Cambridge O Level

MATHEMATICS (SYLLABUS D) 4024/22

Paper 2 May/June 2020
MARK SCHEME
Maximum Mark: 100

Published

Students did not sit exam papers in the June 2020 series due to the Covid-19 global pandemic.

This mark scheme is published to support teachers and students and should be read together with the
question paper. It shows the requirements of the exam. The answer column of the mark scheme shows the
proposed basis on which Examiners would award marks for this exam. Where appropriate, this column also
provides the most likely acceptable alternative responses expected from students. Examiners usually review
the mark scheme after they have seen student responses and update the mark scheme if appropriate. In the
June series, Examiners were unable to consider the acceptability of alternative responses, as there were no
student responses to consider.

Mark schemes should usually be read together with the Principal Examiner Report for Teachers. However,
because students did not sit exam papers, there is no Principal Examiner Report for Teachers for the June
2020 series.

Cambridge International will not enter into discussions about these mark schemes.

Cambridge International is publishing the mark schemes for the June 2020 series for most Cambridge
IGCSE™ and Cambridge International A & AS Level components, and some Cambridge O Level
components.

This document consists of 8 printed pages.

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Generic Marking Principles

These general marking principles must be applied by all examiners when marking candidate answers. They
should be applied alongside the specific content of the mark scheme or generic level descriptors for a question.
Each question paper and mark scheme will also comply with these marking principles.

GENERIC MARKING PRINCIPLE 1:

Marks must be awarded in line with:

• the specific content of the mark scheme or the generic level descriptors for the question
• the specific skills defined in the mark scheme or in the generic level descriptors for the question
• the standard of response required by a candidate as exemplified by the standardisation scripts.

GENERIC MARKING PRINCIPLE 2:

Marks awarded are always whole marks (not half marks, or other fractions).

GENERIC MARKING PRINCIPLE 3:

Marks must be awarded positively:

• marks are awarded for correct/valid answers, as defined in the mark scheme. However, credit is given for
valid answers which go beyond the scope of the syllabus and mark scheme, referring to your Team
Leader as appropriate
• marks are awarded when candidates clearly demonstrate what they know and can do
• marks are not deducted for errors
• marks are not deducted for omissions
• answers should only be judged on the quality of spelling, punctuation and grammar when these features
are specifically assessed by the question as indicated by the mark scheme. The meaning, however,
should be unambiguous.

GENERIC MARKING PRINCIPLE 4:

Rules must be applied consistently e.g. in situations where candidates have not followed instructions or in the
application of generic level descriptors.

GENERIC MARKING PRINCIPLE 5:

Marks should be awarded using the full range of marks defined in the mark scheme for the question
(however; the use of the full mark range may be limited according to the quality of the candidate responses
seen).

GENERIC MARKING PRINCIPLE 6:

Marks awarded are based solely on the requirements as defined in the mark scheme. Marks should not be
awarded with grade thresholds or grade descriptors in mind.

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Mathematics Specific Marking Principles

1 Unless a particular method has been specified in the question, full marks may be awarded for any correct
method. However, if a calculation is required then no marks will be awarded for a scale drawing.

2 Unless specified in the question, answers may be given as fractions, decimals or in standard form. Ignore
superfluous zeros, provided that the degree of accuracy is not affected.

3 Allow alternative conventions for notation if used consistently throughout the paper, e.g. commas being
used as decimal points.

4 Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working
following a correct form of answer is ignored (isw).

5 Where a candidate has misread a number in the question and used that value consistently throughout,
provided that number does not alter the difficulty or the method required, award all marks earned and
deduct just 1 mark for the misread.

6 Recovery within working is allowed, e.g. a notation error in the working where the following line of
working makes the candidate’s intent clear.

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Abbreviations

cao correct answer only

dep dependent
FT follow through after error
isw ignore subsequent working
oe or equivalent
SC Special Case
nfww not from wrong working
soi seen or implied

Question Answer Marks Partial Marks

1(a) 6 2 22790 − 21500

M1 for [×100] oe or
21500
22790
× 100 oe
21500

1(b) 1354.5[0] 2 3 × 2.5

M1 for [1260 + ] × 1260 oe
100

1(c) 63[.00] cao 2 4300

M1 for soi
67.8

2(a) w + 6 + w + w + 6 + w = 37 oe M2 M1 for length = w + 6

6.25 B1

2(b) 295.5 3 B1 for two of 28.5, 15.5, 19.5, 7.5

M1 for their (28.5 × 15.5) – (19.5 × 7.5)

3(a) 33.7 or 33.69… 2 8

M1 for tan =  oe
12

3(b) 8.51 or 8.513… 2 8

M1 for sin 70 =  oe
QL

3(c) 1.3 2 1.6 8

M1 for = oe
TS 6.5

4(a)(i) 15 1

4(a)(ii) Correct histogram 3 B1 for three or more rectangles on

correct base
B1 for three or more correct frequency
densities soi

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Question Answer Marks Partial Marks

4(b)(i) 4.32 oe 3 M2 for

1× 8 + 3 × 16 + 5 × 15 + 7 × 7 + 9 × 4
8 + 16 + 15 + 7 + 4
or M1 for
1 × 8 + 3 × 16 + 5 × 15 + 7 × 7 + 9 × 4

or
 fc
8 + 16 + 15 + 7 + 4

4(b)(ii) Correct cumulative frequency curve 3 B2 for four or five correct points plotted
or B1 for correct cumulative frequencies
soi

4(b)(iii) 4 to 4.2 1

5(a) Correct method to eliminate one variable M2 M1 for attempt to equate coefficients or
attempt to make x or y the subject of one
equation

x = 0.5 oe, y = –2.5 oe A2 A1 for either x = 0.5 oe or y = –2.5 oe

After A0, SC1 for a pair of values that

satisfy either equation or for correct
answers with no working

5(b) 2x2 + 3x – 24 = 0 oe B1

−3 ± 32 − 4 × 2 × −24 B2 FT their 3-term quadratic

2× 2 B1FT for 32 − 4 × 2 × −24
−3 ± p
or for
2× 2

2.79, –4.29 B1

5(c)(i) 36 2 k
M1 for h = oe or [k =] 4.5 × 23
g3 g3

5(c)(ii) 1.5 oe 2 their 36 × 3

M1 for g 3 = oe
32

6(a)(i) 1 1
0.2 or oe
5

6(a)(ii) 23, 43, 53 2 B1 for three correct and one incorrect or

for two correct and none incorrect

6(a)(iii) 6 2 6
0.3 or oe B1 for where k is an integer > 6 or
20 k
for 24, 32, 36, 52, 56 and 64 identified

6(b)(i) 35, 22, 38 1

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Question Answer Marks Partial Marks

6(b)(ii) 77 2 46 + 31
or 0.385 B1 for where k is an integer
200 k
> 77
105
or SC1 for or 0.525
200

6(b)(iii) Large sample B1

46 1 B1
is a lot bigger than oe or
200 6
22 1
is a lot smaller than oe
200 6

7(a)(i) 1, 2 1

7(a)(ii) Correct curve 3 B2FT for 6 or 7 points correctly plotted

or B1FT for 4 or 5 points correctly
plotted

7(a)(iii) Tangent drawn at (2, 16) B1

18 to 27 B1 Dependent on correct tangent or close

attempt

7(a)(iv)(a) a = –60, b = 36 2 B1 for either correct or

3(4x) – 60x + 36 [= 0]

7(a)(iv)(b) y = 20x – 12 ruled line M2 M1 for one correct coordinate soi

0.7 to 0.8, 2.65 to 2.75 B1

7(b) p = 1 B1

q = 9 B2 M1 for [y =] (4 – x)(x + 2) oe or

[y =] q – (x – 1)2 oe or
two correct equations in x and y using
(–2, 0), (4, 0) or (0, 8)
or SC1 for q = –9

8(a) 376.99 to 377.04 2 M1 for π × 102 × figs12

8(b) 767 or 766.5 to 766.6… 3 M2 for

π × 102 + π × 2 × 10 × (3 + 3 + figs 12)
or M1 for π × 102 or
π × 2 × 10 × (3 + 3 + figs 12)

8(c) 28.79 to 28.80… 3 x

M2 for 200 =   × π × 10.32 × 7.5
360
x
or M1 for × kπ used
360

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Question Answer Marks Partial Marks

9(a) 63 B1

MLN ˆ  = 54 angles on a straight line B1

or LMNˆ + MNL ˆ = 126 exterior angle of
a triangle

ˆ =  63 or NML
MNL ˆ = 63 angles in B1
isosceles (triangle)

MNL ˆ = FMN ˆ alternate angles B1

or
GLMˆ = LMF ˆ alternate angles
or
EMLˆ = 54 co-interior with GLMˆ (or
ˆ ) and angles on a
alternate with MLN
straight line

9(b)(i) y 1
oe
2

9(b)(ii) x 1

9(b)(iii) 90 + x oe 2 FT 90 + their (a)(ii)

ˆ = 90° or BDC
B1 for BAC ˆ = 90°

9(b)(iv) y 1
90 – x –  oe
2

10(a) 2
 10   6 
2 M2 M1 for AC 2 = 102 + 6 2 oe
FC 2 =   +   oe
 2  2 2
2 6
6 or EX 2 +    = 122
or EX 2 = 122 –    2
2
2
 10  2
or EY 2 = 122 –     10 
 2 or EY 2 +    = 122
 2

 EF 2 =  122 – their FC 2 oe M1 Dep on M2
 
2
 10 
or  EF 2 =  theirEX 2 –   
 2
2
6
or  EF 2 =  theirEY 2 –   
2

10.49 or 10.488… A1

10(b) 210 or 209.7 to 209.8 2 1

M1 for × 6 × 10 × 10.5
3

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Question Answer Marks Partial Marks

10(c) 29.0 or 28.95 to 28.96 3 122 + 122 − 62

M2 for [cos =]  or
2 × 12 × 12
 3
2sin–1  
 12 
or M1for 62 = 122 + 122 –
 2 × 12 × 12 cosE or
3
sin = 
12

10(d) 34.9 or 34.85 to 34.86 2 FT 72sin(their (c))

1
M1 for  × 12 × 12 × sin(their 29.0) or
2
1
122 − 32 × 6 oe
2

11(a) Reflection y = –1 2 B1 for either

11(b)  0 −1  2 B1 for one correct row or column or

  Rotation, 90o anticlockwise about (0, 0)
1 0

11(c) Triangle with vertices 3 B2 for triangle with correct orientation

(–1, –2) (0, –2) (0, –5) but wrong position
or B1 for triangle reflected in x = 2

After 0, SC1 for triangle with vertices

(–3, –2) (–4, –2) (–4, 1)

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