1. Specification B – Practice Paper A
Unit 3 Higher
Mark Scheme
GCSE
GCSE Mathematics (Modular)
Paper: 5MB3H_01
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Registered Office: One90 High Holborn, London WC1V 7BH

2. GCSE MATHEMATICS UNIT 3
HIGHER PRACTICE PAPER A MARKSCHEME
NOTES ON MARKING PRINCIPLES
1 Types of mark
M marks: method marks
A marks: accuracy marks
B marks: unconditional accuracy marks (independent of M marks)
2 Abbreviations
cao – correct answer only ft – follow through
isw – ignore subsequent working SC: special case
oe – or equivalent (and appropriate) dep – dependent
indep - independent
3 No working
If no working is shown then correct answers normally score full marks
If no working is shown then incorrect (even though nearly correct) answers score no marks.
4 With working
If there is a wrong answer indicated on the answer line always check the working in the body of the script (and on any
diagrams), and award any marks appropriate from the mark scheme.
If working is crossed out and still legible, then it should be given any appropriate marks, as long as it has not been
replaced by alternative work.
If it is clear from the working that the “correct” answer has been obtained from incorrect working, award 0 marks. Send
the response to review, and discuss each of these situations with your Team Leader.
If there is no answer on the answer line then check the working for an obvious answer.
Any case of suspected misread loses A (and B) marks on that part, but can gain the M marks. Discuss each of these
situations with your Team Leader.
If there is a choice of methods shown, then no marks should be awarded, unless the answer on the answer line makes
clear the method that has been used.
Paper: 5MB3H_01 2
Session: Practice Paper A

3. GCSE MATHEMATICS UNIT 3
HIGHER PRACTICE PAPER A MARKSCHEME
5 Follow through marks
Follow through marks which involve a single stage calculation can be awarded without working since you can check the
answer yourself, but if ambiguous do not award.
Follow through marks which involve more than one stage of calculation can only be awarded on sight of the relevant
working, even if it appears obvious that there is only one way you could get the answer given.
6 Ignoring subsequent work
It is appropriate to ignore subsequent work when the additional work does not change the answer in a way that is
inappropriate for the question: e.g. incorrect canceling of a fraction that would otherwise be correct
It is not appropriate to ignore subsequent work when the additional work essentially makes the answer incorrect e.g.
algebra.
Transcription errors occur when candidates present a correct answer in working, and write it incorrectly on the answer
line; mark the correct answer.
7 Probability
Probability answers must be given a fractions, percentages or decimals. If a candidate gives a decimal equivalent to a
probability, this should be written to at least 2 decimal places (unless tenths).
Incorrect notation should lose the accuracy marks, but be awarded any implied method marks.
If a probability answer is given on the answer line using both incorrect and correct notation, award the marks.
If a probability fraction is given then cancelled incorrectly, ignore the incorrectly cancelled answer.
8 Linear equations
Full marks can be gained if the solution alone is given on the answer line, or otherwise unambiguously indicated in
working (without contradiction elsewhere). Where the correct solution only is shown substituted, but not identified as
the solution, the accuracy mark is lost but any method marks can be awarded.
9 Parts of questions
Unless allowed by the mark scheme, the marks allocated to one part of the question CANNOT be awarded in another.
10 Use of ranges for answers
If an answer is within a range this is inclusive, unless otherwise stated.
Paper: 5MB3H_01 3
Session: Practice Paper A

4. GCSE MATHEMATICS UNIT 3
HIGHER PRACTICE PAPER A MARKSCHEME
5MB3H Practice paper A
Paper: 5MB3H_01 4
Session: Practice Paper A

5. GCSE MATHEMATICS UNIT 3
HIGHER PRACTICE PAPER A MARKSCHEME
Question Working Answer Mark Notes
*1
27065 – 25192 = 1873
1873 × 11.45 = 21445.85p
£214.4585
214.46 5
M1 for attempt to subtract the two meter readings
A1 for 1873
M1 for “1873” × 11.45
A1 21445.85
C1 for correctly interpreting their final answer as money
2
(a) 10 1 B1 cao
(b)(i) 10 00 1 B1 cao
(b)(ii) 30 1 B1 cao
(c) 11 20 1 B1 cao
(d) 24 2
M1 for attempt to relate a correct distance and time e.g.
12 km in 30 minutes
A1 cao
3
3x + 3x + x + 2 + x + 2 = 44
8x + 4 = 44
8x = 40
x = 5
Area = 15 × 5
75 5
M1 for attempt to form equation in x equating to 44 with
sight of 3x or x + 2
M1 for establishing 8x + 4 = 44 oe
A1 for x = 5
M1 for 3 × “5” × “5”
A1 cao
4
(a)
Correct
rotation
2
M1 for their rotated triangle in the correct orientation
A1 for their rotated triangle in the correct position
(b)
Correct
translation
2
B1 for triangle moved 6 cm right
B1 for triangle moved 1 cm down
(c)
Reflection
in y = x
2
B1 for reflection
B1 for y = x oe
5
33
+ 2 × 3 = 33
4³ + 2 × 4 = 72
3.53
+ 2 × 3.5 = 49.875
3.63
+ 2 × 3.6 = 53.856
3.7³ + 2 × 3.7 = 58.053
3.8³ + 2 × 3.8 = 62.472
3.75³ + 2 × 3.75 = 60.234375
3.7 4
B2 for a trial between 3.7  x  3.8 inclusive
(B1 for a trial between 3 x 4 evaluated)
B1 for a different trial 3.7 < x  3.75
B1 (dep on at least one previous B1) for 3.7
Accept trials correct to 3 d.p. rounded or truncated.
NB: no working scores, no marks even if answer is correct.
Question Working Answer Mark Notes
Paper: 5MB3H_01 5
Session: Practice Paper A

6. GCSE MATHEMATICS UNIT 3
HIGHER PRACTICE PAPER A MARKSCHEME
6
(a)
–4, –3, –2, –1, 0,
1, 2
2
B2 for –4, –3, –2, –1, 0, 1, 2
B1 –4, –3, –2, –1, 0, 1 with inclusion of –5 or omission of
2
(b) –2 ≤ x < 4 2
B2 for –2 ≤ x < 4
B1 for sight of –2 and 4 with one sign correct
7
85 × 2 = 170
20
170
100
× = 34
Tyres cost £204
Fitting costs 18 +
20
18
100
× = 18+3.60
= 21.60 Total = 204 + 21.60
225.60 4
M1 for attempt to find 20% of 85 or 170 or 18
M1 for adding one 20% to one cost
M1 for complete method
A1 cao
8
2 m³ = 2 × 100 × 100 × 100
= 2 000 000 cm³
2 000 000 ÷ 1000 =
2000 3
M1 for attempt to change 2 m³ into cm³ by multiplying
by 100 × 100 × 100 oe
M1 for attempt to change “2 000 000” into litres by
dividing by 1000
A1 cao
9
8² + 16² = 64 + 256 = 320
√320 = 17.88854382
17.89 3
M1 for 8² + 16²
M1 for √(8² + 16²)
A1 for 17.89 accept 17.8885(4382)
10
186 000 × 60 × 60 × 24 × 365.25
= 5 869 713 600 000
5.87 × 1012
3
M1 for 186 000 × 60 × 60 × 24 × 365.25
A1 for 5 869 713 600 000 oe
A1 for 5.87 × 1012
11
(a) 9, –3, 3 2
B2 for all 3 missing values correct
(B1 for one missing value correct)
(b) Drawing graph 2
B2 for fully correct curved line
(B1 for 6 of their points correctly plotted with attempt
to join with curves)
(c)
x = 4 and
x = –1
2
M1 for attempt to draw line across at y = 3 or attempt
to factorise
A1 for x = 4 and x = –1
Question Working Answer Mark Notes
Paper: 5MB3H_01 6
Session: Practice Paper A

7. GCSE MATHEMATICS UNIT 3
HIGHER PRACTICE PAPER A MARKSCHEME
12
(a)
Ratio of lengths is 3 : 12 or 1:4
2 × 4 =
8 2
M1 for establishing the ratio is 3 : 12 or 1:4 or 2 × 4
A1 for 8
(b)
10 ÷ 4 = 2.5
2.5 + 10
12.5 3
M1 for using the correct ratio or 10 ÷ 4
M1 for “2.5” + 10
A1 cao
13
Half base = 10 × cos 50°= 6.427876
Height = 10 × sin 50°= 7.6604444
Area = ½ × 6.427876 × 7.6604444
= 24.62019383
Alt
Base of triangle = 2 × 10 × cos 50
= 12.85575
Area = ½ 10 × 12.85575 × sin 50°
24.6 6
M1 for attempt to find the height of the triangle or
sight of cos 50°or for attempt to find half the base or
sight of sin 50°
M1 for 10 × cos 50 or 10 × sin 50
M1 for 10 × cos 50 and 10 × sin 50°
A1 for 6.42 or 7.66 or 12.85575
M1 for attempt to find area of triangle either by half
base × ht or by half a × b × sin 50°
A1 cao
14
70 % = 4900
1% = 70
100% = 70 × 100
7000 3
M1 for establishing that 4900 is equivalent to 70% of
original price
M1 for 4900 ÷ 0.7 or 4900 ÷ 70 × 100
A1 for 7000
15
Vol ratio is 27 : 125
Length ratio is 3 : 5
Area ratio is 9 : 25
Large area is 36 ÷ 9 × 25
Alt
Area scale factor is 25 ÷ 9
Large area = 36 × 25 ÷ 9 oe
100 3
M1 for attempt to cube root 27 and 125 or sight of 3 : 5
M1 for establishing the area ratio is 9:25 or scale factor
is 25÷ 9
A1 for 100
*16
(a)(i)
(ii)
6.75
6.65
2
B1 accept 6.749999…
B1 cao
(b)
Upper bound = 26.95 ÷ 6.65 = 4.05263
Lower bound = 26.85 ÷ 6.75 = 3.97778
4.05263
3.97778
3
M1 for attempt to divide Upper bound of area by lower
bound of length
M1 for attempt to divide lower bound of area by upper
bound of length
C1 for upper bound is 4.05 and lower bound is 3.98
(c)
4 with correct
reason 2
B1 for selecting 4
C1 for reason that states that 4.05 and 3.98 are only
the same number to the nearest whole i.e. 4 cm
Question Working Answer Mark Notes
Paper: 5MB3H_01 7
Session: Practice Paper A

8. GCSE MATHEMATICS UNIT 3
HIGHER PRACTICE PAPER A MARKSCHEME
*17 3
M2 for attempt to draw a circle centre (0, 0) with
radius 4
(M1 for attempt to draw a closed curve either with
centre (0, 0) or with radius 4)
C1 for reason that states that the point (1,2) lies inside
the region contained by the curve defined by
x² + y² = 16 and any straight line or line extended
would have to cut the curve twice, once on each side of
the point oe
18
9
5
3
4
3
3
2
−
=
−
−
+ x
x
xx
3(x – 3) – 4(x + 3) = 5x
3x – 9 – 4x – 12 = 5x
–x – 21 = 5x
–21 = 6x
–3.5 4
M1 for multiplying both sides by (x + 3)(x – 3) to get
3(x – 3) – 4(x + 3) = 5x
(M1 for attempt to multiply by 3 by (x +3) or 4 by (x – 3)
M1 dep for attempt to move variables to one side and
constant terms to the other
A1 for –3.5
Paper: 5MB3H_01 8
Session: Practice Paper A