Mathematics
Mark scheme
Ch 1
80 marks
Higher
Set A Paper 3 (Calculator)
Answer
Mark
Mark Scheme
Commentary
1
14.80
1
Award 1 mark for correct answer.
A common error is to ignore the zero in the 2nd
decimal place and write/choose 14.8
2
1
3
1
Award 1 mark for correct answer.
A common error is 3, the enlargement that takes B to
A.
3
1 : 0.625
1
Award 1 mark for correct answer.
Some students do not realise they can have noninteger values in ratios.
4
a
(180 – 32) ÷ 2
= 74°
180 – 74 = 106
(180 – 106) ÷ 2
= 37°
5
a
false
false
true
3
Award 1 mark for each correct answer.
These statements highlight some common
misconceptions. Some students may think there is
some truth in statement B, at least in a “theoretical”
sense. This is an opportunity to emphasise that two
different outcomes do not have to be equally likely.
4
Award 1 mark for 7.5 seen.
Award 1 mark for 152 – 7.52 (= 168).
Award 1 mark for their 168.75
Students may forget to halve the base of the
equilateral triangle. A common error is to add rather
than subtract the squares of the sides.
3
Award 3 marks for 180 (accept without
units).
Award 1 mark for attempt to find LCM
of 20 and 36.
Award 1 mark for systematic listing of
multiples of either 36 or 20.
Some students will not recognise that this is an LCM
question. They may find the solution by writing out
the sequence of time intervals.
b
5
6
7
a
b
c
152  7.52  13 cm
36, 72, 108, 144, 180, 216,...
20, 40, 60, 80, 100, 120, 140, 160, 180
LCM of 20 and 36 is 180
180 milliseconds
© Oxford University Press 2015
Acknowledgements: www.oxfordsecondary.co.uk/acknowledgements
b
Award 2 marks for 74°.
Award 1 mark for
(180 – 32) ÷ 2 or equivalent.
Award 3 marks for 37°.
Award 1 mark for 106 seen.
Award 1 mark for
(180 – „their 106‟) † 2
In part b the method given in the mark scheme is the
one students are most likely to use. A more efficient
method would use „exterior angle of triangle = sum
of opposite interior angles‟, leading directly to
dividing 74 by 2.
Higher Set A Paper 3 (Calculator) – Mark scheme
Answer
8
7x + 6 = 3(x + 4)
13
7x + 6 = 3x + 12
Mark
Mark Scheme
Commentary
4
Award 1 mark for 7x + 6 = 3x + 12
Award 1 mark for correct next step to
collect number or algebra terms
e.g. 4x + 6 = 12 or 7x = 3x + 6 or
7x – 3x = 12 – 6
Award 1 mark for x = 1.5
Award 1 mark for 7x + 6 = 16.5 cm
Often students will stop at the solution x = 1.5 and
forget to work out the side length of the square.
3
a
4x + 6 = 12
4x = 6
x = 1.5
7x + 6 when x = 1.5 = 7 × 1.5 + 6 = 16.5cm
9
a
b
150
 20  7.5 , you can‟t have half a person,
400
and if you round up to 8 boys and 8 girls there
won‟t be enough adults in the sample. If you
round down, there won‟t be enough children.
Change the sample size, have one less/extra
adult in the sample so that there are 16 or 14
i.e. an even number of boys and girls.
b
Award 1 mark for a correct
explanation and 1mark for a
correct supporting calculation
which does not have to be the
1
one above (e.g. of the sample
4
are adults, which is 5. You need
the same number of boys and
girls and 15 is an odd number.)
Award the mark for any sensible
suggestion to get around the
problem, others are possible, e.g.
round the boys up to 8 and the
girls down to 7 (or vice versa).
In part a a common problem is to have incomplete
explanations, e.g. the supporting calculation will be
missing or there will only be a calculation and no
concluding statement.
10
Double (or × 2) 16 32
Add the sequence of even numbers, or add 2, then 4,
then 6, then 8, … 14 22
4
Award 1 mark for each rule and 1
mark for each pair of numbers that
satisfy their rules.
Students often find it difficult to see more than one
possible rule for a sequence.
11
Area of large circle = ᴨ × 32 = 28.27... 1
Area of small circle = ᴨ × 1.52 = 7.06…
28.27… – (2 × 7.06…)
= 14.14…
= 14 cm2
5
Award 1 mark for Area of large circle
= ᴨ × 32 (= 28.27...)
Award 1 mark for Area of small circle
= ᴨ × 1.52 (= 7.06…)
Award 1 mark for „their 28.27‟ – (2 ×
„their 7.06‟)
Award 1 mark for 14.14…
Award 1 mark for sensible degree of
accuracy, e.g. the nearest whole
number.
Students may use the wrong formula for area of a
circle. A common error is to use the diameters 6 cm
and 3 cm instead of the radii. A suitable degree of
accuracy is to the nearest whole number, to match
the measurement given in the question.
© Oxford University Press 2015
Acknowledgements: www.oxfordsecondary.co.uk/acknowledgements
Higher Set A Paper 3 (Calculator) – Mark scheme
Answer
It means there is a charge of 5 Egyptian pounds to
13
exchange money, no matter how much money is
Mark
12
Mark Scheme
Commentary
1
Award any equivalent sensible
interpretation of the intercept.
Students often have difficulty interpreting gradients
and intercepts in real-life contexts.
being exchanged.
13
× 6x × (2y + 9x)
= 6xy + 27x2
2
Award 1 mark for × 6x × (2y + 9x)
Students often find the area of a trapezium difficult,
even when given the formula.
14
B
1
Award 1 mark for correct answer.
Students may not realise that the equation of the
graph rearranges to y = - x2 + 3x, and only one of the
graphs drawn is for a quadratic curve with negative
x2 coefficient.
15
Least:
42.5  7.5
= 0.753 = 0.75 g/cm3 to 2sf
46.5
43.5  6.5
Greatest:
= 0.813 = 0.81 g/cm3 to 2sf
45.5
4
Award 1 mark for each correct
subtraction and each correct division
and rounding.
Some students may not understand that to find the
lower bound, you take the lowest possible starting
point (42.5), subtract the highest possible value (7.5)
and divide by the highest possible value (46.5). And
vice versa for the upper bound.
16
8% of £800 = 64
64
 0.05
x
64 ÷ 0.05 = £1280
2
64
 0.05 or 64 ÷
x
0.05 or equivalent.
A common error here is to calculate 5% of £800.
17
a
b
(2x – 3)(x + 5)
(2x – 7) (2x + 7)
3
Award 1 mark for
a
b
18
Circumference of tin = ᴨ × 8.5 = 26.7...
Length of label 26.7 + 1 = 27.7
27.7 × 8.8 = 243.8 (or 244) cm2
19
a
32 × £4.75 + 3 × £9.50 = £152 + £28.50 =
£180.50 < £210.00
© Oxford University Press 2015
Acknowledgements: www.oxfordsecondary.co.uk/acknowledgements
4
8
Award 2 marks for correct
In b some students may try to write (4x + a)(x + b) or
factorisation.
attempt to take out a single factor of 1 e.g. you may
Award 1 mark for (2x + a)(x + b) see the answer 1(4x2 – 49).
where a + 2b = 7 or ab = -15.
Award 1 mark for correct
factorisation.
Award 1 mark for π × 8.5 (= 26.7...)
Award 1 mark for „their 26.7‟ + 1
Award 1 mark for „their 27.7‟ × 8.8
a
Award 2 marks for £180.50
Award 1 mark for either
In surface area of cylinder problems, students do not
always realise that the circumference of the can is
the length of the rectangular section of the net.
Students often do not work systematically through a
problem like this, where there is a lot of information
Higher Set A Paper 3 (Calculator) – Mark scheme
Answer
b
13
c
d
20
a
b
Mark
£(210.00 – 180.50) ÷ 5 = £29.50 ÷ 5 = £5.90
4 × £4.75 = £19 is returned. There are 40 – 4 =
36 people now on the trip. They each pay 19 ÷
36 = 0.527777… = £0.53 extra.
Fuel cost = (132 ÷ 15) × 4.55 × £1.28 = 8.8 ×
4.55 × £1.28 = 40.04 × £1.28 = £51.25
Driver earns = 8 × £12 = £96
Total cost to company = £96 + 51.25 =
£147.25
Profit = £210 - 147.25 = £62.75
16 : 1
43 : 1 = 64 : 1
Mark Scheme
b
c
d
3
a
b
21
6
a
a
b
b
c
c
Commentary
32 × £4.75 or 3 × £9.50 seen.
to deal with. In d some may divide instead of
Award 2 marks for £5.90. Award multiply when converting gallons to litres, or may
1 mark for £29.50.
miss out this step completely.
Award 2 marks for £0.53 extra
per person.
Award 1 mark for £19 returned.
Award 2 marks for £62.75.
Award 1 mark for £51.25 for
fuel.
Award 1 mark for correct
answer.
Award 2 marks for 64 : 1.
Award 1 mark for 43 seen.
Some students may forget to square for the area
ratio, or may cube the area ratio for the volume ratio.
Award 2 marks for all 4 correct
or 1 mark for 2 or 3 correct.
Award 3 marks for a smooth
curve that passes through all 7
points and is a clear cubic shape.
Or award 2 marks for an attempt
at a cubic curve passing through
at least 5 points.
Or award 1 mark for correctly
plotting at least 4 of their points.
Award 1 mark for line y = -4
showing one intersection with
the curve.
Some students are careless when drawing curves,
common errors are to feather curves, miss points out,
use a blunt pencil. In part c some students may draw
the line x = -4, or y = 4 in error.
A horizontal line drawn at y = -4 showing one
intersection with the curve.
© Oxford University Press 2015
Acknowledgements: www.oxfordsecondary.co.uk/acknowledgements
Higher Set A Paper 3 (Calculator) – Mark scheme
Answer
13
22
17.3°
Mark
3
Mark Scheme
Commentary
Award 1 mark for tan(BAC ) 
2.3
7.4
2.3
)  17.2658...
7.4
Award 1 mark for 17.3
Award 1 mark tan 1 (
23
For the tin, V = π × 5² × h = 25πh
250π
For the bowl V = 23 × π × 53 =
3
250π
25πh =
3
h
24
4
Award 1 mark for 25πh.
250π
Award 1 mark for
3
Award 1 mark for equating their two
expressions for V.
Some students may choose the wrong ratio. Another
2.3
common mistake is for students to find tan( ) .
7.4
Some students may get as far as 25πh =
have difficulty solving for h.
250π
but
3
250  π 10
1
  3 cm (or 3.3 cm)
3  25  π 3
3
a
© Oxford University Press 2015
Acknowledgements: www.oxfordsecondary.co.uk/acknowledgements
5
Award 3 marks for curve fully
correct. Award 2 marks for all 5
correct plots or 1 mark if all
horizontal or vertical distances
are correct or 1 point is plotted
wrongly. Award 1 mark for
joining points with smooth,
increasing curve.
b, c Award 1 mark for each correct.
a
The most common errors are to plot frequencies
instead of cumulative frequencies or to use mid
points, or the start of the intervals, instead of the
ends of the class intervals. In b students may misread
the value from the graph. In c the most common
error is to find the percentage less than 90%, ie 80%.
Higher Set A Paper 3 (Calculator) – Mark scheme
Answer
b
13
c
Mark
Mark Scheme
Commentary
66
10
= 20%
50
Total 80 marks
© Oxford University Press 2015
Acknowledgements: www.oxfordsecondary.co.uk/acknowledgements
Higher Set A Paper 3 (Calculator) – Mark scheme