Cambridge IGCSE™

CAMBRIDGE INTERNATIONAL MATHEMATICS 0607/42

Paper 4 (Extended) March 2021
MARK SCHEME
Maximum Mark: 120

Published

This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the
examination. It shows the basis on which Examiners were instructed to award marks. It does not indicate the
details of the discussions that took place at an Examiners’ meeting before marking began, which would have
considered the acceptability of alternative answers.

Mark schemes should be read in conjunction with the question paper and the Principal Examiner Report for
Teachers.

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

Cambridge International is publishing the mark schemes for the March 2021 series for most Cambridge
IGCSE™, Cambridge International A and AS Level components and some Cambridge O Level components.

This document consists of 9 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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Maths-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.

MARK SCHEME NOTES

The following notes are intended to aid interpretation of mark schemes in general, but individual mark schemes
may include marks awarded for specific reasons outside the scope of these notes.

Types of mark

M Method marks, awarded for a valid method applied to the problem.

A Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. For accuracy
marks to be given, the associated Method mark must be earned or implied.

B Mark for a correct result or statement independent of Method marks.

When a part of a question has two or more ‘method’ steps, the M marks are in principle independent unless the
scheme specifically says otherwise; and similarly where there are several B marks allocated. The notation ‘dep’
is used to indicate that a particular M or B mark is dependent on an earlier mark in the scheme.

Abbreviations

awrt answers which round to

cao correct answer only
dep dependent
FT follow through after error
isw ignore subsequent working
nfww not from wrong working
oe or equivalent
rot rounded or truncated
SC Special Case
soi seen or implied

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Question Answer Marks Partial Marks

1(a)(i) correct triangle B 2 B1 for 90° rotation about wrong centre

(0, 3) (0, 8) (−3, 3)

1(a)(ii) correct triangle C 1

(−3, 0) (−8, 0) (−3, 3)

1(a)(iii) reflection 2 B1 for each

𝑥 + 𝑦 = 0 oe

1(b) enlargement 3 B1 for each

[scale factor] 3
[centre] (0, 0) oe

2(a) 261 000 1

2(b) 5.76 × 10−7 1

2(c) 26.7 2 B1 for 26.68 to 26.69 or answer 26.6

2(d)(i) 303.4[0] cao final answer 2 37 × 820

M1 for oe soi by 303
100

2(d)(ii) 24 1

2(e)(i) 2085 3 695

M1 for soi
5
M1 for (their 139) × (3 + 5 + 7)

2(e)(ii) 295.09 3 M2 for 0.4 × 695 × 1.0125 oe

or M1 for 0.4 × 695 soi by 278
or A × 1.0125

2(f) 2[.00] or 1.998 to 2.001... 3 2663.31

M2 for 12 oe
2100
or M1 for 2100 × r12 = 2663.31 seen

3(a)(i) (0, 3) 1

3(a)(ii) –2 1

3(b) 8 1

3(c) 2 3 2
y= x − 3 oe final answer B2 for answer x −3
3 3

OR

5 − (−1)
M1 for oe
12 − 3
M1 for correct substitution of point into
y = (their m)x + c or e.g. y – 5 = (their m)(x – 12)

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Question Answer Marks Partial Marks

3(d) 2 26 4 2 26
y =− x+ oe final answer B3 for answer − x+ oe
5 5 5 5

OR

5
M1 for gradient
2
−1
M1 for m = or better
their ( 52 )
M1 for (3, 4) substituted into y = (their m)x + c
or e.g. y – 4 = (their m)(x – 3)

3(e)(i) 3 correct ruled lines 3 B1 for each line correct

3(e)(ii) Clear indication of correct 1 FT if appropriate

region

4(a)(i) 110 < m ≤ 130 1

4(a)(ii) 135.2 2 M1 for mid-values seen or implied

4(b)(i) (8) 16 38 77 104 113 120 2 B1 for 4 or 5 correct

FT one error

4(b)(ii) Correct cumulative frequency 3 B2 for 6 points correct

curve
OR

B1FT for 7 heights correct

B1 for plotting at upper boundary of interval

4(b)(iii)(a) 124 to 127 nfww 1

4(b)(iii)(b) 14 to 21 2 B1 for [LQ =] 115 to 118 or [UQ =] 132 to 136

5(a) 103 or 103.3 to 103.4 2 M1 for 492 + 912 oe

5(b) 85.2 or 85.17 to 85.18 2 305

M1 for × π × 2 × 16
360

5(c) 339 or 339.2 to 339.3… 2 1

M1 for × π × 62 × 12
4

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Question Answer Marks Partial Marks

5(d)(i) 1 M1
(x – 3)(x + 1) + (x – 3)(2x + 4)
2
[=11]

x2 – 3x + x – 3 B1 one correct expansion seen

1
or   (2x2 – 6x + 4x – 12)
2
or x2 – 3x + 2x – 6

At least one more line of A1 no errors or omissions

working leading to
2x2 – 3x – 20 = 0

5(d)(ii) (2x + 5)(x – 4) 2 M1 for (2x + a)(x + b) where ab = –20 or a + 2b = –3

or 2x(x – 4) + 5(x – 4)
or x(2x + 5) – 4(2x + 5)

5(d)(iii) 4 , –2.5 1 Strict FT their factors

Dep on factors in part (ii)

5(d)(iv) 12 1 FT 2 × (their positive root (d)(iii)) + 4

6(a)(i) 32 2 k
M1 for y =
x2 x2

6(a)(ii) 2 1 k
FT their k dependent on
x2

6(a)(iii) 1 2 their 32 1
[±] M1 for x 2 = soi by oe
2 128 4

6(b) 250 3 B2 for r = 2(p + 1)3

or M1 for r = k(p + 1)3 oe

OR
r 16
M2 for 3
= oe
(4 + 1) (1 + 1)3

7(a) Correct sketch 3 B2 for correct branches but joined or for ‘correct’ but
with excessive overlap or ‘curl back’
B1 for one correct branch

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Question Answer Marks Partial Marks

7(b) y=0 B2 B1 for each

x=2

7(c) –2.67 < x < 0.524 B2 B1 for x > −2.67 or x < 0.524
or –2.7 < x < 0.52

2 < x < 2.15 B2 B1 for either x > 2 or x < 2.145…

If B0, B0 scored, then SC1 for 2 of the boundaries

–2.67, 0.524, 2.15 seen

8(a) 218 1

8(b) 42 + 172 – 2 × 4 × 17 × cos142 M2 M1 for implicit cosine rule

20.30… A1

8(c) 007 or 006.92 to 006.98 3 4sin142

M2 for sin B = oe
20.3
4 20.3
or M1 for = oe
sin B sin142

OR

17sin142
M2 for sin A= oe
20.3
17 20.3
or M1 for = oe
sin A sin142

8(d) 11.5 or 11.47… 3 B1 for 3 h 36 min or 3.6 h seen

4 + 17 + 20.3
M1 for
their 3.6

9(a)(i) 15 nfww 3 M2 for 8x = 104 or better

or M1 for 3x + x + 2 + 4x + 1 + 8 [=115] oe
If 0 scored, SC1 for 16 as final answer

9(a)(ii) correct shading 1

9(a)(iii) ∈ 1

9(b)(i) 2 1
oe
9

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Question Answer Marks Partial Marks

9(b)(ii) 104 4 M3 for

oe 6 4 6 8 4 6 4 8 8 6 8 4
153 × + × + × + × + × + ×
18 17 18 17 18 17 18 17 18 17 18 17
oe
or M2 for 4 or 5 correct products added
or M1 for 2 or 3 correct products added

OR

6 12 4 14 8 10
M3 for × + × + ×
18 17 18 17 18 17
or M2 for 2 correct products added
or M1 for 1 correct product

OR
 6 5 4 3 8 7 
M3 for 1 –  × + × + × 
 18 17 18 17 18 17 
or M2 for 1 – (two correct products added)
or M1 for 1 – one correct product

52
If 0 scored SC1 for final answer oe
81

10(a) 8 : 19 oe 3
M1 for [Vol A : Vol B =] 23 : 33 oe

M1 for [Vol C =] 27k – 8k k any variable

OR
2
1  3r  3h
M1 for π   ×
3  2 2
1 2 1  19  2
M1 for [VA : VC =] πr h :   πr h
3 3 8 

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Question Answer Marks Partial Marks

10(b) 503 or 502.6 to 502.8 8 3 3

M1 for × 4 oe or ×10
2 2
3 32
or × their l oe if their l is from Pythagoras or 2
2 2

M2 for 4 2 + 10 2 or (their R)2 + (their H )2

or M1 for 4 2 + 10 2 or (their R)2 + (theirH)2

M1 for π× 4 × 116

3 32
M1 for π× 6 × 116 or 2 × π × 4 116
2 2

M2 for CSAa + CSAb + π × (their R)2 – π × 42 oe

or M1 for for CSAa + CSAb or π × (their R)2 – π × 42

oe

11(a) 4 1

11(b) 28 2 B1 for f(32) seen

or M1 for 3 × 3x + 1 oe

11(c) −
1 oe 2 M1 for 3r + 1 = r
2

11(d) 4 3 M1 for (3x + 1)2 – 5

oe , −2 M1 for (3x + 1) = ±5 or [3](x + 2)(3x – 4) = 0 oe
3
or correct substitution in formula for 3x2 + 2x – 8 or
9x2 + 6x – 24
or correct and suitable sketch

11(e) log x 2 M1 for log y = log 3x oe or correct answer seen

log3 x or final answer
log 3 or x = 3 y or log 3 y = x

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