal price (100%) has
been reduced by 20% to
80%:
80%
1%
100%
£680
Boxplot
• The boxplot uses five statistics: least,
lower quartile (Q1), median (Q2),
upper quartile (Q3), greatest.
NOTES: 6·1 pages 56 and 57
✓
680
80
680
× 100
80
✓
= £850
✓
Percentages
• The final price is given after a
reduction so a ‘proportion’ method is
required.
3 marks
Page 60
• Knowing 80% divide by 80 to find 1%
and multiply by 100 to find 100%.
NOTES: 1·5 page 12
Worked Answers to Practice Exam A: Credit Mathematics
Q7 For the dodecahedral dice,
of the 12 possible numbers
4 are multiples of 3:
3, 6, 9, 12
Probability = 4 = 1
12
3
1
3
2 1
=
8 4
✓
1
4
Since > the dodecahedral
dice gives a better chance ✓
3 marks
Q8(a) This gives:
✓
2
x+y
y
x
x+y+y−1=2
x + 2y − 1 = 2
y−1
−1
x + 2y = 2 + 1
x + 2y = 3
Number of outcomes that make the
event happen (favourable outcomes)
Total number of outcomes
✓
For the Octahedral dice, of
the 8 possible numbers only
2 are multiples of 3: 3, 6
Probability =
Understanding probability
• Probability of an event =
✓
2 marks
NOTES: 6·3 page 62
Calculating probability
• Four ‘favourable’ outcomes out
of a total of 12 possible outcomes
(Dodecahedral).
• Two ‘favourable’ outcomes out
of a total of 8 possible outcomes
(Octahedral).
Comparison
• It is essential you state the comparison,
i.e. 31 > 41 along with your conclusion.
Understanding the pattern
• The example A = B + C tells you that
the top circle is the sum of the two
below it: x + y and y + −1
1.
Set up an equation
• Following the pattern for the top
triangle leads to an equation:
x + y + y −1
1 = 2.
Simplify the equation
• Letters on the left, numbers on
the right.
Q8(b) This gives:
13
y+2
y
✓
y + 4 + 2x = 13
2+2x
2
y + 2 + 2 + 2x = 13
2x
y + 2x = 13 − 4
y + 2x = 9
✓
2 marks
Q8(c) Solve simultaneous
equations
• Line up the equations x’s under x’s
and y’s under y’s with numbers on
the right.
Calculation of values
• Aim for one equation with one
unknown.
✓
2x + 4y
4 =6
x + 2y = 3 × 2

2x + y = 9
2x + y = 9 
Subtract
3y = −3 ✓
so y = −1 ✓
Put y = −1 into 2x + y = 9
so 2x − 1 = 9 ⇒ 2x = 10
⇒x=5
Simultaneous equations
• You will recognise this from having
two equations and two unknowns
(x and y).
• Substitute the first calculated value
back into either of the two equations.
• Check using the other equation
Use y = −1 and x = 5 in x + 2y to get
5 + 2 × (−1) = 3.
NOTES: 4·4 page 36
✓
3 marks
Page 61
Worked Answers to Practice Exam A: Credit Mathematics
Amplitude
• y = sin x° has a max or min value of
1 or −1 so y = 5 sin x° has a max or min
value of 5 or −5, true for this graph.
Q9 The amplitude is 5 so
p=5
✓
From 0° to 360° there
would be 3 cycles so q = 3 ✓
2 marks
Q10 A linear equation:
y = px + q
NOTES: 3·4 pages 26 and 27
y-intercept
negative
(q < 0)
gradient
positive
( > 0)
(p
Gradient
• y = px + q is “y = mx + c” with
different letters.
y
• A positive gradient gives an uphill
graph.
✓
x
O
Period
• y = sin x° has period 360° − 1 cycle
every 360°. So y = sin 3x° has period
120° − 3 cycles every 360°; true for
this graph.
y-intercept
• A negative y-intercept means the
graph cuts the y-axis below the x-axis.
✓
NOTES: 5·2 pages 47 and 48
2 marks
Zero index
• a0 = 1 Any non-zero number raised to
the power zero gives 1.
2
Q11(a) 83 − 80
Fractional indices
= ( 8) − 1
✓
= 22 − 1 = 4 – 1 = 3
✓
3
2
m
• an
m
so a n = ( n a )m and in
2
this case: 83
2 marks
Q11(b)
power
root
squared (power 2)
cube root
NOTES: 4·6 pages 41 and 42
Simplifying surds
• Always attempt to get the smallest
number under the square root by
factorising out squares: 4, 9, 16,…
2+ 8
= 2 + 4×2
= 2+ 4× 2
✓
= 2 + 2 2 =3 2
✓
Adding Surds
• Compare 2 + 2 2 = 3 2 with
x + 2x = 3x
2 marks
Page 62
NOTES: 4·6 page 40 and 41
Worked Answers to Practice Exam A: Credit Mathematics
Q12 In 1 hour the tip of
the minute hand travels
the whole circumference
of the clock face
✓
so π D = 27
✓
27
π
⇒D=
✓
The diameter, D, of the
clock face is the same
length as the side of the
square frame
=
=
108
π
Set up equation
• C = 2πrr or C = πD : the second
version is the best one for this
question.
NOTES: 2·4 page 17
Change the subject
• Divide both sides by π.
Calculation
• 4 sides make up the perimeter, so × 4.
so Perimeter = 4 × D
27
4×
π
Strategy
• Evidence that you knew to use the
circumference formula to set up an
equation will gain you the strategy
mark.
✓
• The working 4 × 27
is essential
π
especially since the answer 108
is given.
π
metres
4 marks
Practice Exam A: Paper 2 Worked Answers
Q1 Use T =
so T =
D
S
✓
5 ⋅ 6 × 107
120
✓
= 466666⋅66… hours
=
466666 ⋅ 66…
24
•
days
✓
Calculator
• 5⋅6 × 107 is entered into your calculator
like this:
5
·
6
7
× 10x
=
Conversion
• Change hours to days: divide by 24.
= 19444⋅44…
= 1⋅9 × 104 days
(to 2 sig figs)
D,S,T Formula
You are asked to find the
D
Time taken. Covering T in
S T
the triangle gives
D , the Distance divided by the Speed.
S
✓
4 marks
Scientific notation
• Convert your answer to scientific
notation:
a × 10n
a number between The decimal
1 and 10 in this
point in 1·9
case 1·9
moves 4 places to
the right so n = 4
NOTES: 1·4 page 10
Page 63
Worked Answers to Practice Exam A: Credit Mathematics
Q2(a) Compare y = mx + c
m = gradient
=
245 − 120
5
=
Linear equation
• You will gain a mark for evidence if
you realise “y = mx + c” is involved.
✓
125
5
Gradient
= 25
“y-intercept” is 120
• Gradient =
✓
✓
distance along
• Read the scales carefully, they are not
the same on the two axes.
Equation is C = 25t + 120
3 marks
Q2(b) C = 282·5
distance up or down
Correct equation
• C is used instead of y and t is used
instead of x. y = 25x + 120 will not
gain the final mark.
NOTES: 5·2 page 48
✓
so 25t + 120 = 282·5
Strategy
• Knowing to use C = 282·5 in the
equation will gain you the strategy
mark.
⇒ 25t = 282·5 − 120
= 162·5
⇒t=
162 ⋅ 5
25
= 6·5
✓
The conference lasted
6 21 hours.
✓
3 marks
The mean
• The total of the numbers divided
by the number of numbers gives the
Mean.
Σx
Q3 x (mean) = n
=
7 + 10
10 + 8 + 9 + 1
13
3 + 7 + 10 + 8
8
so x =
x
7
10
8
9
13
7
10
8
72
8
=9
x − x (x − x )2
4
−2
1
1
1
−1
0
0
4
16
4
−2
1
1
1
−1
∑ (x − x )2 =
Solving the equation
• Setting up the correct equation and
solving it correctly gain you two more
marks.
Standard deviation
• There are two equivalent formulae:
✓
s=
s=
Σ (x − x )
n −1
=
2
s=
28
7
= 4
2
Σ(( x − x )
n −1
2
and
2
Σ x − ( Σ x) / n
n −1
The solution uses the first formula. If
you are used to using the 2nd formula
then stick with what you are used to!
✓
=2
• Use a table of values when calculating
as it is easier to read and leads
to fewer errors.
28
✓
Mean = 9
standard deviation = 2
✓
• Working is expected and you will lose
marks for not showing working. If
you know how to calculate s on your
calculator you should do this to check
your calculation is correct.
4 marks
Page 64
NOTES: 6·2 pages 60 and 61
Worked Answers to Practice Exam A: Credit Mathematics
Q4 Reduction factor is 0·94 ✓
and is applied three times.
So final altitude
= 340 × 0·94 × 0·94 × 0·94 ✓
.
= 282·39… =. 282 km
(to nearest km)
✓
3 marks
Q5 Use the Sine Rule
c = 1·78
A
50°
b
a
first calculate
angle C
C
Angle C = 180° − (46° + 50°)
= 180° − 96° = 84°✓
so
b
sin50
⇒b=
=
1⋅ 7
78
sin 84
.
37 metres
= 1⋅371… =. 1⋅ 3
angle C =
=
A
80
67·5°
C
B
1
(180°
2
1
2
✓
− 45°)
× 135° = 67·5° ✓
Now use
“SOHCAHTOA”
in triangle ABC ✓
sin 67·5° = 80
AC
✓
⇒ AC × sin 67·5° = 80
⇒ AC =
Angle sum
• The angle sum in a triangle is 180°.
✓
Q6 The large triangle is
isosceles
So angle A =
angle C
Strategy
• If you know two angles and one side
then you should use the Sine Rule.
Calculation
• Always make sure your calculator is in
degree mode. Make sure you know
how to do this on your calculator.
NOTES: 3·3 pages 22 and 23
4 marks
C
Rounding
• Unless accuracy is stated in the
question any correct rounding is
acceptable.
Substitution
• Correctly substituting the values into
the Sine Rule will gain you a mark.
(to 3 sig figs)
45° D
Time calculation
• Shorter version is 340 × 0·943.
✓
1⋅ 78
78 × sin 5
50
sin 84
A
NOTE: 1·5 page 11
✓
a
b
c✓
=
=
sin A ✓ sinB ✓ sinC
B
46°
Reduction factor
• Reducing by 6% is equivalent to
calculating 94%, i.e. multiplying by 0·94.
80
sin 67 ⋅ 5
= 86·59…
=.. 86 ⋅ 6 cm ✓
(to 3 sig figs)
5 marks
Strategy
• Evidence that you knew to use
“SOHCAHTOA” in triangle ABC by
finding angle C would gain you the
strategy mark.
Angle sum
• Angle A and angle C between them
make up 135° of the 180° angle sum
in the large triangle.
Correct ratio
• Working from angle C: the opposite
AB is known and the hypotenuse AC
is required, so use “SOH” i.e. sin.
Correct equation
80
• Sin 67·5 = AC
or its equivalent will
gain you a further mark.
Calculation
• Multiply both sides of the equation
by AC and then divide both sides by
sin 67·5°.
NOTES: 3·1 page 20
Page 65
Worked Answers to Practice Exam A: Credit Mathematics
Q7 Use the Cosine Rule
b2 = a2 + c2 − 2ac cos B
Strategy
• Two sides and the included angle are
given so the Cosine Rule is used.
✓
B
c=10·5
95°
a=18
A
C
= 182 + 10·52 − 2 × 18 × 10·5 × cos 95° ✓
= 467·19…
so b = 467 ⋅ 19 … = 21·61…
.
so AC =. 21⋅ 6 cm
(to 1 dec. place)
Calculation
• In most calculators just key in the
complete calculation as written in the
solution.
✓
✓
4 marks
Q8
A
x
B
x
D
5
C
x +x =5
⇒ 2x2 = 25
2
2
⇒ x2 =
25
=
2
12·5
Strategy
• Constructing the right-angled triangle
and using Pythagoras’ Theorem is the
crucial strategy.
✓
Pythagoras’ Theorem
• Marking BC and BD with letter x
helps.
✓
Strategy
• Realisation that AC is a radius, length
5 m, is the 2nd crucial strategy step.
So x = 12 ⋅ 5 . AC is a radius
so AC = 5
✓
AB = AC − BC = 5 − 12 ⋅ 5
= 5 − 3·53… = 1·464…
.
46 metres
So AB =. 1⋅ 4
(to 3 sig figs)
✓
Calculation
• Never round answers until the final
answer.
Rounding
• Rounding to 3 significant figures is
essential.
5 marks
Q9
C
a=1·4 50°
b=1·6
B
A
1
2
=
ab sin C
1
× 1·4 × 1·6 × sin 50° ✓
2
= 0·857…
Volume = 0·857… × 3
✓
✓
=.. 2 ⋅ 5
57 m3
Substitution
• The area formula uses two sides and
the included angle.
Calculation
• At this stage do not round your answer.
Substitution
• Always check the units match—in this
case they do.
(area of end) (length)
= 2·573…
NOTES: 2·4 page 19
Formula
• This formula is true for all prisms.
Volume = Area of end × length✓
Use Area =
Rounding
• 1 mark is allocated for correct
rounding.
NOTES: 3·3 page 23
Triangle BCD
(see diagram)
is isosceles
with BC = BD
=x
✓
By Pythagoras’ Theorem:
2
Substitution
• Correct substitution of the values into
the formula will gain you a mark.
✓
5 marks
Answer
• Any correct rounding will do for the
final answer.
NOTES: 3·3 page 22 and 2·1 page 14
Page 66
Worked Answers to Practice Exam A: Credit Mathematics
Q10 3cos x° + 4 = 3
⇒ 3cos x° = −1
1
3
⇒ cos x° = − (negative)✓
So x° is in the 2nd or 3rd
quadrant
1st quadrant angle is given
by:
 1
1
cos x° = ⇒ x = cos−1  
 3
3
So x = 70·52...
✓
This gives:
.
x = 180 − 70·52... =. 109·5
(2nd quadrant)
x = 180 + 70·52... =.. 250·5
(3rd quadrant)
(correct to 1 dec. place) ✓
Rearrangement
• You solve this equation as you
would solve 3c + 4 = 3 to get c = − 31.
1st quadrant angle
• Initially ignore the negative sign and
use cos−1 button with ( 31) to get an
angle in the 1st quadrant—between
0° and 90°.
The quadrants
• Use
S A
T C
or
to
determine the quadrants: cosine is
negative in 2nd and 3rd quadrants.
• 2nd quadrant: 180 − (1st quad angle).
3rd quadrant: 180 + (1st quad angle).
NOTES: 3·5 page 27
3 marks
Starting
• You will gain a mark for either of the
two area expressions appearing.
Q11(a)
Use Area = length × breadth
Pool:
x+2
Strategy
• Any correct method for finding an
expression for the area of path and
then setting up an equation will
guarantee the strategy mark.
Outside of Path:
x
Area
A
x+6
x+2
Area
B
Area = x(x + 2) Area = (x + 2)(x + 6)
= x2 + 2x
= x2 + 8x + 12
✓
✓
Area of Path =
Area B – Area A
= x2 + 8x + 12 − (x2 + 2x)
= x2 + 8x + 12 − x2 − 2x
= 6x + 12
✓
Area of Pool = Area of Path
⇒ x2 + 2x = 6x + 12
• There are other methods than the one
shown in the solution for finding the
areas.
Simplification
• Simplifying expressions as you proceed
through your solution makes the
subsequent working easier.
Equation rearrangement
• Since the answer is given
(x2 − 4x − 12 = 0) it is important that
your working is clear.
⇒ x2 + 2x − 6x – 12 = 0
⇒ x2 – 4x – 12 = 0
✓
4 marks
Page 67
Worked Answers to Practice Exam A: Credit Mathematics
Q11(b) x2 − 4x − 12 = 0
Solving a quadratic
• Each factor could be zero:
x − 6 = 0 or x + 2 = 0 giving the two
solutions x = 6 and x = −2
⇒ (x − 6) (x + 2) = 0
⇒ x = 6 or x = − 2
✓
Invalid solution
• You must state clearly that x = −2
makes no sense. There is a mark for
this.
x = −2 is not sensible for
a length
✓
So x = 6 is only solution.
Pool dimensions:
6m×8m
✓
3 marks
Page 68
Answer
• The dimensions of the pool are x
metres by (x + 2) metres with x = 6.
NOTES: 4·7 page 42
Worked Answers to Practice Exam B: Credit Mathematics
Practice Exam B: Paper 1 Worked Answers
Q1 8·82 − 2·9 × 3
= 8·82 − 8·7
Order of operations
• Multiplication is done before
subtraction.
✓
= 0·12
Decimal subtraction
✓
2 marks
0⋅1
12
Q2 2x2 + 3x − 2
= (2x − 1)(x + 2)
Factorisation
• You should always check your answer
by multiplying out using “FOIL”.
✓✓
NOTES: 4·3 page 33
2 marks
Subtraction of fractions
• The lowest common denominator is
6 as this is the smallest number 2
and 3 divide into exactly.
Q3 2 (1 21 − 31 )
3
2  3 1
=  − 
3  2 3
=
2  9 2
−
3  6 6 
=
2 7
×
3 6
=
8⋅8
82
• Line up the decimal points: −88⋅7
=
14
18
2 9−2
×
3
6
=
7
9
Multiplication of fractions
a×c
• “Multiply across”: ab × dc = b × d .
✓
• Cancel if possible: in this case divide
top and bottom of the fraction by 2 to
get 97 .
✓
NOTES: 1·2 pages 8 and 9
2 marks
Brackets
• Take care with a negative multiplier:
−x(2 − 3x) = −2x + 3x2 (2nd term
positive).
Q4 x(2x − 1) − x(2 − 3x)
= 2x2 − x − 2x + 3x2
= 5x2 − 3x
✓✓
Simplify
• Gather ‘like terms’: 2x2 + 3x2
and −x −2x.
✓
3 marks
NOTES: 4·1 page 30
Q5(a)
f x) = 6
f(
Substitution
• From f(
f x) to f(
f −3) involves replacing
x by the value −3 and calculating the
result.
x
so f(
f −3) =
6
−3
= −2
✓
1 mark
Q5(b)
f a) = 3 ⇒ 6 = 3
f(
✓
so 6 = 3a ⇒ a = 2
✓
a
NOTES: 4·1 page 30
Equation
• f(
f a) is replaced by
6
a
for the 1st mark.
Solving
• Divide both sides by 3.
2 marks
Page 69
Worked Answers to Practice Exam B: Credit Mathematics
Q6(a)
y = − 1x + 3
Equation of straight line
• The general equation of a straight
line is:
✓✓
2
2 marks
y = mx + c
gradient y-intercept
In this case m = − 21 and c = 3
NOTES: 5·2 page 48
Q6(b)
For A(−2, a)
x = −2 and y = a
Substitution
• Since A(−2, a) lies on the line its x
and y coordinates satisfy the equation
of the line. So substitution is the
strategy here.
Substitute these values in
the line equation:
1
2
a = − × (−2) + 3
✓
⇒a=1+3 =4
✓
Calculation
• Negative × Negative = Positive. No
calculator is allowed in Paper 1 work.
2 marks
NOTES: 5·2 page 48
Q7(a) Let u be the charge of
an ‘up’ quark and let d
be the charge of a ‘down’
quark:
2u + d = 1
✓
Setting up an equation
• It is important to be clear what the
letters you use are representing. Here
it is the value of the electric charge.
1 mark
Q7(b) u + 2d = 0
Setting up an equation
• ‘Algebraic’ mentioned in a question
means that algebra should be used, e.g.
equation solving, letter substitution etc.
✓
1 mark
Q7(c) Solve simultaneous
equations
Strategy
• Two equations with two unknowns
(letters) indicates simultaneous
equations need to be solved to find the
values of the unknowns.
✓
Method
• Line up the equations: in this case ‘u’s
are lined up and ‘d’s are lined up.
4 u + 2d = 2
2u + d = 1  × 2

u + 2d = 0 
u + 2d = 0 ✓
=2
subtract 3u
⇒u=
2
3
• The aim is always to eliminate one of
the unknowns, in this case d.
✓
Calculation
• Divide both sides by 3.
2 marks
NOTES: 4·4 page 36
Page 70
Worked Answers to Practice Exam B: Credit Mathematics
Probability
• Careful reading of the table is required
to extract the right information.
Q8(a) Total no. of pupils in
class 3M1 = 12 + 16 = 28
Probability of choosing a
boy =
12
28
=
3
7
• Probability =
No. of boys in class 3M1
total no. of pupils in class 3M1
✓
1 mark
Q8(b) Total no. of pupils
Probability
• The totals are now different as all the
classes have been put into one large
group.
Total of all boys
• Probability =
Total of all pupils
= 12 + 16 + 10 +14 +14 + 6 = 72
Total no. of boys
= 12 + 10 + 14 = 36
Probability of choosing a
boy = 36 =
72
1
2
NOTES: 6·3 page 62
✓
1 mark
Boxplot
• The following diagram shows the
percentage of data in each part of the
plot:
Q9 75% were less efficient than
38 mpg.
✓
1 mark
25%
Q10(a)
25%
25% 25%
• 75% of the data occurs below 38 and
25% occurs above.
3
:
1
(nitrogen) (potassium)
12 kg
4 kg
NOTES: 6·1 page 57
His lawn would get 4 kg
potassium.
✓
Ratio
• 3 parts = 12 kg so 1 part = 4 kg.
This is a ‘proportion’ calculation.
1 mark
Q10(b)
Strategy
• The total weight is required so the
weight of Phosphorus has to be
calculated.
3
:
2
:
1
(nitrogen) (phosphorus) (potassium) ✓
12 kg
8 kg
4 kg
Total weight
= 12 + 8 + 4 = 24 kg
Ratio
• 3 parts = 12 kg, 1 part = 4 kg
so 2 parts = 2 × 4 kg = 8 kg.
✓
3 × 7 kg packets = 21 kg
(3 kg short)
Solution
• 24 kg are required. Dividing by 7
4 × 7 kg packets = 28 kg
(4 kg spare)
So 4 packets are
required.
gives: 7
)
3· 4 2
3 2
24· 0 0
. You must however
round up to 4 packets otherwise you
will be short. The context of the
problem tells you when to round up
in this way.
✓
3 marks
NOTES: 1·5 page 12
Page 71
Worked Answers to Practice Exam B: Credit Mathematics
Q11(a) S3 = t1 + t2 + t3
= 1 + 7 + 19 = 27
Pattern completion
• Completing the S3 row gains 1 mark
and completing the S4 row gains the
2nd mark.
✓
S4 = t1 + t2 + t3 + t4
• When solving ‘pattern’ questions it is
important that you compare each line
with the previous lines to see how the
pattern is building up.
= 1 + 7 + 19 + 37
= 64
✓
2 marks
Generalisation
• Recognition of the ‘Cubes’ is crucial
to solving this question:
Q11(b) S3 = 27 = 33
S4 = 64 = 43
Sn = n3
✓
13,
23,
33,
43,
53,…
1,
8,
27, 64, 125,…
1 mark
Strategy
• This difficult question requires
insight into the relationship between
Sn, Sn+1 and tn+1: adding the first (n + 1)
terms involves adding the first n terms
and then the (n + 1)th term. This, in
symbols, is: Sn+1 = Sn + tn+1.
Q11(c) Sn+1 = (t1 + t2 + … + tn) + tn+1
⇒ Sn+1 = Sn + tn+1
✓
• An alternative explanation is that
removing the first n terms from the
first (n + 1) terms will leave only the
(n + 1)st term:
⇒ (n + 1)3 = n3 + tn+1
so tn+1 = (n + 1)3 − n3
✓
2 marks
1
1
Q12(a) 9− 2 =
t1+ t2 + t3 + … + tn + tn+1
remove these (S
Sn )
Sn +1
Negative indices
1
• a−n = n is the general Index Law used.
a
✓
1
92
=
1
9
=
1
3
Fractional indices
✓
m
• an =
2 marks
( a)
n
m
is the general Index Law
used. This may help:
m
an
power
root
1
so a 2
power 1
square root
1
So 9 2 means the square root of 9 to
the power 1.
Q12(b)
t × t2
1
= t 2 × t2
=t
1
+2
2
=t
✓
5
2
Simplifying indices
• a m × a n = a m+n is the Index Law used.
✓
2
1
• Note t 2 is not acceptable.
2 marks
NOTES: 4·6 pages 41 and 42
Page 72
Worked Answers to Practice Exam B: Credit Mathematics
Volume of a cuboid
• 5 × 2 × 2 = 20 and x × x × x = x3
Q13
Volume of Cuboid
= length × breadth × height
• You will not lose a mark for forgetting
to use the units cm3.
= 5x × 2x × 2x
= 20x3 cm3
Volume of a prism
• Volume of prism
= Area of end × length.
✓
Blocked-off Volume
• The triangular end is right-angled so
use 21 × base × height. Base is AB and
height = 2x cm.
= Area of Triangle × breadth
=
1
2
× AB × 2x × 2x
= 2 × AB × x2 cm3
Blocked Volume =
2 × AB × x =
2
1
4
Set up equation
• “One quarter of the volume” is the
clue to setting up this equation.
✓
1
× Cuboid Volume
4
✓
Solving the equation
• Since x ≠ 0 both sides can be divided
2 × AB × x2 5x3
2
by 2x :
=
.
2x2
2x2
× 20x
3
⇒ 2 × AB × x2 = 5x3
⇒ AB =
5x3
2x2
=
5x
2
cm
✓
4 marks
Practice Exam B: Paper 2 Worked Answers
Multiplication factor
• An increase of 1·2% is the equivalent
of calculating 101·2%. Multiplying by
1·012 calculates this amount.
Q1 The yearly increase factor
is 1·012. The factor is
✓
applied three times (07–10)
Total amount in 2010
✓
= 7·2 × 1·012 × 1·012 × 1·012
= 7·46...
✓
=.. 7·5 million tonnes
✓
(correct to 2 significant
figures).
Time calculation
• From 2007 to 2010 involves 3
complete years. Since each year’s
increase is 1·2% three multiplications
by 1·012 are necessary.
• A shorter version is 7·2 × 1·0123.
4 marks
Calculation
• ∧ 3 These keys mean “raise to the
power 3” or “cubing”.
Rounding
• When accuracy is mentioned in the
question there will be 1 mark allocated
for the correct rounding.
NOTES: 1·5 pages 11 and 12
Page 73
Worked Answers to Practice Exam B: Credit Mathematics
Q2(a) Mean
=
Mean
• Mean =
44 + 47 + 38 + 97 + 40 + 52
6
= £53
total of the numbers
the number of numbers
NOTES: 6·2 page 59
✓
1 mark
Standard deviation
• You should always double check your
calculations and, if possible, confirm
your answer using the STAT mode on
your calculator.
Q2(b) There are 6 pieces of
data so n = 6
From part (a) x = 53
x x − x (x − x )
44 −9
81
47 −6
36
38 −15 225
97 44 1936
40 −13 169
52 −1
1
• It is important that all your working
is shown. You will not gain the marks
by calculating the standard deviation
on your calculator and just writing the
answer down.
2
s=
=
Σ (x − x )2
n −1
2448
5
✓
Calculation
• Any correct rounding will gain the mark
so long as your calculation is correct.
= 489 ⋅ 6
= 22⋅126…
2 ⋅1
13 ✓
=.. £22
Σ((x − x ) = 2448
2
NOTES: 6·2 pages 60 and 61
(to 2 decimal places)
Comparison using statistics
• The standard deviation, s, measures the
variation of the data about the mean. A
greater value of s means more variation,
a lesser value means less variation.
2 marks
Q2(c) There was less variation
in the amounts spent on
a weekday (s = 8·4) than
on a Saturday (s = 22·13,
greater than 8·4)
✓
Diagram
• In ‘Bearing’ questions diagrams are
essential.
1 mark
• “due west” gives the clue that the
Leven to Elie line LE is at rightangles to the North line.
Q3 Angle CEL = 320° − 270°
C
= 50°
✓
N
4°
E
320°
L
C
e=
11·9
L
44°
86°
50°
c
E
Angle CLE
= 90° − 4° = 86°
c
sin 44
=
NOTES: 3·2 page 21
Angle sum in triangle
• To use the Sine Rule the 3rd angle
must be calculated.
Angle ECL
= 180° − (50° + 86°)
= 180° − 136°
= 44°
✓
Use the Sine Rule:
so
• Bearings are always measured
clockwise from the North line.
N
11⋅ 9
sin 50
c
sinC
⇒c=
=
Sine Rule
• Multiply both sides by sin 44° to
calculate c so that sin 44° will appear
at the top of the fraction.
e
sinE
11⋅ 9 × sin 4
44
sin 50 ✓
Calculation
• Always check that your answer,
10·8 km, appears reasonable in
the given context. In this case it is
comparable to 11·9 km, the other
length, and so seems reasonable.
= 10·79…
.=
So required distance . 10·8 km ✓
(correct to 1 dec place)
NOTES: 3·3 pages 22 and 23
4 marks
Page 74
Worked Answers to Practice Exam B: Credit Mathematics
Q4(a)
Volume of Cylinder = πr2 × h ✓
Volume of cylinder
• This formula is not given and you are
expected to know it.
In this case
r=
1
2
• As for any prism the volume is given
by the area of the end (πr2) × the
length (h).
× 58
= 29 cm, h = 97 cm
V = π × 292 × 97
= 256281·7… cm3
Calculation
• Care should be taken that the radius
is used in the formula, not the given
diameter.
✓
= 256·2… litres
=.. 256 litres
(to the nearest litre)
Units
• 1000 cm3 = 1 litre so division by 1000
is necessary to convert to litres.
✓
3 marks
NOTES: 2·1 page 14
Q4(b)
256·2… − 64 = 192·2…litres
so πr2 × h = 192281·7…
⇒h=
Strategy
• Using the volume formula with the
remaining volume and changing the
subject to h will gain you this mark.
192281⋅ 7
π × 292
= 72·77…
=.. 72·8 cm
(to 1 dec place)
• Notice the volume 192281·7… is back
in cm3 to allow h to be calculated in cm.
✓
✓
Calculation
• Brackets are essential in the calculator.
Calculation: 192281·7… ÷ (π × 292)
2 marks
• Using the rounded answer (256 litres)
would not be penalised although the
calculated height would be 72·7 cm
not 72·8 cm.
Q5 nth number = 1 − 3n + 3n2
Find n where nth number
= 61
so 1 − 3n + 3n2 = 61
✓
⇒ 3n2 − 3n − 60 = 0
✓
Strategy
• Setting the formula for the nth term
of the sequence equal to 61 gains the
strategy mark.
⇒ 3(n2 − n − 20) = 0
Quadratic equation
• You must recognise this as a quadratic
equation and arrange it in the standard
form: in this case an2 + bn + c = 0.
⇒ 3(n + 4)(n − 5) = 0
⇒ n + 4 = 0 or n − 5 = 0
⇒ n = −4 or n = 5
✓
Solving the equation
• Removal of the common factor first
makes the factorising easier.
In this case n = −4 makes no
sense for the position of a
number in a sequence
So n = 5 is the only
solution
• Both solutions should be given at this
stage.
✓
NOTES: 4·7 page 42
61 is the 5th number in the
sequence
Solution in context
• Rejection of n = −4 should be clear.
• You should check that substitution of
n = 5 in 1 − 3n + 3n2 does produce 61.
4 marks
Page 75
Worked Answers to Practice Exam B: Credit Mathematics
Q6 In triangle ABC
x
23°
9 cm
C
sin 35° =
35°
9·77...
C
‘SOHCAHTOA’
• Care should be taken when the
unknown is on the bottom of the
fraction. First multiply by x on both
sides of the equation and then divide
by cos 23°.
9
In triangle ACD
A
y
⇒ x cos 23° = 9
⇒ x = cos 23 = 9.77…
B
D
✓
✓
cos 23° = 9x
A
Strategy
• This question involves a two stage
‘SOHCAHTOA’: first in triangle
ABC and then in triangle ACD.
y
9⋅7
77…
✓
⇒ 9·77… × sin 35° = y
⇒ y = 5·607…
.
So CD =. 5·61 cm
(to 3 sig figs)
• For the 2nd ‘SOHCAHTOA’ use the
full calculator value for AC. Don’t
round decimals in the middle of a
problem, only at the end.
✓
Calculation
• Any correct rounding is acceptable
since accuracy is not mentioned in the
question.
4 marks
NOTES: 3·1 page 20
Q7
q=2·5
Area of triangle
P
r=3·4
R
1
2
= qr sin P
✓
Strategy
• Normally the formula
Q
so 4·2 =
1
2
× 2·5 × 3·4 × sin P
“Area = 21 ab sin C” is used to find the
area but in this case the area is given so
an equation is set up and solved first
for sin P.
⇒ 4·2 = 4·25 × sin P
⇒
4 ⋅2
4 ⋅2
25
= sin P
so sin P = 0·9882…
Calculation
• The aim is to find a value for sin P.
✓
⇒ angle P = sin (0·9882…) = 81·2…°
−1
.
so angle P =. 81°
(to 2 significant figures)
✓
Angle
• Use sin−1 ans EXE on your calculator
where “ans” uses the previous answer,
0·9882..., on the calculator display.
3 marks
• EXE and = are the same – it depends
on your calculator.
NOTES: 3·3 page 22
Intercepts
• There is a correspondence between
factors and x-axis intercepts:
Q8(a) Set y = 0 for x-axis
intercepts
so k(x − p)(x − q) = 0
⇒ x − p = 0 or x − q = 0
⇒ x = p or x = q
Intercept
x−p
( , 0)
(p
x−q
(q, 0)
• Since p = −2 the factor x − p = x − (−2)
= x + 2 so x + 2 as a factor gives (−2, 0)
as an x-axis intercept.
From the given graph
this gives:
✓
p = −2 and q = 6
Factor
NOTES: 4·6 page 42
✓
2 marks
Page 76
Worked Answers to Practice Exam B: Credit Mathematics
Substitution
• If a point (a, b) lies on a graph then
values x = a and y = b will satisfy the
equation of that graph. Substituting
these values into the equation gives an
equation with only one unknown k in
this case.
Q8(b) The equation is
y = k(x + 2)(x − 6) and
from the graph (0, 6) lies
on the parabola.
So substitute x = 0 and
y = 6 into the equation of
the parabola.
✓
y
Calculation
6
• Some graphs with
equation
x
6
y = k(x + 2)(x − 6) are −22
shown. They all have
x-intercepts (−2, 0) and (6, 0) and are
all parabolas. Only one graph passes
through (0, 6): y = − 21 (x + 2)(x − 6).
6 = k × (0 + 2) × (0 − 6)
⇒ 6 = k × 2 × (−6)
⇒ 6 = k × (−12)
⇒k=
6
−12
=−
1
2
✓
2 marks
Q8(c)
x-value
• The graph is symmetrical with axis of
symmetry x = 2.
The x value that gives
the maximum is half way
between −2 and 6
i.e. x =
−2 + 6
2
=
4
2
=2
• Required value is the mean of −2
and 6.
✓
y-value
• The question asks for the coordinates
of a point, i.e. (2, 8) not the separate
values.
The equation is
y = − 1(x + 2)(x − 6)
2
NOTES: 5·4 page 52
now substitute x = 2
1
2
so y = − (2 + 2)(2 − 6)
=−
1
2
× 4 × (−4) = 8
Scale factor
• If the length scale factor is k,
the area scale factor is k2
and the volume scale factor is k3.
The maximum turning
point is (2, 8)
✓
2 marks
Calculation
• For a reduction, the scale factor
should lie between 0 and 1. You
should check that your final answer is
smaller than the given volume of 43
litres.
Q9 The smaller volume is
required so a reduction
scale factor is needed.
Rounding
• 1 mark is allocated for correct
rounding to the nearest litre since the
required accuracy is mentioned in the
question.
Length scale factor:
6
8
=
3
4
= 0·75
Volume scale factor: 0·753 ✓
Required volume
= 43 × 0·753
= 18·14...
=.. 18 litres
(to the nearest litre)
NOTES: 2·2 page 15
✓
✓
3 marks
Page 77
Worked Answers to Practice Exam B: Credit Mathematics
Strategy
• The construction of triangle ACE and
the use of Pythagoras’ Theorem are
the essential steps in this solution.
Q10 AC is a radius
so AC = 2·5 cm
A
CD is also a
radius
E D
F
so CD = 2·5 cm
✓
ED = DF − EF = 5 − 4·8 = 0·2 cm
C
Use of radius
• All radii in a circle are equal. The
diameter in this case is 5 cm so radius
= 21 × 5 cm.
and CE = CD − ED
= 2·5 − 0·2 = 2·3 cm
Calculation
• The symmetry of the diagram (CE is
an axis of symmetry) means that the
gap AB is double the length AE
✓
Use Pythagoras’ Theorem in
A triangle ACE
2·5
C
2·3
Answer
• The value 2·0 cm seems reasonable
in this context. The value of 0 ⋅ 9
96
should not be rounded before the final
answer is reached.
AE2 = AC2 − CE2
E = 2·52 − 2·32 = 0·96
B
So AE = 0 ⋅ 9
96 ✓
• Any correct rounding is acceptable
in the final answer as accuracy is not
mentioned in the question.
So the gap AB = 2 × AE
96
= 2 × 0 ⋅9
= 1·959…
=.. 2·0 cm (to 1 dec place)
✓
NOTES: 2·4 page 19
4 marks
•
Q11(a) Using S =
D
T
Average Speed
180
= x m.p.h.
Using algebra
• You should be aware that “2 less
than x” translates to the algebraic
expression x − 2.
✓
1 mark
• Note the distance has changed to 60
miles.
Q11(b) Again using S = D
T
“2 hours less” gives x − 2 hours
So Average speed =
D.S.T. formula
Covering up S reveals D
as
T
D
the formula for calculating
Average Speed.
S T
60
x−2
m.p.h. ✓
1 mark
Page 78
Worked Answers to Practice Exam B: Credit Mathematics
Q11(c)
The two average speeds
are equal so:
180
x
=
Set up equation
• The strategy here is to equate the two
expressions for the average speeds.
60
x−2
• You would still gain this mark even if
you equated “wrong” expressions.
✓
⇒ 180(x − 2) = 60x
Solving
• Multiply both sides of the equation
by x and then by x − 2 (or “crossmultiply”).
⇒ 180x − 360 = 60x
⇒ 120x = 360
⇒x=3
Interpretation
• What does x = 3 mean? You must look
back to x hours and x − 2 hours and
substitute.
✓
1st stage took x = 3 hours.
2nd stage took x − 2 = 1 hour.
Total journey took 4 hours.
NOTES: 4·4 page 34
✓
3 marks
Page 79
Worked Answers to Practice Exam C: Credit Mathematics
Practice Exam C: Paper 1 Worked Answers
Q1 1⋅3 × (4⋅92 + 0⋅08)
= 1⋅3 × 5
✓
Order of operations
• Calculations in brackets are always
done first.
= 6⋅5
✓
Calculation
• No calculator is allowed in Paper 1.
2 marks
• When adding decimals remember to
4⋅9
92
line up the decimal points: +0⋅008 .
5⋅0
00
Division of fractions
• Change mixed fractions to ‘top-heavy’.
1
Q2 2 ÷ 1
3
3
=
2
3
÷
4
3
=
2
4
=
1
2
=
2
3 ( × 3)
4
3 ( × 3)
• An alternative method to the one
given is to ‘invert then multiply’
6 = 1.
so 2 ÷ 4 = 2 × 3 = 12
2
3
4
3
3
✓
✓
Calculation
• Remember to ‘cancel down’ if possible.
NOTES: 1·2 page 8
2 marks
Substitution
• For f(
f −1), every occurence of x is
replaced by the value −1. No further
calculation is needed to gain the 1st
mark.
Q3 f(
f x) = x(2 − x)
⇒ f(
f −1) = −1 × (2 − (−1)) ✓
= −1 × (2 + 1)
Calculation
• The 2nd mark here is for the subsequent
calculation after the substitution.
= −1 × 3 = −3
✓
2 marks
• Remember: brackets first and that
subtracting a negative is the same as
adding the positive value.
NOTES: 4·1 page 30
Q4 3 + 2x < 4(x + 1)
⇒ 3 + 2x < 4x + 4
Brackets
• Both terms are multiplied by 4. A
common mistake here would be 4x + 1.
✓
(now subtract 4 from each side)
Simplification
• The ‘balancing process’ aims to get
letters on one side of the inequality
sign and numbers on the other side.
⇒ −1 + 2x < 4x
(now subtract 2x
x from each side)
⇒ −1 < 2x
⇒
−1
<
2
✓
• Best to avoid −2x < 1. The next step
here would be divide by −2 and ‘swap
the sign round’: x > −12 .
x
so x > −
1
2
✓
Answer
• Notice if a < b then b > a.
3 marks
NOTES: 4·4 page 35
Page 80
Worked Answers to Practice Exam C: Credit Mathematics
Q5(a) 9y2 − 4
Difference of squares
• The pattern is: A2 − B2 = (A − B)(A + B).
= (3y)2 − 22
= (3y − 2) (3y + 2)
• You should always check that your
answer multiplies out giving the
original expression (use ‘FOIL’).
✓
1 mark
NOTES: 4·3 page 32
Factorising
• When simplifying algebraic fractions the
1st step is to factorise both expressions.
Q5(b)
9 y2 − 4
15 y − 10
(3 y − 2)(
)(3 y + 2)
=
5(3 y − 2)
• ‘Hence’ appearing in a question means
you should use the answer that you got
in the previous part of the question: in
this case 9y2 − 4 = (3y − 2) (3y + 2).
✓
= 3y + 2
✓
5
Cancelling
• Any factor, e.g. 3y − 2, that appears
on the top and the bottom can be
cancelled.
2 marks
NOTES: 4·5 page 37
Q6 P =
1st rearrangement
• The subject of the formula, A, in this
case appears on the bottom of the
fraction, i.e. W is divided by A. The
inverse process is multiplication so
multiply both sides of the formula by A.
W
4A
(×A) (×A)
⇒ PA =
W
4
✓
2nd rearrangement
• A is multiplied by P. The inverse
process is division so divide both
sides by P.
(÷P) (÷P)
⇒Α=
W
4P
✓
NOTES: 4·5 pages 39 and 40
2 marks
Cosine Formula
• The formula as given in the exam is:
b2 + c2 − a2
2bc
and you have to be able to adapt this
to the particular letters used in the
example.
cos A =
Q7 Use the Cosine Rule
C
s=10
l=8
c=12
S
cos C =
s 2 + l 2 − c2
2sl
=
102 + 82 − 1
122
2 × 10
10 × 8
=
100 + 64 − 144
160
=
20
160
L
=
1
8
✓
✓
Substitution
• Be very careful with order of the
letters. Swapping values, e.g. l and c,
will give you the wrong answer.
✓
Calculation
• It is important to show all your
working since the answer, 81 , is given.
3 marks
NOTES: 3·3 page 24.
Page 81
Worked Answers to Practice Exam C: Credit Mathematics
Choice of diagram
• Stem-and-leaf (back to back) would
not be appropriate in this case since
most of the data is single digit values.
Q8
1
1
2
3
Q1 = 2
0
0
1
5
5
6
7
9 11
Q2 = 5 Q3 = 7
1
Q1 = 1
1
3
4
5
8
✓
• To calculate Q1 (Lower quartile), Q2
(median) and Q3 (upper quartile) each
of the data sets has to be ordered least
to greatest.
✓
NOTES: 6·1 page 56
9
Q 2 = 2 Q3 = 5
Boxplots:
company
A
company
B
minutes
9 10 11
Boxplots
• For comparison purposes the two
boxplots are shown on the same scale.
✓
0
1
2
3
4
5
6
7
8
3 marks
• Here is a reminder of how each
statistic is used to construct the
boxplot:
least
Q9 f(
f x) = g(x)
Q2
Q3
greatest
NOTES: 6·1 page 57
⇒ 4 + 3x − x2 = 10 − 2x
✓
⇒ 0 = 10 − 2x − 4 − 3x + x2
⇒ x2 − 5x + 6 = 0
Set up equation
• f(
f x) is replaced by 4 + 3x − x2 and g(x)
is replaced by 10 − 2x.
✓
⇒ (x − 2)(x − 3) = 0
Standard form
• When you recognise a quadratic
equation then rearrange it to the form:
ax2 + bx + c = 0
⇒ x − 2 = 0 or x − 3 = 0
⇒ x = 2 or x = 3
✓
3 marks
Q10
Q1
Solving quadratic equation
• Solving a quadratic equation will
not require “the formula” unless the
question asks for the roots “to 1 dec.
place” or similar.
NOTES: 4·7 page 42
( 18 )2 + ( 6 )2
= 18 + 6 = 24
✓
= 4×6 = 4 × 6
1st simplification
=2× 6 =2 6
• ( 18 )2 = 18 × 18 = 18. In general you
✓
should know: a × a = a.
2 marks
2nd simplification
• 24 is simplified when all “square
factors” have been removed: in this
case 4.
NOTES: 4·6 pages 40 and 41
Q11 x3 × (x −1)−2
= x3 × x−1 × (−2) = x3 × x2
✓
= x3+2 = x5
✓
Laws of indices
• The 1st law used here is (am)n = amn.
2 marks
• The 2nd law used is am × an = am+n.
NOTES: 4·6 pages 41 and 42
Page 82
Worked Answers to Practice Exam C: Credit Mathematics
Q12 Gradient = 24 − 20 = 4 = 1✓
16
16
Gradient
4
• Gradient =
The intercept is (0, 20) ✓
distance up (or down)
distance along
• Take care when reading the scales as
they are not the same on the two axes.
Equation is L = 1W + 20 ✓
4
“ = mx + c”
“y
• The equation follows the pattern:
3 marks
y = mx + c
this is L gradient this is (0, 20) is the
W
intercept
= 21
Q13(a) Let k euros be the cost
of 1 kg of Kenyan and
r euros be the cost of
1 kg of Rwandan
so 2k + 3r = 6·1
NOTES: 5·2 page 48
✓
1st equation
• ‘Algebraic’ refers to using letters.
1 mark
Q13(b) 3k + 2r = 5·9
• It is important to state clearly what
each letter stands for: in this case the
cost of 1 kg.
✓
1 mark
2nd equation
• Notice that there are no units
appearing in the equations.
• When simultaneous equations are
anticipated it is useful to write the
equations in the form: ax + by = c
where a, b and c are the given values,
e.g. 2, 3, 6·1 or 3, 2, 5·9 and x and y
are the variables, in this case k and r.
Q13(c) Solve simultaneous
equations:
2k + 3r = 6 ⋅ 1× 3 6k + 9r = 18·3

3k + 2r = 5 ⋅ 9 × 2 6k + 4r = 11·8✓
Subtract:
5r = 6·5
⇒
r = 1·3 ✓✓
Substitute r = 1·3 in
2k + 3r = 6·1
Method
• The ‘simultaneous equation’ strategy
will be rewarded with a ‘strategy mark’.
⇒ 2k + 3 × 1·3 = 6·1
1st variable
• A further mark is awarded for the
correct calculation of one of the
variables, either k or r.
⇒ 2k + 3·9 = 6·1
⇒ 2k = 2·2 ⇒ k = 1·1 ✓
• In the working shown ‘k’ has been
eliminated. ‘r’ could have been
eliminated by multiplying the
top equation by 2 and the bottom
equation by 3 and then subtracting.
So 1 kg of Kenyan costs
1·10 euros and 1kg of
Rwandan costs 1·30 euros
So 4 kg of Kenyan and
1 kg of Rwandan costs
4 × 1·10 + 1 × 1·30
= 5·70 euros
5 kg of Blend C costs
5·70 euros
✓
2nd variable
• Either of the two equations can be
chosen for substitution of the 1st
value to calculate the 2nd value.
Answer
• The question did not ask for just the
values of k and r!
4 marks
NOTES: 4·4 page 36
Page 83
Worked Answers to Practice Exam C: Credit Mathematics
Practice Exam C: Paper 2 Worked Answers
05 × 1
10
Q1 No. of letters = 1⋅ 05
−8
−1
3×1
10
Strategy
• This is a division, it is sometimes useful
to try to solve a simpler similar problem
to see how the solution is found:
• If each letter were 2 mm wide and the
line was 6 mm long there would be 3
letters: 6 divided by 2 gives 3.
✓
= 35 00000
= 3·5 × 106
✓
2 marks
Calculation
• 1·05 × 10−1 is entered in the calculator as
follows: 1 . 0 5 × 10x (−) 1
• On your calculator, you may have EXP
not × 10x
NOTES: 1·4 page 10
Strategy
• The ‘original’ price has to be found. It
will be 100%. This is now a ‘proportion’
problem: find 1% then find 100%.
Q2 It is now 106 1% of the
2
original cost so:
106·5%
£87·82
1%
£87
7 ⋅8
82
106 ⋅ 5
✓
£87
7 ⋅8
82
× 100
106 ⋅ 5
100%
Calculation
• It is sometimes best not to calculate
intermediate answers, e.g. £87·82 ÷
106·5, but build up the calculation on
paper and then do the calculation at
the end. The solution does just that.
✓
= £82·46 ✓
NOTES: 1·5 page 12
3 marks
Substitution
• Care should be taken over negative
values.
Q3 3x2 − x − 5 = 0
Compare ax2 + bx + c = 0
⇒ a = 3, b = −1, c = −5
• It is safer to state the values of a,
b and c that you are going to use.
Correct substitution into the formula
will gain the 1st mark.
2
x = − b ± b − 4ac
2a
=
1 ± (−1)2 − 4 × 3 × (−5)
2 ×3
✓
=
1 ± 1 + 60
6
✓
so x =
1+ 6
61
6
=
1± 6
61
6
or x =
Simplification
• Before ‘going decimal’ it is clearer,
and avoids mistakes, to separate the
two solutions (roots) as is done in the
3rd last line.
Calculation
• Giving a clear indication of each decimal
before it is rounded will allow you to
gain the last mark for rounding even if
your answers are wrong before rounding.
1− 6
61
6
x = 1·468… or x = −1·135... ✓
x =.. 1·5 or x =.. −1·1
✓
Rounding
• Always follow precisely any rounding
requests.
4 marks
NOTES: 4·7 page 43
Page 84
Worked Answers to Practice Exam C: Credit Mathematics
Strategy
• You need to calculate the three angles
in triangle DEL and then use the Sine
Rule.
Q4
N
35°
E
N
F
158°
35°
D 118°
L
L
angle EDL
= 118°− 35°
= 83°
✓
✓
1st angle
• Remember bearings are always
measured clockwise from the North line.
angle FEL
= 158°− 35°
= 123°
2nd & 3nd angles
• Extending the line DE (EF in the
diagram) allows you to move the
information at angle D up to angle E
(35° is a corresponding angle).
So angle DEL
= 180° − 123° = 57°
angle ELD = 180°− (83°+57°)
= 180°− 140° = 40°
✓
D 83°
NOTES: 3·2 page 21
Now use the Sine
Rule
E
l=343
57°
d
sinD
d
40°
⇒
L
⇒d=
=
d
sin83
343 × sin 83
sin 40
The Sine Rule
• Substitution of the values into the
Sine Rule will gain you a mark.
l
sinL
=
343
sin 40
✓
Calculation
• Multiply both sides of the equation
by sin 83°.
= 529·63…
• Any correct rounding is acceptable.
✓
=.. 530
The distance from
Edinburgh to London is
approximately 530 km to the
nearest km.
NOTES: 3·3 pages 22 and 23
Strategy
• Removal of the smaller circle area from
the area of the larger outer circle is the
crucial idea for the strategy mark.
5 marks
Substitution
• The given diameter measurements
should be halved to give the radius.
This is necessary as the area formula
A = πr2 uses the radius value and not
the diameter value.
Q5(a) Use A = πr2.
Outside circle diameter
= 30 cm ⇒ radius = 15 cm
Inside circle diameter
= 25 cm ⇒ radius
= 12·5 cm
Calculation
• The use of common factor and
difference of squares is not expected—
but it is fun!
Area of = Outside − Inside
plastic
circle area circle area ✓
= π(152 − 12·52)
• The alternative is to enter π × 152 − π
× 12·52 as written straight into your
calculator.
= π(15 − 12·5)(15 + 12·5)
NOTES: 2·4 page 17
= π × 152 − π × 12·52
✓
= π × 2·5 × 27·5
= 215·98…
=.. 216 cm2
✓
(to the nearest 1 cm2)
3 marks
Page 85
Worked Answers to Practice Exam C: Credit Mathematics
Q5(b)
Volume = area of end × length
Volume
• This is a typical “prism formula”:
Volume = area of the end × length.
= 215·98... × 100
= 21598·44... cm
3
• Note that 216 cm2 is not used. Do not
use rounded answers from previous
parts of a question in subsequent
parts of the question.
✓
Weight = 21598·44... × 0⋅12
Weight
• Rounding is not crucial and
conversion to kg not essential.
= 2591⋅81... g
= 2⋅591... kg
=.. 2⋅6 kg
(to 2 sig figs)
NOTES: 2·1 page 14
✓
2 marks
Strategy
• Triangle ABC is right-angled and
therefore Pythagoras’ Theorem can
be used.
Q6(a)
C
B
2·3
A
1·2
The radius of
the fuselage is
2·3 m
Pythagoras’ Theorem
• Notice that CB is a radius and that CA
is part of a radius. Any radius in this
circle has a length 1 × 4·6 = 2·3 m.
ΑC = 2·3 − 1·2
= 1·1 m (see
diagram)
✓
2
Calculation
• AB is one of the smaller sides in
triangle ABC so the calculation
involves a subtraction.
Use Pythagoras’ Theorem in
triangle ABC:
✓
C
AB2 = BC2 − AC2
2·3m
B
1·1m
A
• Remember that rounding off decimal
answers should only be done at the
end of the problem when the final
answer is given.
= 2·32 − 1·12
= 4·08
Solution
• By symmetry AB is half the width of
the compartment floor.
so AB = 4 ⋅ 0
08 = 2·019…✓
By symmetry: width of floor
= 2 × AB
so x = 2 × 2·019... = 4·039...✓
⇒ x =.. 4·04 metres.
4 marks
Solution
• Two equal and parallel
chords in a circle will be
the same distance from
the centre because the
diagram is symmetric.
Q6(b) By symmetry (see diagram)
the ceiling is as far
above C as the floor
1·1
C is below C.
1·1
NOTES: 2·4 page 19
Height = 2 × 1·1
= 2·2 metres ✓
1 mark
Page 86
C
Worked Answers to Practice Exam C: Credit Mathematics
Q7 9 parts : 2 parts
(honey)
(cocoa)
Strategy
• In a question like this you will know
that some of the ingredients will be
left over when the mixing is complete.
This is because, in this case, 49 21 : 12
is not the same as 9 : 2. The strategy
is to determine, with working, which
ingredient will be left over.
Try to use all the cocoa:
So 2 parts cocoa weighs
12 kg
So 1 part weighs 6 kg
• Alternative working here might
be that 9 : 2 is the same as 54 : 12
(multiplying by 6) and comparing
49 21 : 12
So 9 parts honey weighs
9 × 6 = 54 kg
There is not enough
honey.
✓
Calculation
• Calculation of the amount of cocoa
required, 11 kg, to go with 49 21 kg
honey will gain you this mark.
Try to use all the honey:
So 9 parts honey weighs
49·5 kg
Answer
• Round down not up—you need
complete jars!
So 1 part weighs
49·5 ÷ 9 = 5·5 kg
NOTES: 1·5 page 12
So 2 parts cocoa weighs
2 × 5·5 = 11 kg
✓
leaving 1 kg cocoa unused.
Strategy
• At each of the points of intersection
the y-coordinate is 0·2 and the
x-coordinate satisfies cos x° = 0·2.
The appearance of the equation
cos x° = 0·2 and an attempt to solve it
will gain you this mark.
Total weight used
= 49·5 + 11 = 60·5 kg
So 60 1–kg jars can be
made.
✓
3 marks
1st quadrant angle
• Using cos−1 on your calculator will
always give you the 1st quadrant angle
(don’t use this key with a negative
value).
Q8 For the points of intersection
solve: cos x° = 0·2
✓
x° is in the 1st or 4th quadrants
1st quadrant angle is given by:
x = cos−1(0·2) = 78·46... =.. 78·5 ✓
Other angles
• You should identify the 4 quadrants on
the graph that is given:
4th quadrant angle is given by:
x = 360 − 78·46... = 281·54...
=.. 281·5
y
A
y 1st 2nd 3rd 4th 1st 2nd 3rd 4th
+360 +360 +360 +360
x
B
x
78·5
281·5
It should then be clear that the
x-coordinate of B is given by the 4th
quadrant value + 360°.
281·5+360
The x-coordinate of A is 78·5
• The 4th quadrant value
is obtained using this
diagram:
From the graph above
B lies 1 cycle on from the 4th
quadrant intersection
The x-coordinate of B is
281⋅5 + 360 = 641⋅5
✓
3 marks
NOTES: 3·5 page 27
Page 87
180º – 1st
quad
180º + 1st
360º – 1st
quad
quad
Worked Answers to Practice Exam C: Credit Mathematics
Q9(a) For 2 hours:
Calculation
• Both values correct will gain you this
mark.
Earthmove: 64 + 2 × 30 = £124
(delivery) (2 hours)
Trenchers: 28 + 2 × 34 = £96
(delivery) (2 hours)
✓
Comparison
• The two calculations have to be clearly
shown.
1 mark
• A comparison should be made between
the two costs, e.g. “£44 cheaper” or
“£664 is less than £708” etc.
Q9(b) For 20 hours:
Earthmove: 64 + 20 × 30
= £664
Trenchers: 28 + 20 × 34
= £708
1st formula
• Either one of the two expressions
correctly stated will gain you the 1st
mark here.
✓
So Earthmove is cheaper in
this case by £44
1 mark
2nd formula
• Two correct expressions: two marks
gained.
Q9(c) For n hours:
Earthmove: 64 + n × 30
= £30n + 64
✓
Trenchers: 28 + n × 34
= £34n + 28
✓
Strategy
• From parts (a) & (b) you should have
realised that at some point Trenchers
becomes the more expensive option
having been cheaper when the number
of hours was low.
2 marks
• Using the formulae you worked out
in (c) and setting them equal to each
other would gain you this mark.
Q9(d) For a 2 hour hire
Trenchers are cheaper
but not for a 20 hour hire.
To find the ‘cross-over
value’:
NOTES: 4·4 page 34.
Solution & interpretation
• What does n = 9 mean in the context of
the question? A statement is necessary.
Solve 34n + 28 = 30n + 64
⇒ 4n = 36 ⇒ n = 9
✓
Trenchers and
Earthmove cost the same
for a 9 hour hire.
Variation Formula
• For direct variation the variable letter,
L, goes on the top of the fraction, for
inverse variation the letter, T, goes on
the bottom.
Any hire less than 9
hours is cheaper with
Trenchers.
✓
• In general if A varies directly as B
then A = k × B. If A varies inversely
as B:
then A = k × B1 where k is a constant.
2 marks
Q10(a) D ∝ L2
T
so D = k ×
L
T2
✓
where k is a constant.
1 mark
Page 88
Worked Answers to Practice Exam C: Credit Mathematics
Q10(b) Replace L by 4L
and T by 2T:
Strategy
• The method here is to take the formula
for D and replace the right-hand side
of the formula by a new expression: one
where each of the letters is replaced by
the results of the changes described in
the question. This new expression is
then simplified and compared with the
original expression for D.
So D = k × L2
T
becomes:
k × 4L
(2T
2T)2
=2×
4
=
1
2
=
✓
k× 2 L
4T2
k L
T2
Answer
• The result of the simplification should
be described—D is now multiplied by 1.
D
2
NOTES: 4·8 pages 44 and 45
So the Diameter, D, is
halved.
✓
Pattern extension
• The order of writing the line cannot be
altered.
2 marks
Q11(a)
4th line: 8 + 1 = 4 × 6 − 5 × 3✓
1st expression
• For finding one of 2n + 1 or n(n + 2) or
(n + 1)(n − 1) you will gain the 1st mark
here.
1 mark
Q11(b)
1st line: 2 + 1 = 1 × 3
− 2×0
2nd line: 4 + 1 = 2 × 4
3rd line: 6 + 1 = 3 × 5
4th line: 8 + 1 = 4 × 6
− 3×1
Remaining expressions
• Careful use of brackets is essential:
n × n + 2 (= n2 + 2) is not the same as
n(n + 2).
n + 1 × n − 1 (= 2n − 1) is
not the same as (n + 1)(n − 1).
− 4×2
− 5×3
• Always check: e.g. substituting n = 3
should produce the 3rd line.
✓
nth line: 2n + 1 = n(n + 2) − (n + 1) (n − 1)
✓
Simplification
• Again great care should be taken with
brackets, especially expanding
(n + 1)(n − 1) because − (n2 − n + n − 1) is
not the same as −n2 − n + n − 1.
2 marks
Q11(c) Simplify the right-hand
side of the nth line:
• Proving 2n + 1 = n(n + 2) − (n + 1)(n − 1)
using ‘algebra’ means that all lines are
true for instance n = 10 gives:
n(n + 2) − (n + 1)(n − 1)
= n2 + 2n − (n2 − n + n − 1)
20 + 1 = 10 × 12 − 11 × 9 which is line
10.
= n2 + 2n − n2 + n − n + 1
= 2n + 1
NOTES: 4·2 page 31
✓
This is the same as the
left-hand side of the nth
line.
So the pattern always
holds.
1 mark
Page 89
Worked Answers to Practice Exam C: Credit Mathematics
Strategy
• Similar triangles calculation to find DE
followed by Pythagoras’ Theorem in
triangle DEH.
Q12 Triangles ABC and ADE are
similar (equiangular)
✓
AE = AC + CE
A
A
3
2·7
C
=5m
3
is
5
3
D
E
2·5
H
• You should always ‘disentangle’ the
two similar triangles, drawing them
separately and writing in each of the
lengths that you know.
5
3
NOTES: 2·2 page 15
so DE = × 2·7 = 4·5 m
4·5
Similar triangles
• Enlargements have scale factor greater
than 1.
The
enlargement
scale factor
E
×5
=3+2
D
B
5
✓
Pythagoras’ Theorem
• Finding a smaller side involves a
subtraction.
Use Pythagoras’
Theorem in triangle
DEH:
✓
Calculation
• Remember to use x2 key in this
calculation.
EH2 = ED2 − DH2
= 4·52 − 2·52
= 14
Answer
• Always check that your final answer
seems reasonable in the context of
the question. In this case 3·74 m
compares well to just over 5 m for the
‘king post’ so seems reasonable.
✓
so GF = EH = 14 = 3·741…
The gap between the vertical ✓
posts is 3·74 m (to 3 sig figs).
5 marks
NOTES: 2·3 page 16
Page 90
Practice Exam D: Paper 1 Worked Answers
Q1
35% of £351
10% ↔ £35.10
30% ↔ £105.30
and 5% ↔ £17.55
so 35% ↔ £122.85
Answer is £122.85
Strategy
•You should be familiar with common
% equivalents
1
e.g., 25% = , 50% = 1 , 75% = 3 ,
4
2
4
1
1
10% = , 20% = etc.
10
5
These are useful for non-calculator
papers.
Calculations
Multiply by 3 to get 30%
Divide 10% by 2 to get 5%
Add 30% and 5% to get 35%
3
3
2 marks
NOTES: 1·5 page 11
Order of operations
• Bracket calculations should be done
first.
Q2
4·25 ÷ (4·3 + 0·7)
= 4·25 ÷ 5·0
= 0·85
• You may be familiar with BODMAS
Division
Brackets → Of → M ultiplications
A ddition
→ S ubtraction
This gives the order of precedence
with performing numerical
calculations.
3
3
2 marks
Calculations
0⋅ 8 5
4⋅25
4 2
• Notice
gives the working 5 4 ⋅ 2 5
5⋅0
)
Order of operations
• Note: of changes to ×
Q3
2
1 3
1 − of 1
3
2 5
1  3 2
= 1 −  ×1 
2  5 3
• Multiplication is performed before
subtraction.
Multiplication
2
2
2
5
3
• Note 1 is 1 and i.e. and so
3
3
3
3
3
• The rule is a × c = a × c i.e. multiply
b d b×d
the numerators and multiply the
denominators.
3
1  3 5
=1 − × 
2  5 3
1 15
=1 −
2 15
1
1
= 1 −1=
2
2
NOTES: 1·2 page 8
3
Subtraction
3
• Alternative is
3 2 1
− = since 1 1 = 3
2 2 2
2 2
2
and 1 = .
2
NOTES: 1·2 page 9
3 marks
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Worked Answers to Practice Exam D, Credit Mathematics
Q4
Method 1 (Plotting)
From the graph:
gradient = 3
3
y-intercept is −2. 3
Equation is
y = mx + c
with m = 3 and
c = −2
3
i.e. y = 3x − 2.
Gradient
• After plotting the points this can be
read off the graph: 1 along 3 up is
3
= 3 for the gradient.
1
NOTES: 5·1 page 46
y
0
-2
-intercept
• This is where the graph crosses the
y-axis. In this case the value is −2.
x
Equation
• General form is y = mx + c
3 marks
gradient y-intercept
NOTES: 5·2 pages 47, 48
Gradient & -intercept
• The substitution method uses the fact
that a straight line graph has equation
of the form y = mx + c and that the
values of the x and y-coordinates
of any point on this line satisfy the
equation. So (0,−2) lies on the line
which means that x = 0 and y = −2
satisfy y = mx + c. These values are
substituted for x and y to give a true
statement.
Q4
Method 2 (Substitution)
Suppose the equation is
y = mx + c
Use point (0, −2): x = 0, y = −2
Substitute in y = mx + c
⇒ −2 = m × 0 + c ⇒ c = −2 3
The equation now is y = mx − 2
Use point (1,1): x = 1, y = 1
Substitute in y = mx − 2
⇒1=m×1−2⇒1=m−2
⇒ m = 3
3
So the equation is y = 3x − 2. 3
• In general 2 points are used and in
many cases simultaneous equations
would then be needed to find m and c.
Equation
• Substitute m = 3 and c = −2 in
y = mx + c.
3 marks
Fraction
• Each term should be multiplied by k.
Q5
2
3 + = 1 (×k)
k
⇒ 3k + 2 = k
⇒ 3k − k = −2
⇒ 2k = −2
⇒ k = −1
2
• × k = 2, the ‘k’s cancel.
k
3
Like terms
• Subtract k from both sides and
subtract 2 from both sides.
3
3
• Alternative method: subtract 3 from
2
both sides giving = −2 then
k
3 marks
multiply both sides by k.
Solution
• Divide both sides by 2: 2k = −2
2
2
NOTES: 4.4 page 34
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Worked Answers to Practice Exam D, Credit Mathematics
Q6
All possible outcomes are shown
in this table:
2nd Spinner
1
2
3
4
5
1st 1 (1,1) (1,2) (1,3) (1,4) (1,5)
S
P 2 (2,1) (2,2) (2,3) (2,4) (2,5)
I
N 3 (3,1) (3,2) (3,3) (3,4) (3,5)
N 4 (4,1) (4,2) (4,3) (4,4) (4,5)
E
R 5 (5,1) (5,2) (5,3) (5,4) (5,5)
The favourable outcomes (total 5
or less) are shown shaded in the
table.
Outcomes
• This mark is for stating either that
there are 10 favourable outcomes or
that there are 25 possible outcomes in
total.
• You should take care to be systematic
in your listing of outcomes. It is very
easy to miss a possibility and lose half
the marks!
Probability
• Probability
No. of favourable outcomes
=
mes
Total No. of outcom
There are 10 favourable
outcomes out of 25 possible
outcomes.
3
10 2
= .3
Required probability =
25 5
• Cancelling down is not essential to
gain full marks.
NOTES: 6·3 page 54
2 marks
Q7(a)
Card 4: 3 × 4 − 4 × 0
Correct term
• To gain this mark you must be careful
to follow the exact pattern. Writing
any of the following down will not
gain this mark:
3
1 mark
12, 12 − 0, 3 × 4 − 0 etc.
Correct expression
• Use of brackets is crucial:
n − 1 × n − n × n − 4 is wrong
• You should check by substituting,
e.g. n = 3 to make sure you get the
pattern on card 3.
Q7(b)
Card n: (n − 1) ×
n − n × (n − 4) 3
This gives n (n − 1) − n (n − 4)
= n2 − n − n2 + 4n 3
= 3n
3
Removing brackets
• Careful with the negative signs:
−n (n − 4) =
−n × n − n × (−4)
= − n2 + 4n
(negative × negative gives positive)
3 marks
Simplify
• n2 terms cancel.
• Note −n + 4n is same as 4n − n.
NOTES: 4·2 page 30
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Worked Answers to Practice Exam D, Credit Mathematics
Start table
• Note the headings of the table.
Important for clarity.
• Each new frequency number is added
to the running total as you go down
the last column.
Q8
Cumulative frequency table:
Time
Cumulative
spent Frequency
frequency
(min)
3
 0
 1
 1
10
 7
 8
20
11
19
30
22
41
40
19
60
50
 8
68
60
 2
70
3
Complete table
• You should always check that the total
of all the frequency numbers (middle
column) is the same as the last entry
(in this case 70) in the cumulative
frequency column.
Median
• If all the data were put in increasing
order the median is the middle number
(or the mean of the middle two).
• There are 70 numbers to put in
increasing order: 1 ‘0’ followed by
7 ‘10’s’ followed by 11 ‘20’s’ and so on:
0 10 10 10 ···· ···· 50 50 60 60
70 numbers
Now think of these split into two
groups of 35:
0 10 10 10····30 30 30 30····50 50 60 60
Placed in increasing order of
time spent, the median would
be the mean of the 35th and 36th
pupils’ time spent.
From the table, both the pupils
are in the ‘Time spent: 30 min’
category.
So the Median time is
30 minutes.
median
NOTES: 6·2 page 59
3
3 marks
Q9
10
By Pythagoras’
(radius)
Theorem
x2 = 102 − 82
8
= 100 − 64
= 36
⇒ x = 36 = 6 feet
x
The distance
from the centre
to the ceiling is
14 − 6 = 8 feet.
By Pythagoras’ Theorem
y2 = 102 − 82
y y
⇒ y = 6 feet
(same
8
10
calculation
(radius)
as above)
so, by symmetry,
ceiling = 2 × 6 = 12 feet.
Strategy
• Evidence that you know to use a rightangled triangle (i.e. drawing a triangle)
is enough to gain you this 1st mark.
Calculations
• Correct use of Pythagoras’ Theorem
gains this mark.
• Remember, all radii in this circle are
the same length, i.e. 10 feet.
3
8
Height difference
• At this stage you should use the
information that between the floor and
the ceiling is 14 feet.
• Note that the centre of the circle is
not halfway between the floor and the
1
ceiling. If you use × 14 = 7 you will
2
lose the 1st three marks!
3
8
6
14
3
Further calculation
• This right-angled triangle is the
same size as the previous triangle but
rotated 90° so this calculation has
already been done!
3
Solution
• Multiplying by 2 produces the width
• Doubling any wrong answer obtained
previously would gain you this mark.
3
5 marks
NOTES: 2·4 page 19
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Worked Answers to Practice Exam D, Credit Mathematics
Substitution
• Replacement of A by 27 gains this
mark.
Q10(a)
V = 4 3 + 3 A with A = 27
V = 4 3 + 3 27
• It is important to write all steps down
so that the marker can understand
your thinking.
3
= 4 3 +3 9×3
Simplification
• This mark is for changing 3 27 to 9 3
= 4 3 +3×3× 3
= 4 3+9 3
3
= 13 3
3
• When simplifying surds you should
look for square number factors (e.g. 4,
9, 16, 25 etc.): in this case the factor 9.
Solution
• Compare 4x + 9x = 13x: it is the same
process.
3 marks
NOTES: 4·6 page 40
Equation
• Replacing V by 5 3 gains this mark.
Q10(b)
V = 4 3 + 3 A with V = 5 3
⇒5 3 = 4 3 +3 A
Stating solving
• You should be aware that ‘like terms’
5 3 and 4 3 should be gathered
on one side of the equation and then
simplified.
(
3
⇒5 3−4 3 =3 A
⇒ 3=3 A
⇒
• Compare 5x − 4x = x.
3
( 3 ) = (3 A )
2
2
Solution
• The same action should be performed
to both sides of an equation to retain
the balance.
⇒ 3 = 9A
⇒
1
3
= A ⇒A=
3
9
)
(
3
(3 A )
2
)
2
• 3 A = 3 A × 3 A = 3 × A × 3 × A = 3 × 3 × A × A
= 3 A × 3 A = 3 × A × 3 × A = 3 × 3 × A × A
3 marks
= 9 × A = 9A
NOTES: 4·6 page 40, 41
Q11(a)
Fixed Fee = £51
Strategy
• The strategy is a ‘subtraction’ then a
division. You should be clear in your
working that you understood this.
Money available for printing
= £100 − £51 = £49 = 4900p
Calculation
• You have to take care over units. Use
49
4900
pence in both:
or pounds:
0⋅7
70
but don’t mix pounds and pence.
No. of ‘25 leaflets’ he can afford 3
4900
=
= 70
70
Total No. of leaflets = 70 × 25
= 1750
3
• Deal in batches of 25 leaflets but
remember that leaflets are required.
70 is not the correct answer!
2 marks
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Worked Answers to Practice Exam D, Credit Mathematics
Expression for money available
• You should mirror your numerical
calculation in part (a) above but now
using the given letters.
Q11(b)
Fixed fee = £A
• Changing £49 to 4900p involved
multiplying by 100 hence 100(B − A).
Money available for printing
= £B − £A = £(B − A)
= 100(B − A)p
3
Expression for ‘25 leaflet’ number
B−A
• Note that
would be equivalent
c
49
to
in part (a) which mixes units
70
(£ and p) and so is wrong.
No. of ‘25 leaflets’ he can afford
100(B − A)
× 25
c
Total No. of leaflets
=
T =
3
100(B − A)
× 25
c
So T =
2500(B − A)
c
Formula
• You should simplify if possible.
• “T = ” is important since you are
2500(B − A)
asked for a formula.
is
c
not a formula.
3
3 marks
Area
• Obtaining x(3x + 2) gains this mark.
Q12
Area = length × breadth
So l = (3x + 2) × x
⇒ x(3x + 2) = 1
Equation
• This mark is gained by setting up the
equation: Area expression = 1.
3
3
Standard form
• Once a quadratic equation is
recognised you should rearrange the
terms into the standard form:
ax2 + bx + c = 0, i.e. x2 term, followed
by ‘x’ term then the constant.
⇒ 3x2 + 2x = 1
⇒ 3x2 + 2x − 1 = 0
3
⇒ (3x − 1)(x + 1) = 0
3
⇒ 3x − 1 = 0 or x + 1 = 0
Factorising
• Having factorised the quadratic
expression you should multiply
the brackets out to check that the
factorisation is correct:
⇒ 3x = 1 or x = −1
1
⇒ x = or x = −1
3
x = −1 is not a valid solution as
lengths are positive
1
So x = is the only solution. 3
3
(3x − 1)(x + 1) =
3x2 + 3x − x − 1
= 3x2 + 2x − 1
Solution
• In many cases a mark will be assigned
for rejecting an invalid solution like
x = −1 and so this statement should
be clear.
5 marks
NOTES: 4·7 page 42
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Practice Exam D: Paper 2 Worked Answers
Q1
Mean
• Recall
that the mean is calculated
10 + 11 + 12 + 12 + 8 + 10 + 14
from:
Σx
=
7
n
total of all the data (numbers)
x 1210
+ 11
Mean =
Σx 7710 + 11Σ+
10++12
14+ 8 + 10 + 14
+ 12
+ 8++12
11
x
⇒
=
=
hours
Mean
=
x
=
=
number of pieces of data
Mean = x =
=
7
n
7
n 7
• Calculators may have a ‘Statistical
77
77
=
= 11 hours
3
11xhours
⇒x =
=⇒
Mode’ where numbers are entered and
7
7
various statistics are then calculated.
Mean = x =
This is a useful checking facility but
it is not a substitute for showing the
steps of your calculations – otherwise
you may lose marks.
Standard Deviation (Two methods)
Method 1:
Method 2:
x
10
11
12
12
8
10
14
x−x̄ (x−x̄)2
−1
1
0
0
1
1
1
1
−3
9
−1
1
3
9
Σx =
77
Σ(x−x̄)2 = 22
s=
=
Σ(x − x )2
n −1
22
= 1⋅914
6
.
=. 1·9 hours
(to 2 sig figs)
x
10
11
12
12
8
10
14
x2
100
121
144
144
64
100
196
Starting st.d. formula
• There are two formulae given on your
formulae page in the exam. You should
use the one you are most familiar with.
Both formulae are illustrated in the
solutions.
• Producing 22 (or 869) will gain you
this mark.
Σx =
77
Σx2 = 869
s=
3
Σx2 − (Σx)2 / n
n −1
2
=
869 − 77
73
Calculations
22
• Obtaining
gains this further
6
mark
• Notice in ‘Method 2’ both n and n − 1
are used, i.e. 7 and 6. This can be the
cause of some confusion.
Value
• A standard deviation does have a
unit attached (in this case: hours)
though you will not be penalised for
omitting it.
6
22
6
.
=. 1·9 hours 3
(to 2 sig figs)
=
• Any correct rounding is acceptable.
NOTES: 6·2 page 60, 61
4 marks
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Worked Answers to Practice Exam D, Credit Mathematics
Strategy
• Knowing to use the area formula gains
you this 1st mark.
Q2
1
Area = ab sinC
2
with a = 1, b = 1·62
A
• You should also be familiar with
different forms of this formula, e.g.
1
1
1
bc sin A , ac sinB or pq sinR
2
2
2
in ΔPQR etc.
3
1·62
and ∠C = 72°
3
Correct version of formula
72°
B
• In this case Area =
C
1
1
ab sinC is the
2
appropriate form to use.
1
× 1 × 1⋅62 × sin 72° 3
2
= 0 ⋅ 7703
 0 ⋅ 770
3

(to 3 sig figs)
So Area =
Substitution
• Correct replacement of the letters with
the given values will gain you this mark.
• Do not miss this step out – you lose
this mark if your final answer is wrong
and this step is not clearly written out.
4 marks
Value
• Any correct rounding is acceptable
here.
NOTES: 3·3 page 22
Substitution
• This first mark is obtained for replacing
the letters by their given values.
Q3
a=
V2
r
• It is important to write
4 × 108
down in your solution. If it is not
on the page and your final answer is
wrong you will not gain the first mark.
With V = 1·2 × 103 and
r = 4 × 108
So a =
(or a =
(1⋅2 × 103 )2
4 × 108
(1 ⋅ 2 × 103 )2
= 3⋅6 × 10 −3
Brackets
• Dealing with 1·2 × 103 squared gains
this mark.
3
3
3
12002
1440000
=
400000000 400000000
Calculation
• It is recommended that these standard
form numbers are entered into your
calculator using the EXP button.
e.g. 1·2 EXP 3
= 0 ⋅ 0036 = 3⋅6 × 10 −3 )
3 marks
NOTES: 1·4 page 10
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Worked Answers to Practice Exam D, Credit Mathematics
Strategy
• Pythagoras’ Theorem states:
given a right2
2
2
c
a angled triangle ⇒ a + b = c
Q4
b
A
AB = 5·5 cm
BC = 13·2 cm
B
C
This is equivalent to saying that if
a2 + b2 ≠ c2 then the original triangle
cannot be right-angled. So you are
checking AB2 + BC2 and AC2 to see if
they are equal or not.
AC = 14·4 cm
• The cosine rule could be used to
calculate angle ABC.
AB2 + BC2 = 5·52 + 13·22 = 204·49
3
also AC2 = 207·36
3
So AB2 + BC2 ≠ AC2 and
Pythagoras’ Theorem
3
conditions are not met.
ΔABC is not right-angled.
Calculation
• This mark is for correct calculation of
the values 204·49 and 207·36.
• A pproximations are not appropriate
here.
3 marks
Conclusion
• A clear statement ‘AB2 + BC2 ≠ AC2’ is
required.
NOTES: 2·3 page 16
3
Strategy
• For this mark there should be evidence
in your solution that you knew to use
similar triangles.
2m C
Scale factor or equation
• All lengths in the large triangle are
3 times the lengths in the small triangle.
Equivalently small triangle lengths are
one third those in the large triangle.
Q5
Here are two similar triangles:
A
D
4m
B
6m
C
E
BC = half length of attic floor
=
1
×
=6
pq12
sinR
2
m
• A ratio method is also valid:
DE 2 Since ratios of corresponding
=
4
6 sides are equal.
The triangles are similar as
corresponding angles are equal.
Method
Reduction scale factor from
EC 2 1 3
ΔABC to ΔDEC is
= =
BC 6 3
1
4
1
1
⇒ DE = × AB = × 4 = = 1 3
3
3
3
3
1
×4
3
appearing or for DE being isolated in
2
the equation: i.e. DE = × 4
6
• This mark is for either
Solution
• This is for the final calculation.
The small supports have length
1
3
1 m
3
• 1·3 m or 1·33 m would also be acceptable.
NOTES: 2·2 page 15
4 marks
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Worked Answers to Practice Exam D, Credit Mathematics
Coefficients
• It is essential to rearrange this
quadratic equation into standard form:
ax2 + bx + c = 0
• Take care over negatives. The x-term
−6x has coefficient −6 not 6, so b = −6
not b = 6.
Q6
2x2 + 3 = 6x
⇒ 2x2 − 6x + 3 = 0
Substitution and calculation
• The ‘Quadratic Formula’ is given to
you on the formula page in your exam.
• It is important that you write out the
formula and state the values of a, b
and c.
• Take care over negatives (−6)2 = 36:
positive. Also −b when b = −6 gives
−(−6) = 6: positive.
Compare ax + bx + c = 0
2
⇒ a = 2, b = −6 and c = 3
3
−b ± b2 − 4ac
So x =
2a
=
−( −6) ± ( −6)2 − 4 × 2 × 3
2×2
Values
6 ± 36 − 24
=
4
6 ± 12
3
=
4
6 + 12
6 − 12
So x =
or x =
4
4
= 2 ⋅366 or 0 ⋅633
3
 2 ⋅ 4 or 0 ⋅6

(correct to 1 decimal place)
6 ± 12
giving
4
the two possibilities i.e. 6 + 12
4
6 − 12
and
before attempting the
4
calculations on your calculator.
• In this case rounding to the correct
accuracy is important. If the question
states the accuracy, e.g. 1 decimal
place, then you will lose the mark if
you do not follow this instruction.
• It is useful to rewrite
3 marks
NOTES: 4·7 page 43
Expression
• Any cuboid has three sets of two
identical rectangles for faces:
Q7(a)
Top and bottom:
Here is the net of the cuboid:
(back)
4x cm × x cm
(end)
(bottom)
(end)
x cm × x cm 4x cm × x cm x cm × x cm
(front)
4x cm × x cm
(top)
4x cm × x cm
So total surface area
3
= 4 × (4x × x) + 2 × (x × x)
= 16x2 + 2x2
= 18x2 cm2
1 mark
h
l
b Area
=l×b
b
l
The two ends:
h
b
Front and back:
l
Area
=b×h
h Area
=l×h
So total surface area.
=2×l×b+2×b×h+2×l×h
In the example in this question the
‘Top and bottom’ rectangles are
identical to the ‘Front and back’
rectangles.
• All your working is important since
the expression 18x2 is given in the
question.
NOTES: 4·1 page 29
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Worked Answers to Practice Exam D, Credit Mathematics
Volume expression
• This mark is for writing down 4x3.
A common mistake is to write 4x2 so
be careful with the index 3.
Q7(b)
Volume = length × breadth × height
= 4x × x × x
= 4x3 cm3
3
Equation
• Ignoring units, you will gain this mark
for equating the area expression from
part (a) and the volume expression.
3
x Value
• An alternative to factorising would be
to divide both sides by x2 giving
4x = 18 which can then be solved. The
assumption here is that you are not
dividing by zero. There are situations
(not this question) where x = 0 would
be a valid solution so care must be
taken.
3
So 4x3 = 18x2
⇒ 4x3 − 18x2 = 0
⇒ 2x (2x − 9) = 0
2
⇒ 2x2 = 0 or 2x − 9 = 0
⇒ x = 0 or 2x = 9
9
⇒ x = 0 or x = = 4 ⋅5
2
(x = 0 is not a valid solution)
length = 4 × 4·5 = 18
Dimensions are
18 cm × 4·5 cm × 4·5 cm
Dimensions
• The dimensions are
4x cm × x cm × x cm so x is replaced
by 4·5 giving the answer.
3
4 marks
NOTES: 4·4 page 34, 4·7 page 42
Angles of triangle
• No further calculation can be done
without the angles of ΔPQR being
calculated first.
Q8
N
P
N
120°
114°
54°
N
60°
7
36°
R
30°
3
• The right-angles appear since
‘R lies due west of rock Q’ and so RQ
is perpendicular to the North lines.
120°
• Extending line PQ allows the 120°
angle to appear at Q, a crucial step in
finding ∠PQR.
Q
The diagram above shows that in
ΔPQR, ∠R = 36° and ∠Q = 30°
Use of sine rule
• Just writing down the sine rule will
gain this mark.
so ∠P = 180° − (36° + 30°)
= 180° − 66° = 114°
Substitution
• This mark is for replacing the letters
in the Sine rule with the appropriate
values.
Now use the sine rule:
p
r
p
73
=
⇒
=
sin P sin R
sin 114° sin 36°
⇒p=
7 × sin 114°
= 10 ⋅879  3
sin 36°
3
So distance RQ =·· 10·9 km
(to 3 sig figs)
Rearrangement
• The aim is to isolate p so that
calculation can then be done on your
calculator.
3
Solution
• Key in 7 × sin 114° ÷ sin 36°.
5 marks
• Any valid rounding will be accepted.
NOTES: 3·3 pages 22, 23
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Worked Answers to Practice Exam D, Credit Mathematics
Value
• Initial amount of petrol is when D = 0
(no distance has been travelled). This
is shown on the graph where it crosses
the P-axis, i.e. at (0, 56).
Q9(a)
56 litres
3
1 mark
Strategy
• Dividing by 5 is the crucial idea here.
The gradient of the graph (ignoring the
negative value) is also an appropriate
measure provided 500 is replaced by
5, i.e. 5 units of 100.
Q9(b)
Over 500 km, 56 − 16 = 40
3
litres were used.
40
= 8 litres
So over 100 km,
5
3
were used.
Calculation
• For this mark, 8 should appear.
Consumption therefore was at a
3
rate of 8 litres per 100 km.
Answer
• A statement concerning rate is
required here.
3 marks
4cosx° − 1 = 0
Simplification
• Always the aim initially in solving
these trig equations is to isolate cosx°
(or sinx° or tanx°).
⇒ 4 cos x° = 1
1
⇒
cos x° =
4
1st solution
• Quadrants can be determined either
from the diagram: S A which
Q10(a)
3
T C
(cosx° is positive when x° is a 1st
quadrant or 4th quadrant angle)
3
−1  1 
1st quadrant: x = cos   = 75⋅5
 4
4th quadrant: x = 360 − 75·5
= 284·5
3
indicates which of sinx°(S), cosx°(C)
or tanx°(T) are positive in the various
quadrants (they are all (A) positive in
the 1st quadrant) or from the various
trig graphs.
• Use INV COS 0·25 or INV COS
(1 ÷ 4), the brackets are vital in this
last case.
So for 0 ≤ x < 360 the two
solutions are x = 75·5, x = 284·5.
2nd solution
• The 4th quadrant angle is always
obtained by: 360° – (related 1st quad
angle).
3 marks
• It is always important to look closely
at the range of allowed angles. In this
case the range is 0 ≤ x < 360.
Q10(b)
NOTES: 3·5 page 27
Using part (a) solutions, the
 x °
solutions of 4 cos   − 1 = 0
 2
x
are = 75⋅5 or 284 ⋅5
2
⇒ x = 2 × 75·5 or 2 × 284·5
Solution
• From part (a), you have calculated
angles that satisfy the equation
4 cos (angle) − 1 = 0 to get: angle =
75·5 or angle = 284·5. In this case
x
the ‘angle’ is
so you know possible
x 2
values for . Doubling then gives you
2
values for x.
⇒ x = 151 or 569
However 0 ≤ x < 360
3
So x = 151 is the only solution.
1 mark
28740_pp049-pp064.indd 62
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Worked Answers to Practice Exam D, Credit Mathematics
Strategy
• For this mark your working should
show that you have realised a rightangled triangle is key to the solution
of this problem.
Q11(a)
Q
P
0·25 m x°
C
0·75 m
(radius)
1m
Q
0·25
C
P
x° 0·75
3
Correct ratio
• All radii in this circle are length
0·75 m. This is crucial for finding the
lengths of two of the sides of the
right-angled triangle.
5m)
0·7 dius
(ra
• Use cosine since the adjacent (A) and
Hypotenuse (H) are known
(SOH CAH TOA).
in ΔCPQ
3
0 ⋅25 1
=
cos x° =
0 ⋅ 75 3
3
 1
⇒ x = cos −1   = 70 ⋅528
 3
.
=. 70·5
(to 3 sig figs)
Calculation
1
• or 0·333··· for cosx°.
3
Solution
• Key in: INV COS (1 ÷ 3)
3
• Any correct rounding is acceptable.
NOTES: 3·1 page 20
4 marks
Strategy
• You should be attempting to find the
fraction of the whole circumference
that arc PR makes. So the appearance
109.4 
of
will gain you this mark.
360
Q11(b)
P
C
The angle at the
centre
Start of calculation
• You use C = πD with D = 1·5 m
(double the radius of 0·75 m).
= 180° − 70·528···°
= 109·471···°
(shaded angle)
R
3
109 ⋅ 471
× Circumference
360
3
109 ⋅ 471
× π × 1⋅ 5 (diameter)
=
360
3
= 1⋅ 4329 
3
=⋅⋅ 1⋅ 43
Arc PR =
So the pedal travels 1·43 metres
(to 3 sig figs)
4 marks
• This mark is gained from
109 ⋅ 4 
× π × 1⋅ 5
360
Calculation
• Always use the π button as 3·14 is only
an approximation to 3 sig figs and may
introduce errors into your calculation.
• Do not use a rounded answer from
part (a), i.e. use 70·528···. (full version
in calculator) not 70·5.
Rounding
• The correct rounding is 1·43, any other
rounding will not gain this mark. Also
use of rounded answers in the course
of the calculations in this question may
introduce rounding errors.
NOTES: 2·4 page 17
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Practice Exam E: Paper 1 Worked Answers
Multiplication
•T
he common rule is ‘turn upside down
a
b
and multiply’ so ÷ becomes ×
b
a
Q1
1
11
1
÷ 2 9 ÷ 26
9 x −63
2
2×2
= x − 31and6 × 6 =
=4
1 × 613
1 =13
6 ÷ 1 =6 × 3
1
3
= ÷ 9= 6× 9 13
9 6 9 13
2 = 2
3
=
39 39
2 marks
Q2
9·9 − 7·2÷30
0 ⋅ 72
7 ⋅2
= 9 ⋅9 −
= 9 ⋅9 −
3
30
= 9 ⋅9 − 0 ⋅24 = 9 ⋅66
Calculation
•C
ancellation can be done (when
multiplying) before multiplying. This
1× 6
6
avoids larger numbers, e.g.
=
9 × 13 117
Notes: 1·2 page 8
Order of operations
•D
ivision is performed before
subtraction – remember BODMAS.
Calculations
7 ⋅2
•N
otice
: you divide top and
30
0 ⋅ 24
bottom by 10 to get 0 ⋅ 72 : 3 0 ⋅ 72
3
which is an easier division to perform.
3
3
)
2 marks
Notes: 4·5 page 39
Fraction
•Y
ou should always get rid of fractions
first as the question inevitably
becomes easier to deal with.
Q3
a = 1 + 3b
bc
(× bc) (× bc)
⇒ abc = 1 + 3b
⇒ abc − 3b = 1
⇒ b(ac − 3) = 1
1
⇒ b =
ac − 3
Like terms
• I n this case the letter b occurs on both
sides of the equation. These two terms
need to be brought together on the
same side.
3
3
• Subtract 3b from both sides.
3
Solution
• b is a common factor.
3 marks
•D
ivide both sides by ac – 3. Compare
1
for example b × 5 = 1 ⇒ b =
5
Strategy
• Multiplication by
Q4
2
3
Expected No. =
× 9010 = 203
19
2
gains this mark.
9
Solution
• Divide ‘top’ and ‘bottom’ by 9 to give
2 × 10
= 20
1
Notes: 6·3 page 62
2 marks
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Worked Answers to Practice Exam E, Credit Mathematics
Gradient
5
10
Q5
The usual y = mx + c
becomes B = mA + c
5 1
=
gradient m =
10 2
c = –3 (B-intercept)
1
Equation is B = A − 3
2
gradient =
distance up (or down)
distance along
Y-intercept
•T
he graph crosses the B-axis at
(0, −3).
3
3
33
Form of equation
•T
his mark is for knowing to use
y = mx + c as the form of the equation
of a straight line graph.
4 marks
Equation
•N
otice A and B are used for x and y.
1
y = x − 3 would not gain this mark.
2
Notes: 5·2 page 48
Q6
(3 – 2a)(2 + a) – (3 – a2)
= 6 + 3a – 4a – 2a2 – 3 + a2
= 3 – a – a2
First brackets
•T
here are 4 terms multiplying out the
brackets: F irsts: 3 × 2
O utsides: 3 × a I nsides: –2a × 2
3
3
3
Lasts: –2a × a
You may remember this using FOIL.
3 marks
Second brackets
•T
hink of this as –1 (3 – a2)
giving –1 × 3 and –1 × (–a2)
for the two terms.
Like terms
•T
here are: 6 – 3, 3a – 4a and –2a2 + a2
giving the final three terms.
Notes: 4·2 pages 30, 31
Scale factor
7
• or 3 ⋅ 5 appearing will gain you this
2
mark.
Q7
The enlargement scale factor from
small triangle to large triangle
7
= = 3⋅5 3
2
So lamp post height = 3·5 × 1·6
= 5·6 metres3
No, her estimate was not correct
because the lamp post is 0·4 m
short of 6 m.
3
• Remember: Enlargement scale factors
are greater than 1 and reduction scale
2
factors are between 0 and 1. So
7
would be a reduction factor.
Calculation
• 1 mark for multiplying by the scale factor.
•N
o calculator: try 3 × 1·6 followed by
0·5 × 1·6 then adding the results.
Notes: 2·2 page 15
3 marks
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28740_pp065-pp080.indd 66
Statement
•1
mark is given for this statement
which must show that you have
compared 5·6 m with 6 m.
11/11/10 4:28:50 PM
Worked Answers to Practice Exam E, Credit Mathematics
Square root
Q8
a ×a
=a
=
1
2
1
3
• a = a 2 . This knowledge gains
1 mark.
2
×a
1 +3
a 2 2
3
2
=a
3
4
2
2
=a
Multiplication
•T
he Index Law is am × an = am + n
3
•Y
ou should not leave a4/2 −
simplification to a2 is required.
2 marks
Notes: 4·6 pages 41, 42
Q9(a)
x + y = 30
Equation
•H
e has x coins of one kind and y coins
of another kind making 30 coins in
total. This is an addition: x + y = total
no. of coins.
3
1 mark
Terms
•N
otice 1 20p-coins weighs 5 g
2 20p-coins weigh 5 × 2 g
3 20p-coins weigh 5 × 3 g
Q9(b)
He has x coins each 5 g and y
coins each 8 g, totalling 210 g 3
So 5x + 8y = 210
3
x 20p-coins weigh 5 × x g
= 5x g
•T
his mark is for 5x and 8y appearing.
2 marks
Equation
•T
otal weight is 5x + 8y, again – an
addition.
Strategy
• The evidence for this mark would be
multiplying one equation by a suitable
factor: e.g. ×5 to get 5x + 5y = 150
Q9(c)
Use simultaneous equations
5x + 8y = 210 5x + 8y = 210
x + y = 30 × 5 5x + 5y = 150
Subtract 3y = 60 3
⇒ y = 20 3
Substitute y = 20 in x + y = 30
so x + 20 = 30 ⇒ x = 10
3
He therefore has 10 twenty pences
and 20 fifty pences.
•A
n alternative would be ×8 to get:
8x + 8y = 240 then: 8x + 8y = 240
5x + 8y = 210
subtract: 3x
= 30
Value of
•A
lternatively x may be found first (see
alternative working above).
Value of x
• Substitute in the ‘easier’ equation, in
this case x + y = 30
3 marks
•Y
ou would be wise to finally check
that your values for x and y satisfy the
other equation. So with x = 10 and
y = 20:
5x + 8y = 5 × 10 + 8 × 20
= 50 + 160 = 210.
Notes: 4·4 page 36
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Worked Answers to Practice Exam E, Credit Mathematics
Equation
•N
otice that 2x2 is not the same as (2x)2:
(2x)2 = 2x × 2x = 4x2 which is not 2x2.
Q10
Use Pythagoras’ Theorem:
2x
Calculation
•T
he alternative from 5x2 = 1002 is:
x
100
3
x2 =
x + (2x) = 100
⇒ x2 + 4x2 = 1002 ⇒ 5x2 = 1002
2
2
2
Simplification
• If you chose the
1002
1002 100
⇒x =
⇒x=
=
5
5
53
2
20 = 4 × =
5 =20
2 5.
so
x = 2000.
2000 route then:
2000 = 100 × 20 = 100 × 4 × 5
= 100 × 4 × 5
Now simplify:
100 100 × 5 100 5
=
=
5
5
5× 5
10000
= 2000
5
= 10 × 2 × 5 = 20 5.
• Simplification for surds means
an answer of the form a b where
b has no more square number
factors (4, 9, 16,…). So notice
that 2000 = 100 × 20 = 10 20
would not gain this mark since
20 = 4 × 5 = 2 5.
3
3 marks
Notes: 4·6 page 41
Effect on B
•T
his mark is for dealing with the B term
to get 4B = 4 × B = 2 × B = 2 B.
Q11
Multiplying B by 4 gives:
3
1000 1000 1 1000 1
=
= ×
= A
2
B
4B 2 B 2
3
So multiplying B by 4 has the
effect of halving A.
Effect on
B
• This mark is for making clear that 2
appearing on the denominator means
1
multiplying by .
2
Notes: 4·8 page 45
2 marks
Q12
1
a = and b = 3
2
1000
Value of a
• The value of a gives the amplitude.
3
Value of b
• The value of b gives the number of
cycles for each 360°. In this case 3
complete cycles of 120° make up the
360°.
3
2 marks
Notes: 3·4 pages 26, 27
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Worked Answers to Practice Exam E, Credit Mathematics
Q13
Strategy
•Y
ou are creating a right-angled
triangle by drawing in the axis of
symmetry of the whole diagram.
D
8m
A
8m
B
w
10 m
(radius)
Method
•C
orrectly applying Pythagoras’
Theorem to the right-angled triangle
will gain you this mark.
C
Calculation
• I t is useful to memorise some of the
smaller ‘Pythagorean triangles’. These
are triples of whole numbers that
create lengths of sides of right-angled
triangles, e.g. 3, 4, 5; 6, 8, 10; 9, 12, 15,
also 5, 12, 13; 10, 24, 26 etc.
raw CB, the perpendicular 3
D
bisector of AD, the 10 m chord.
In ∆ABC use Pythagoras’
Theorem:
3
w + 8 = 10
⇒ w2 = 102 – 82 = 100 – 64 = 36 3
so w = 36 = 6
3
The width of the road is
6 metres.
2
2
2
Solution
• Statement of width is required here.
Notes: 2.4 page 19
4 marks
Value
• Sequence notation is as follows:
1st term 2nd term 3rd term nth term
Q14(a)
u1
n
un = 2 – 1
so u3 = 23 – 1 = 8 – 1 = 7
u2
u3 …
un
•T
o find the 3rd term u3, you should
replace n by 3 in the nth term formula.
3
• 2n is not the same as 2n:
1 mark
3 = 2 × 2 × 2 = 82n with n = 3 gives
2
2×3=6
Q14(b)
We need to find n
so that 2n – 1 = 127
⇒ 2n = 128
⇒n=7
Value of n
•P
aper 1 is the non-calculator paper
so in this case you need to be able to
calculate the powers of 2:
2 = 2, 22 = 4, 23 = 8, 24 = 16, 25 = 32,
26 = 64 leading to 27 = 128
3
We now have this fact:
27 – 1 = 127 is a prime number.
Statement
•Y
ou have to be very clear in what you
say to earn this mark. Basically you are
restating the theorem with n replaced
with 7: ‘If 27 – 1 is a prime number
then 7 is prime’ and also state that
127 is prime.
The theorem then states that
therefore 7 is a prime.
This is certainly true and is
therefore an example of the
truth of the theorem.
3
2 marks
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Worked Answers to Practice Exam E, Credit Mathematics
Practice Exam E: Paper 2 Worked Answers
Method
•T
he question asks for rounding
(2 significant figures). This is the clue
that factorising is not the method to be
used. Solving a quadratic equation by
factorising will not produce values that
need rounding.
Q1
5x2 + x – 1 = 0
Compare ax2 + bx + c = 0
⇒ a = 5, b = 1 and c = – 1
•T
his mark will be gained from
correctly substituting the values of a, b
and c into the quadratic formula.
3
− b ± b2 − 4ac
Substitute in x =
2a
•T
he quadratic formula will appear on
your formula sheet during your exam.
−1 ± 12 − 4 × 5 × ( −1)
⇒x=
2×5
Calculation
• Notice in this case that c is negative
and so – 4ac = – 4 × 5 × (– 1) = 20 is
therefore positive.
−1 ± 1 + 20 −1 ± 21
=
10
10
3
−1 + 21
−1 − 21
so x =
or x =
10
10
= 0 ⋅ 358… or − 0 ⋅ 558…
3
⋅= 0 ⋅ 36 or − 0 ⋅ 56
3
⋅
(to 2 sig figs)
=
• In a question like this, if you get a
negative value under the square root
sign you have made a mistake.
• This mark would be gained for
reaching 21 .
Solutions
•M
ake sure you write down the nonrounded values from your calculator
first.
4 marks
Rounding
•T
his mark may be gained from
correctly rounding a wrong answer.
Notes: 4·7 page 43
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Worked Answers to Practice Exam E, Credit Mathematics
Mean
•C
orrect calculation of the mean gains
this mark.
Substitution
• Mark is gained for reaching 80.
Q2(a)
Σx
= x = Σx
Mean
=x =
Mean
n n
92 + 89 + 88 + 95 + 96 + 86
= = 92 + 89 + 88 + 95 + 96 + 86
6 6
546
3
= = 546
= 91= 91
6 6
For standard deviation s:
x
92
89
88
95
96
86
x − x̄ (x − x̄)2
1
1
−2
4
−3
9
4
16
5
25
−5
25
• Alternative: x
92
89
88
95
96
86
∑ x = 546
and the mark is gained for reaching
49766.
=
Calculation
298116
49766 −
49766
49686
80
6 is =gained
• This
for−reaching
=
5
5
5
• Alternatively:
Σ(x − x )2 = 8080 3
=
=4
s=
5
2 n −1
Σ(x − x )
80
3
=
=4
s=
n −1
5
Standard Deviation = 4 metres 3
=
x2
8464
7921
7744
9025
9216
7396
2
∑ x = 49766
s=
=
2
2
Σx − ( Σx ) / n
=
n −1
298116
6
=
5
49766 −
4 marks
298116
49766 − 49686
80
6
=
=
5
5
5
49766 −
5462
6
5
49766 − 49686
80
=
5
5
49766 −
Solution
• This mark is for finding the square root.
Notes: 6·2 pages 60, 61
Comparing means
•A
clear statement should be made
concerning the heights (on average).
Q2(b)
On average the heights at the 2nd
site are less (by 15 metres) than
those at the 1st site.
3
• J ust saying the mean at the 2nd site is
less than the mean at the 1st site will
not gain this mark.
However at the 2nd site there
was greater variation among
the heights than at the 1st site
(compare s = 15 for 2nd site with
3
s = 4 for the 1st site).
Comparing standard deviations
•Y
our statement here must mention
‘variation’ or ‘spread’of the heights.
• Stating that one standard deviation is
less than the other will not gain this
mark.
2 marks
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Worked Answers to Practice Exam E, Credit Mathematics
Multiplying factor
6 ⋅5
• Notice that 6 ⋅ 5% =
= 0 ⋅ 065
100
•A
fter 1 year the value is the original
100% value plus 6·5% hence 1 + 0·065 =
1·065.
Q3
The yearly multiplication factor
is 1·065 (6·5% increase)
3
After 3 years the estimated
worth is 15500 × 1·0653
= 18723·219….
At the start of 2014 the plot
should be worth £18723 (to
the nearest £1).
Correct power
•B
e careful to count the years correctly:
from 2011 to 2014 is 3 increases not 4:
3
2011
2012
2013
2014
1st increase2nd increase3rd increase
(×1·065)
(×1·065)
(×1·065)
3
Solution
• Any reasonable (correct) rounding is
acceptable here to gain the mark.
3 marks
Notes: 1·5 page 11
Q4
Strategy
•T
o gain this mark you will have to
show clearly in your solution that you
understand that £15·64 is 115% of his
original rate.
After a 15% rise he will receive
115% of his old rate of 100%. 3
So 115% ↔ £15⋅64
15⋅64
⇒
1% ↔ £
115
15⋅64
⇒ 100% ↔ £
× 100
115
= £13⋅60
Process
•T
his is a ‘direct proportion’ question:
divide by 115 to obtain 1% then
multiply by 100 to obtain 100%.
3
Calculation
•T
he correct calculation produces
£13·60.
3
3 marks
Notes: 1·5 page 12
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Worked Answers to Practice Exam E, Credit Mathematics
Fraction
•A
s with solving equations you should
first attempt to get rid of the fractions:
in this case multiply both sides of the
inequality by 6.
Q5
x −3 2
>
6
3
(×6) (×6)
x −3
2
2×2
− 322and
=4
x − 3 x − 3 × 6 =×x62−=3xand
× 2 ×26×=2
× 6 =6 x − 63 and × 6 = 3 × 63== 4 1 =14
6
3
1
⇒x–3>4
3
⇒x>7
3
Solution
• Add 3 to both sides.
Notes: 4·4 page 35
2 marks
Fraction
•T
he areas of sectors
are proportional to
the angles at the
centre:
48°
area of sector
360°
area of whole circle
Q6
angle at centre
48
of the whole
× area of whole circle So required area =
Area
angle= atangle
centre
at
centre
360
360
×
area
of
whole
circle
Area
=
×
area
of
whole
circle
Area
=
circle
area.
3
angle at centre
360
360 of whole circle
Area =
48 × area
360
=
×
π × 202
48
48
2
Correct formula
× 20
× π × 202
3
48 = 360 ×=2π360
•
A
rea of a circle = πr2 where r is the
=
× π × 20= 360
167 ⋅551…
radius. This formula is not given to
360 = 167 ⋅551
… ⋅551
= 167
2…
you in the exam.
= 167 ⋅551… =2168 m2
= 2168 m
3
= 168 m
48
(to the nearest 1 m2)
• You are using:=
× πr 2 with r = 20 m
= 168 m

360
3 marks
Calculation
•R
emember for accuracy to use the π
button on your calculator and not just
3.14.
•A
ny correct rounding is acceptable
here.
Notes: 2·4 page 17
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Worked Answers to Practice Exam E, Credit Mathematics
Angles of triangle
•T
he right angles appear in the diagram
due to the fact that K is “due East” of
C and so line CK is perpendicular to
the North lines.
Q7
L
N
N
48°
66°
C
•Y
ou also know the angle sum in a
triangle is 180°.
L
Strategy
• This mark is for using the Sine rule.
K
Substitution
Angle LCK = 90° – 66° = 24°
Angle CKL = 90° + 48° = 138°
So Angle CLK = 1
80° – (24°
+ 138°)
= 180° – 162°
= 18°
3
a
b
c
=
=
sin A sin B sin C
changes in this case to
c
k
l
=
=
sin C sin K sin L
•C
L = k and you know CK = l = 11.
This indicates that you use
k
l
c
and ignore
.
=
sin K sin L
sinC
∠K = 138° and ∠L = 18°
•N
otice that
L
Use the 3
18°
Sine
Rule
24° 138°
C
11
K
k
l
=
sinK sinL
k
11
11sin138°
⇒
=
⇒k=
sin138° sin18°
sin18°
⇒ k = 23 ⋅ 81…
3
The distance between
Cowdenbeath and Leven is
23.8 km (to 1 dec place).
•T
his mark is for correctly replacing
l, K and L by 11, 138° and 18°
respectively.
Solution
•M
ultiply both sides by sin 138° to
isolate k.
•A
ny correct rounding is acceptable
here.
3
•W
hen using sin, cos,or tan always
check your calculator is in the correct
MODE. There should be ‘DEG’ or
‘D’ on the screen. A good check is
testing sin 90 – you should get 1, if not,
reset your calculator.
4 marks
Notes: 3·3 pages 22, 23
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Worked Answers to Practice Exam E, Credit Mathematics
Rearrangement
•T
he aim in this type of trig equation is
to isolate tanx° (or sinx° or cosx°).
• Subtract 7 from both sides and then
divide both sides by 3.
Q8
3 tanx° + 7 = 0
⇒ 3 tanx° = −7
7
3
⇒ tan x° = −
3
(tanx° is negative if x° is a
2nd quadrant angle or a 4th
quadrant angle.
The related 1st quadrant angle
7
is tan −1   = 66 ⋅8° )
 3
2nd quadrant: x = 180 – 66·8
3
= 113·2
4th quadrant: x = 360 – 66·8
= 293·2
3
Solutions are x = 113·2,
x = 293·2
1st solution
•F
or determining the correct quadrants
S A
you can use the diagram: T C
giving 2nd and 4th quadrants as
negative for tanx°.
•F
or the 1st quadrant angle always
ignore any negative sign so, in this
7
case, use positive .
3
•2
nd quadrant angle is
180° – (1st quadrant angle).
2nd solution
•4
th quadrant angle is
360° – (1st quadrant angle).
•A
ngle values are usually approximated
to 1 decimal place.
3 marks
Notes: 3·4 page 27
Q9
1
Area = bc sin A
2
where Area = 6·3, c = 3 and
∠A = 110°
1
so 6 ⋅3 = b × 3 × sin 110°
2
⇒ 2 × 6·3 = 3b sin 110°
2 × 6 ⋅3
⇒
=b
3 sin 110°
⇒ b = 4·469...
so AC =·· 4·47 cm
(to 3 sig figs)
Strategy
•T
his is for the correct use of the Area
formula.
• On your formulae page is:
1
Area = ab sin C . You have to be able to
2
1
change this to: Area = bc sin A to fit
2
this situation.
3
3
Substitution
•T
his mark will be gained for
replacing the letters (and area) by the
appropriate values.
Solution
• I n the given working, both sides were
first doubled to get rid of the fraction.
Then both sides were divided by
3 sin 110°.
1
•A
n alternative is to work out
×3×
2
sin 110° giving: 6·3 = 1·409... × b so
6 ⋅3
b=
etc
etc.
1⋅ 409…
•A
ny correct rounding is acceptable
here.
3
3 marks
Notes: 3·3 page 22
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Worked Answers to Practice Exam E, Credit Mathematics
Strategy
•Y
ou should be aware that some of one
of the ingredients will be left over.
The strategy is to find whether there is
too much oil or too much vinegar. The
initial working in the solution is to test
this.
Q10
Vinegar : Oil
4 parts : 9 parts
Attempt to use all of the oil:
9 parts ↔ 600 ml
600
ml
⇒ 1 part ↔
9
600
⇒ 4 parts ↔
× 4 = 266.6…
9
There is only 248 ml of vinegar –
not enough.
Now attempt to use all of the
3
vinegar:
4 parts ↔ 248 ml
248
ml
⇒ 1 part ↔
4
248
⇒ 9 parts ↔
× 9 ml
4
3
= 558 ml
Calculation
•T
he method shows a proportion
calculation. For example, knowing the
volume of 4 parts you can calculate the
volume of 1 part by dividing by 4 and
so on.
•A
n alternative method would be to
build towards the 248 ml and 600 ml
using the 4 and 9 ratio numbers:
Vinegar (248) Oil (600)
× 60:
add 1 part:
add 1 part:
4 ml
240 ml
244 ml
248 ml
9 ml
540 ml
549 ml
558 ml
As can now be seen: all the vinegar is
used but only 558 ml of oil.
So 558 ml oil will be required.
(This will leave 600 – 558 = 42 ml
oil unused)
Solution
•N
otice that it is the total volume of the
two ingredients that is required, not
just the individual quantities.
Therefore use 248 ml of vinegar
and 558 ml of oil,
3
i.e. 248 + 558 = 806 ml of
vinaigrette can be made.
Notes: 1·5 page 12
3 marks
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Worked Answers to Practice Exam E, Credit Mathematics
Dimensions
•N
otice that since there are two
borders each of 0·5 m then the
outside dimensions of the frame are
reduced by 2 × 0·5 m = 1 m to find the
dimensions of the picture:
Q11(a)
0·5
m
x metres
(x–1) metres
0·5
m
so x metres reduces to (x – 1) metres
and (x – 1) metres reduces to (x – 2)
metres.
0·5 m
Area
•Y
ou must take care to show your full
working since the expression x2 – 3x + 2
is given in the question. So it would be
wise not just to write (x – 1) (x – 2)
but to include the step x2 – 2x – x + 2
which shows all the terms when
multiplying out the brackets.
x–1
x–2
metres metres
0·5 m
As shown in the diagrams
Width of picture = x – 0·5 – 0·5
=x–1m
•H
ere’s a much more complicated route
to the answer:
All the 4 shaded
rectangles have area
0·5 × (x – 1) =
0·5 (x – 1) m2
Breadth of picture = x – 1 – 0·5
– 0·5
=x–2m
3
Area of picture = (x – 1) (x – 2)
= x2 – 2x – x + 2
= x2 – 3x + 2 m2 3
Total shaded area = 4 × 0·5 (x – 1)
2 marks
=2 (x – 1) m2
Area of picture = x (x – 1) – 2 (x – 1)
(picture + frame) (frame)
= x2 – x – 2x + 2
= x2 – 3x + 2 m2
Notes: 4·2 pages 30, 31
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Worked Answers to Practice Exam E, Credit Mathematics
Equation
•T
his mark is for writing down the
relationship between the picture area and
total area correctly in algebraic form.
Q11(b)
Total Area = x (x – 1)
= (x2 – x) m2
3
•A
common error is to put the factor
on the wrong side of the equation. 5
Picture Area = (x – 3x + 2) m
3
Now Picture Area = × Total Area
5
3 2
2
⇒ x − 3x + 2 = (x − x)
3
5
2
2
⇒ 5(x – 3x + 2) = 3(x – x)
⇒ 5x2 – 15x + 10 = 3x2 – 3x
⇒ 5x2 – 3x2 – 15x + 3x + 10 = 0 3
⇒ 2x2 – 12x + 10 = 0
⇒ 2(x2 – 6x + 5) = 0
⇒ 2(x – 5) (x – 1) = 0
⇒ x – 5 = 0 or x – 1 = 0
⇒ x = 5 or x = 1
3
2
2
Standard form
• On recognising the equation as a
quadratic you should then rearrange
into ‘standard form’: ax2 + bx + c = 0
ready for solving.
•T
here are many steps in this
rearrangement and therefore many
places for errors. You should practice
this rearrangement until you can
perform it confidently with no
mistakes.
Solutions
•W
hen a quadratic equation arises in
a context problem like this it will be
solvable by factorising. Quadratic
formula use is only tested on a given
equation in the credit exam – never in
context.
Now x = 1 is not a valid solution
as this would give zero and
negative lengths for some of the
dimensions.
So x = 5 is the only valid
solution
(giving a 5 m × 4 m frame).
3
Valid solution
•T
his mark is for clearly rejecting x = 1
as a valid solution.
4 marks
Notes: 4·7 page 42
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Worked Answers to Practice Exam E, Credit Mathematics
Strategy
•T
his is the graph of a quadratic
function and is in the shape of a
parabola. It has an axis of symmetry
(dotted line). Fold the parabola along
this axis of symmetry and the two
x-axis intercepts will coincide. The
axis is midway between these intercepts
and it also passes through the
minimum turning point of the graph.
Q12
The x-axis intercepts are given
by solving f(x) = 0
3
⇒ (x – 5) (x + 2) = 0
⇒ x – 5 =0 or x + 2 = 0
⇒ x = 5 or x = –2
3
3
2
–2
5
x
x-axis intercepts
• This mark is for obtaining –2 and 5.
A
−2 + 5 3
=
Mid point is given by
2
2
3
So x-coordinate of A is
3
2
 3  3   3 
f   =  − 5  + 2
 2  2   2

Mid point
• The mid point is the mean of –2 and 5.
Solution
•T
he heights of points on the graph,
above or below the x-axis are given
by substituting the appropriate
x-coordinate value into the formula
of the graph, i.e. when x = a the point
on the graph (above or below) is
(a, f(a)). For point A in this example
3
3 
its coordinates are  , f    .
 2  2 
1
• −12 and –12·25 are acceptable.
4
Notes: 5·4 page 52
 3 10   3 4 
= −  + 
 2 2   2 2
−7 7
49
× =−
2 2
4
The required y-coordinate
49
is −
4
=
3
4 marks
Formula
•V
olume of a Prism =
Area of end × length.
These ‘row covers’ are prisms with
ends that are semicircles.
Q13(a)
Volume = Area of end × length
1
= πr 2 × length
3
2
1
= × π × 22 × 8
2
= 50 ⋅265… =⋅⋅ 50 ⋅ 3 m3 3
(to 3 sig figs)
•Y
ou may know the volume of a
cylinder formula: Volume = πr2h. In
this case h is the length and you need
to halve the result.
Solution
•R
emember to use the π button on your
calculator not 3·14.
2 marks
•A
ny correct rounding is acceptable
here.
Notes: 2·1 page 14
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Worked Answers to Practice Exam E, Credit Mathematics
Equation
•T
he working shows the ‘changing
the subject’ although setting up the
equation:
2
50 ⋅ 265 = πr 3 and
3
proceeding to isolate r3 involves the
same steps.
Q13(b)
2
V = πr 3 with V = 50·265…
3
Change the subject to r:
3V
= r3
3V = 2πr 3 ⇒
2π
3
×
⋅
…
3
V
3
50
265
⇒r = 3
=3
3
2π
2π
•B
oth sides are multiplied by 3 then
divided by 2π to isolate r3.
Substitution
•R
eplacing V with 50·265... gains the
mark.
= 3 24 = 2 ⋅884 …
Height is 2·88 metres
(to 3 sig figs)
Solution
•T
he cube root is required. On most
calculators there is a x key. (Some
have a 3 key). For the cube root key
3
3 marks
in: 3 x followed by the number
whose cube root is required.
•Y
ou may wonder at exactly 24
appearing. There is a reason
for this. Look at part (a):
1
2
V = × π × 2 × 8 = 16π . Now replace
2
1
3V
2
V = × π ×2 V
× 8by
= 16
16π in r = 3
. This gives
2π
2
3 × 16π 3
= 24 after cancelling
2π
1
2
V = × π ×2 ×8 =
by162π.
2
Notes: 2·1 page 13, 4·5 page 39
r=
3
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Practice Exam F: Paper 1 Worked Answers
Common denominator
40
27
or
• This
mark is for finding
15
15
2
•N
OTE: 2 is 6 thirds and 2 thirds
3
4
giving 8 thirds in total 1 is 5 fifths
5
and 4 fifths giving 9 fifths in total.
Q1
4 8 9
2
2 −1 = −
5 3 5
3
8×5 9×3
−
=
3×5 5×3
40 27 13
=
−
=
15 15 15
3
3
Calculation
40 27 40 − 27 13
=
= .
• −
15 15
15
15
2 marks
NOTES: 1·2 pages 8, 9
Order of operations
•B
racket calculations are done first.
Q2
Calculations
•T
his is a non-calculator paper so
it is important that you know how
to handle divisions like this which
involve a decimal point. Treat the
division as a fraction. You are allowed
to multiply top and bottom by the
same number: in this case 10. This
clears the decimal point problem in
the division.
82·8 ÷ (1·8 − 0·9)
= 82·8 ÷ 0·9
=
3
82 ⋅8 82 ⋅8 × 10 828
=
=
0 ⋅9
0 ⋅9 × 10
9
= 92
3
2 marks
Factorising
•A
lways check by multiplying out
(using FOIL for instance):
Q3(a)
2x2 + x − 6
= (2x − 3)(x + 2)
(2x − 3)(x + 2) =
2x2 + 4x − 3x − 6
= 2x2 + x − 6 as required
3
NOTES: 4·3 page 33
1 mark
Factorise numerator
•W
hen simplifying algebraic fractions
you are looking for a shared factor in
the ‘top’ and the ‘bottom’ so you will
need to factorise.
Q3(b)
3
2x + 4
2(x + 2)
2
=
=
2x2 + x − 6 (2x − 3)(x + 2) 2x − 3
3
Simplification
• ‘ Hence’ in part (b) means you must
use the answer to the previous part (a).
• ( x + 2) is a shared factor and may be
cancelled.
2 marks
1
x −3
by cancelling the 2’s is not permitted.
•N
o further cancelling is possible:
NOTES: 4·5 page 37
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Worked Answers to Practice Exam F, Credit Mathematics
Substitution
•F
or f(−5): every occurrence of x in the
formula should be replaced by −5.
Q4
f (x) =
6
3+ x
⇒ f ( −5) =
6
6
=
= −3
3 + (−5) −2
Calculation
•P
ositive divided by negative gives
negative.
3
3
NOTES: 4·1 page 30
2 marks
Gradient
•R
emember:
distance up or down
gradient =
distance along
Q5
•A
downhill graph (left to right) has a
negative gradient.
The equation is of the form
y = mx + c
gradient
m=
y
−6
3 10
=−
8
4
-intercept
-intercept
•T
his is where the graph crosses the
y-axis. In this case the value is 10.
8
6
3
Equation
•T
he equation of a linear graph is of
the form:
x
0
c = 10
3
3
Equation is y = − x + 10
4
y = mx + c
3
gradient y-intercept
•T
o gain this mark you do not need to
get rid of the fraction in the equation.
3 marks
NOTES: 5·2 page 48
Linear scale factor
6
3
or or 0.6
•T
his mark is for
10
5
• I n similar shape problems you work
from the known (large nut) to the
unknown (small nut). In this case
from large nut to small nut. This is a
reduction so you would expect a scale
factor less than 1 (but larger than 0)
Q6
The weights are proportional to
the volumes of the nuts
3
6
Length reduction
=
= 0.6
scale factor
10
Volume scale factor
3
3
 6
 3
•T
his mark is for   or   or 0.63
 10 
 5
⇒v
olume reduction
= (0·6)3
scale factor
3
3
weight of smaller nut =
15 × (0·6)3
= 3·24 g 3
•A
s an example:
length scale factor ×2
×2
(length)
×8
(volume)
volume scale factor ×8 i.e. ×23
4 marks
Method
•M
ultiply the large volume by 0·63.
Solution
•A
ny rounding is acceptable here.
NOTES: 2·2 page 15
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Worked Answers to Practice Exam F, Credit Mathematics
Brackets
•T
hink of −(2 − a) as −1(2 − a) giving
−1 × 2 and −1 × (−a) as the two terms.
Q7
3a – (2 – a) = 5a
⇒ 3a – 2 + a = 5a
Like terms
•G
ather all the terms with the letter a
on one side of the equation.
3
⇒ 3a + a – 5a = 2
⇒ –a = 2
3
⇒ a = –2
3
Solution
•N
otice – (−2) = 2 or multiply both
sides by −1.
NOTES: 4·3 page 34
3 marks
Area
•A
rea of a right-angled triangle
1
= base × height
2
1
In this case= ××22××1.5
1·5
2
•A
lternatively:
Q8(a)
Area ∆ADC = Area ∆ABC
So Area of kite =
2 × Area ∆ADC
1
== 2 × × 22 ×× 11·5
.5
2
3
=
3mm2 2
=3
D
1·5
A
B
2
1 mark
1·5
C
2
flip ABC
upside down
The area of the resulting rectangle is
1·5 × 2 = 3 m2.
Strategy
•F
or this mark you will need to show
you understood the relationship
between the diagonals of the kite and
the area of the kite.
Q8(b)
A
D
Area of
surrounding
B rectangle
3
= 2 × Area of kite
Equation
•P
roduct of the two diagonals =
2 × Area of kite leads to DB × 2·5 = 6
which gains this mark.
= 2 × 3 = 6 m2
C
so DB × AC = 6
so DB × 2·5 = 6
⇒ DB =
12
6
=
= 2⋅4 m
2 ⋅5 5
3
Solution
•D
ivide both sides by 2·5.
3
•D
oubling top and bottom gets rid of
the problem of division by 2·5.
NOTES: 2·1 page 13
3 marks
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Worked Answers to Practice Exam F, Credit Mathematics
Horizontal displacement
•S
uppose a is positive then y = (x − a)2
shifts y = x2 horizontally a units to
the right and y = (x + a)2 shifts y = x2
horizontally a units to the left. So
y = (x − 3)2 is a shift horizontally of
3 units to the right.
Q9
y
2
0
x
3
Vertical displacement
•S
uppose b is positive then y = x2 + b
shifts y = x2 vertically b units upwards
and y = x2 – b shifts y = x2 vertically
b units downwards. So y = x2 + 2 shifts
y = x2 vertically 2 units upwards.
3
3
2 marks
Equation
•x
coins of one type and y coins of
another making a total of 18 coins.
This is an addition.
Q10(a)
x + y = 18
3
1 mark
Terms
•T
his mark is for obtaining 10x and 5y.
Q10(b)
3
3
10x + 5y = 125
Equation
•C
areful with units: change £1·25 to
125 pence.
2 marks
Strategy
•W
ith 2 different letters (x and ) and
2 different equations you should know
to use simultaneous equations.
Q10(c)
10x + 5y = 125
•A
lways look for a common factor in all
terms in an equation (5 in this case).
Both sides of 10x + 5y = 125 can
be divided by the factor 5 to give an
easier equation to deal with.
Simplifies to 2x + y = 25
Use simultaneous equations:
subtract
2x + y = 25 

x + y = 18 
x
= 7
3
Value of x or
•A
n alternative would have been to
double all terms in x + y =18:
2x + 2y = 36
2x + y = 25
subtract
y = 11
3
Substitute x = 7 in x + y = 18
⇒ 7 + y = 18 ⇒ y = 11
She has 11 5p-coins
3
Statement
•A
clear statement is required here.
3 marks
NOTES: 4·4 page 36
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Worked Answers to Practice Exam F, Credit Mathematics
Expression
•N
ote that the question refers to
pence. So the expression 0·6 x is not
acceptable.
Q11(a)
60x pence
3
1 mark
Calculation
•T
his is the calculation that gives you
the pattern for the general formula
that is asked for in the next part of the
question. Be clear what each stage of
the calculation involves.
Q11(b)(i)
800 + (25 − 10) × 45 pence
= 800 + 15 × 45 pence
•A
crucial feature is dealing with the
10 free downloads. Only further
downloads are charged. This gives
25 − 10, i.e. 15 downloads that are
paid for at 45 pence each.
= 800 + 675 pence
= 1475 pence = £14·75
3
1 mark
Dealing with free downloads
•S
ince 10 downloads are free, only
the remaining x − 10 downloads
will be charged. This mark is for the
appearance of the expression x − 10.
Q11(b)(ii)
3
Cost = 800 + (x − 10) × 45 pence
Expression
•S
implification is not necessary to gain
this mark, so 800 + 45 (x − 10) would
be acceptable. Simplification, however,
will be needed to complete the rest of
this question.
= 800 + 45x − 450 pence
= 350 + 45x pence
3
2 marks
NOTES: 4·1 page 30
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Worked Answers to Practice Exam F, Credit Mathematics
Inequality
•C
areful reading of the question
implies that members eventually get
the best deal if they are buying a large
number of tunes. You could check this
with ‘easy’ numbers:
Q11(c)
Cost, to a member, for x tunes
(x > 10)
= 350 + 45x pence
Cost, to a non-member, for x
tunes
non-member member
1 download:
£0·60
£8·00
110 downloads:
£66·00
£53·00
So at some number of downloads:
= 60x pence
Cost to member < Cost to non-member.
For membership to be cheaper
we require: 350 + 45x < 60x 3
•T
his mark is gained for writing down
the algebraic inequality.
⇒ 350 < 60x − 45x
⇒ 350 < 15x
350
⇒
<x
15
70
⇒
<x
3
1
70
3
= 23
so x >
3
3
3
The minimum no. of songs is 24.
Solving inequality
•S
ubtract 45x from both sides and then
1
divide both sides by 15 to get x > 23 .
3
Solution
1
• You need to interpret x > 23 carefully.
3
As the check shows, 23 songs is still
cheaper for a non-member. The
required solution is the smallest whole
1
number that satisfies x > 23 i.e., 24.
3
Let’s Check:
non23
member
member
Songs: 350 + 45 × 23 23 × 60
= 1385 p
= 1380 p
(cheaper by 5 p)
Note: Alternative method:
• I t is possible to do a Trial and Check
method (similar to the check in
the working on the left). The three
available marks would be given as
follows:
24
350 + 45 × 24 24 × 60
Songs: = 1430 p
= 1440 p
(10 p more expensive)
1st mark: Use at least 3 trials.
2nd mark: Two of the trials must be
for 23 songs and 24 songs.
3rd mark: Clear statement of solution
obtained.
3 marks
NOTES: 4·4 page 35
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Worked Answers to Practice Exam F, Credit Mathematics
Practice Exam F: Paper 2 Worked Answers
Strategy
• I n questions where you know the
quantity after a percentage increase
or decrease, you should match the
original quantity (before the increase or
decrease) to 100%. In this case the price
before VAT is matched to 100%. Now
add on VAT at 15%: 100% + 15% =
115% which is matched to £273.70.
Q1
115% of the correct price is
£273·70. You need to find 100%
of the correct price.
3
115% ↔ £273·70
273.70
1% ↔ £
115
273.70
× 100
100% ↔ £
3
115
= £238
School should have been
charged £238.
Calculation
•T
his now is a proportion problem:
divide by 115 to get 1% of the
correct price, then multiply by 100
to get 100% of the correct price. An
indication that you know this process
will gain this mark.
3
Solution
•T
he final answer is awarded this mark.
3 marks
Notes: 1·5 page 12
Substitution
•C
orrect substitution in the
circumference formula will gain you
this work.
Calculation
• Key in: π × 2 × 2·36 EXP 17
Q2
C = π × D = π × 2r
= π × 2 × (2.36 × 1017)
= 1.48 × 1018 km
(to 3 sig figs)
3
33
•T
he alternative is to calculate
π × 2 × 2·36 to get
14·828··· × 1017 = 1·482··· × 1018
•A
ny correct rounding is acceptable
here.
3 marks
Scientific notation
•T
his final mark is for writing your
answer in scientific notation.
•A
ny correct rounding is acceptable
here.
Notes: 1·4 page 10
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Worked Answers to Practice Exam F, Credit Mathematics
Q3(a)
Σx
n
Mean = x =
Mean
•1
mark for correct calculation of 28.
27 + 35 + 24 + 23 + 29 + 30
6
168
3
=
= 28 thousands
6
=
Formula
(x − x )2
n −1
Σx2 − (Σx)2 n
or s =
gains this mark.
n −1
•T
he alternative formula gives:
•C
orrect substitution into s =
For the standard deviation s:
x
27
35
24
23
29
30
– (x – x–)2
x–x
–1
1
7
49
–4
16
–5
25
1
1
2
4
–)2 = 96
Σ(x – x
s=
=
4800 − 4704
96
=
5
5
Solution
•A
common error is to forget to take
the square root.
•A
ny correct rounding is acceptable
here.
2
(x − x )
96
=
= 4.381 3
n −1
5
.
so s =. 4.38 thousands
3
(to 3 sig figs)
s=
4800 − (28224 6)
5
Notes: 6·2 pages 60, 61
3 marks
Comparing means
•Y
our statement concerns circulation
figures. It is not sufficient to state that
one mean is less than the other.
Q3(b)
On average the Aberdeen
paper has lower circulation
(mean = 28) than the Dundee
3
paper (mean = 35).
Comparing standard deviations
•T
he larger the standard deviation
then the more the data values are
spread out around the mean value.
So: larger standard deviation – more
variation, smaller standard deviation –
less variation. Again, you are making
statements in this case about the
circulation figures being more or less
varied.
However, there is less variation
(about the mean) in the
circulation figures for the
Dundee paper (s = 1·2) than for
the Aberdeen paper (s = 4·38) 3
2 marks
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Worked Answers to Practice Exam F, Credit Mathematics
Strategy
•F
or this mark you should have
attempted to use ‘SOHCAHTOA’ in
triangle ADE.
Q4
A
3 cm
x°
D
8 cm
E
Calculation
•Y
ou know the Opposite side AE =
3 cm and the Adjacent side DE = 8 cm
which leads you to ‘TOA’ and the use
of tanx°.
3
Use trig in right-angled
∆ADE.
3
3
3


tan x° = ⇒ x = tan −1  
 8
8
= 20.556
The wedge makes an angle of 3
20.6° with the horizontal (to
1 dec. place)
•K
ey in: INV tan (3 ÷ 8) The brackets
are vital here (try leaving them out and
compare the answer!)
•1
mark for correctly choosing ‘tan’ and
1 mark for reaching 20·556···°
Rounding
•1
mark for correct rounding to
1 decimal place.
3 marks
Notes: 3·1 page 20
Expression
•U
se the pattern:
Q5(a)
(2y − 1)(y − 2)
= 2y2 − 4y − y + 2
= 2y2 − 5y + 2
3
(·
(.
(.
(.
·)(·
.)(.
.)(.
.)(.
·)
.)
.)
.)
Firsts
Remember
Outsides
this using
Insides
FOIL
Lasts
• Take care with negatives, e.g. the ‘Lasts’
give −1 × (−2) a positive result of 2.
1 mark
Notes: 4·2 page 31
Q5(b)
1st term
−1
x 2 (x
=x
=
− 1 +1
− 3)
− 1 +1
2
1
x 2
1
•x
= x1 so you have: x 2 × x and use
the index law: am × an = am + n
− 3x
−1
2
−1
− 3x 2
2nd term
3
3
•T
his mark is for obtaining −3x −
1
2
Notes: 4·6 pages 41, 42
2 marks
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Worked Answers to Practice Exam F, Credit Mathematics
Q5(c)
108 − 5 3
= 36 × 3 − 5 3
= 6 3 −5 3 = 3
Simplify surd
•Y
ou may have noticed 108 = 9 × 12
leading to 108 = 9 × 12 = 9 × 12 = 3 12
108 = 9 × 12 = 9 × 12 = 3 12 . 108 is, in this case,
not fully simplified 12 still has a
square number factor of 4. So:
3
12 = 4 × 3 = 4 × 3 = 2 3 so that
3
3 12 becomes 3 × 2 3 = 6 3 which is
now fully simplified.
2 marks
Subtraction
•C
ompare 6 3 − 5 3 = 3 with
6x – 5x = x
Notes: 4·6 page 40
Volume
•V
olume of prism = Area of end
× length.
Q6(a)
Volume = Area of end × length
= 1·6 × 4·2
3
= 6·72 m3
• Units are m3 for volume.
1 mark
Q6(b)
Again, Volume = A
rea of end
× length
1
⇒ 6.72 = × πr 2 × length
2
⇒ 2 × 6.72 = π × 12 × length
2 × 6.72
⇒
= length
π
So length = 4·278···
.
=. 4·28 metres
(to the nearest centimetre)
Strategy
•F
or this mark there needs to be
evidence that you are finding the area
of a semicircle times the length of
the trough and equating this with the
previous known volume.
Substitutions
•U
se the radius = 1 m in the expression
1 2
πr .
2
3
3
•T
his mark is gained for the correct
1
equation: 6.72 = π × 12 × length.
2
3
Rearrangement
•C
orrect isolation of ‘length’ gains this
mark.
1
•D
ouble both sides to get rid of
then
2
divide both sides by π.
3
4 marks
Solution
•U
se π button.
•A
ny correct rounding is acceptable
here.
Notes: 2·1 pages 13, 14
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Worked Answers to Practice Exam F, Credit Mathematics
Strategy
•Y
ou have to use the 136° angle at
P and so have to move the 136°
information to the point Q. One of the
ways to do this is to extend the line PQ
to create ‘corresponding angles’:
Q7(a)
N
136º
P
N
136º
44º
136º
Q
P
3
258º
Calculation
•∠
PQN = 180° − 136°
R
∠PQR = 360° − (258° + 44°)
= 360° − 302° = 58°
136º
Q
•T
here are 3 angles around Q totalling
360°: ∠PQR, ∠PQN and reflex
∠NQR.
3
2 marks
Notes: 3·2 page 21
Strategy
•T
his mark is for using the Cosine
Rule.
•T
he formula given to you during your
exam is: a2 = b2 + c2 − 2 bc cos A
You have to be able to change this
form into one that is appropriate to
∆PQR and in particular, the form for
finding q(RP).
Q7(b)
The three forms are:
P
10·2 km
p2 = q2 + r2 – 2qr cos P
q2 = p2 + r2 – 2pr cos Q ← use this
r2 = p2 + q2 – 2pq cos R
58º Q
R
12·8 km
se the Cosine Rule
U
q2 = p2 + r2 − 2prcosQ
Substitution
•N
otice that p (RQ) = 12·8 km and
r (PQ) = 10·2 km and ∠Q = 58°.
3
• I t is important to show all your
working. If the substitutions line is
not written down and you then make a
calculation error you will also lose this
mark, not just the calculation mark.
= 12·82 + 10·22 − 2 × 12·8 × 10·2 × cos 58°
3
⇒ q2 = 129·50···
so q = 129.50  = 11.38
.
So RP =. 11·4 km (to 3 sig figs)
3
3
Solution
•R
emember the square root!
4 marks
Rounding
•T
here is a mark for the correct
rounding here.
Notes: 3·3 page 23
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Worked Answers to Practice Exam F, Credit Mathematics
Q8(a)
Substitution
• I t is important to fully understand
what the variable n stands for in a
question like this. The order of stating
the dimensions is important: 3 squares
× 2 squares gives n = 3 not n = 2.
n squares × 2 squares
becomes 3 squares × 2 squares
So put n = 3 into the formula:
3
r = × 3 × (3 + 1)
2
3
3
= × 3 × 4 = 18
2
18 different rectangles can be
3
drawn.
•T
his mark is for
3
3
r = × 3 × (3 + 1) or × 3 × 4.
2
2
Calculation
•T
his mark is for the appearance of 18.
2 marks
Q8(b)
3
n(n + 1) = 198
2
⇒ 3n(n + 1) = 396
⇒ 3n2 + 3n – 396 = 0
⇒ n2 + n – 132 = 0
Equation
• I n this case n is unknown but the
result of applying the formula is
known, i.e. 198.
3
3
•T
his mark is for equating n(n + 1)
2
with 198.
Rearrangement
•M
ultiply both sides by 2 to get rid of
the fraction 3 . Now multiply out the
2
brackets, then subtract 396 from both
sides.
3
2 marks
Notes: 4·1 page 30
Factorising
• I t is important to note the word
‘hence’ in the question. You must
use the previous answer, namely the
equation n2 + n − 132 = 0 as a means
of solving the question.
Q8(c)
n2 + n – 132 = 0
3
⇒ (n – 11)(n + 12) = 0
⇒ n – 11 = 0 or n + 12 = 0
3
⇒ n = 11 or n = – 12
n = – 12 is not a valid solution
since n is positive.
So n = 11 is the only solution. 3
The grid is 11 squares × 2 squares.
A trial and improvement method is
possible where various sizes of grid are
explored leading to the discovery of
11 squares × 2 squares. This strategy
is ruled out due to the use of the one
word ‘hence’!
•T
his mark is for (n − 11)(n + 12) or
(n + 12)(n − 11).
3 marks
Solutions
• I nitially there are two solutions,
11 and −12 that satisfy the given
equation.
Valid solution
•T
his mark can only be gained if you
explicitly reject the value n = −12
as not valid in the context of the
question.
Notes: 4·6 page 42
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Worked Answers to Practice Exam F, Credit Mathematics
Strategy
•T
he angle at the centre of the circle is
in direct proportion to the length of
the arc. So we have
Q9
Length of arc AB
96
3
=
× circumference
360
96
=
× π × 8.6 = 7.204  km
360
3
so 15 km ↔ 51 microsec
51
km ↔
microsec
⇒ 1 51
microsec
⇒ 1 km ↔
15
3
15
51
⇒ 7.204
51 ↔. × 7.204  microsec
⇒ 7.204  ↔
× 7 204
15  microsec
15
= 24 ⋅ 496
=
24
⋅
496
The proton takes
24·5 microsec
3
to travel from A to B in the
tunnel.
96°
arc AB
360°
whole circumference.
96
as the required
360
fraction of the circumference that
gives the length of arc AB.
This leads to
•T
his mark is for the appearance of
96
or equivalent.
360
Arc length
•C
= πD where D = 2 × radius
= 2 × 4·3 km.
Method
•A
proportion problem: Divide by 15
then multiply by 7·204···. Evidence
that you knew this will gain you this
mark.
4 marks
Solution
•C
orrect calculation gains the final
mark.
Notes: 2·4 page 17
Q10(a)
(x − 1) cm
Expression
•N
otice that, since AEDF is a rhombus,
then ED = AF = 1 cm.
3
1 mark
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Worked Answers to Practice Exam F, Credit Mathematics
Q10(b)
B
F x−1
x
Strategy
•A
lternatively the ratios of
corresponding sides may be used
leading to:
C
1
D
E
1
D
D = x cm since it is a diagonal
B
of the pentagon.
DC = 1 cm since it is one of the
edges of the pentagon
The enlargement scale factor 3
from ∆DFC to ∆BDE is x
(BD = x × DC)
So x × CF = DE
⇒ x(x – 1) = 1
⇒ x2 – x = 1
⇒ x2 – x – 1 = 0.
x −1 1
x − 1 1both ⇒ x(x − 1) = 1 sides by x) ⇒ x(x − 1) = 1
x
1 = (multiply
= (multiply
= 1 (multiply
both
sides by xboth
− 1) ⇒ x(x − 1) = 1.sides by x) ⇒ x(x − 1) = 1
x
x
1
x
1 x 1− 1
=
(multiply both sides by x − 1).⇒ x(x − 1) = 1.
1 x −1
x
1
x
1
sides
or = both(multiply
=
(multiply
sides by xboth
− 1) ⇒
x(xby
− 1x) =− 1). ⇒ x(x − 1) = 1.
x
1
1 xboth
− 1 sides by x − 1).⇒ x(x − 1) = 1.
1 = x − 1 (multiply
1 x −1
Process
•T
his mark is for correctly applying
the scale factor (or the above working)
leading to x(x − 1) = 1.
Proof
• I t is vital that the step x2 − x = 1 is
written out in full. When you have a
‘show that’ question where the final
answer (in this case x2 − x − 1) is
given, all steps in the working should
be shown. The marker has to know
that you understood all the steps.
3
3
Notes: 2·2 page 15
3 marks
Q11(a)
d = 7 – 3sin(30h)° with h = 3
⇒ d = 7 – 3sin90°
⇒d=7–3×1=7–3=4
The depth is 4 metres.
Substitution
•Y
ou have to translate 3 pm into
h = 3 – although the wording of the
question makes this easier: “3 hours
after noon”.
3
Solution
•T
his mark is for correct calculation
of 4.
3
2 marks
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Worked Answers to Practice Exam F, Credit Mathematics
Q11(b)
You require d = 5
So 7 – 3sin(30h)° = 5
⇒ 7 – 5 = 3sin(30h)°
⇒ 3 sin(30h)° = 2
2
⇒ sin (30h)° =
3

−1 2 
So 30h = sin  3 
Equation
•T
his is the reverse process of part
(a). You now know the answer after
substitution, i.e. d = 5 but you do
not know the value of h that was
substituted to give you 5.
3
•T
his mark is for 7 − 3sin(30h)° = 5.
Rearrangement
•C
ompare 7 – 3x = 5 ⇒ 7 − 5 = 3x
3
⇒ 2 = 3x
⇒ 30h =41·81...
41.81
⇒h=
= 1.3936
30
3
This is 1·3936··· hours after noon
now 0·3936··· × 60 = 23·62...
.
=. 24 minutes
This is 1 hr 24 minutes after noon,
i.e. 0124 hours.
⇒x=
2
3
Solution
•T
his mark would be gained for
1·39, changing to minutes is not
necessary (but does produce a more
understandable answer!)
 2
•T
he ‘angle’ is 30 h and sin−1   gives
 3
41·81··· so to find h, division by 30 is
necessary.
3 marks
Notes: 3·5 page 27
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