Mathematics 0580/43 October/November 2020: question paper and mark scheme
Cambridge IGCSE Mathematics (0580), Paper 4 (Extended), October/November 2020. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
(a) The Earth has a surface area of approximately 510 100 000 km2.
(i) Write this surface area in standard form.
…… km2 [1]
(ii) Water covers 70.8% of the Earth’s surface.
Work out the area of the Earth’s surface covered by water.
…… km2 [2]
(b) The table shows the surface area of some countries and their estimated population in 2017.
Country Surface area (km2) Estimated population in 2017
Brunei 5.77 × 103 433 100
China 9.60 × 106 1 388 000 000
France 6.41 × 105 67 000 000
Maldives 3.00 × 102 374 600
(i) Find the total surface area of Brunei and the Maldives.
…… km2 [1]
(ii) The ratio surface area of the Maldives : surface area of China can be written in the form 1 : n.
Find the value of n.
n = …… [2]
(iii) Find the surface area of France as a percentage of the surface area of China.
…… % [2]
(iv) Find the population density of the Maldives.
[Population density = population ÷ surface area]
…… people/km2 [2]
(c) The population of the Earth in 2017 was estimated to be 7.53 × 109.
The population of the Earth in 2000 was estimated to be 6.02 × 109.
(i) Work out the percentage increase in the Earth’s estimated population from 2000 to 2017.
…… % [2]
(ii) Assume that the population of the Earth increased exponentially by y% each year for these
17 years.
Find the value of y.
y = …… [3]
Mark scheme
1(a)(i) 5.101[00…] × 108 final answer 1
1(a)(ii) 361 150 800 oe 2 FT their (a)(i)
70.8
M1 for × 510 100 000
100
70.8
or for × their a(i)
100
1(b)(i) 6070 oe 1
1(b)(ii) 32 000 oe 2 B1 for figs 32
1(b)(iii) 6.68 or 6.677 … 2 6.41× 105
M1 for [× 100] oe
9.6[0] × 106
1(b)(iv) 1250 or 1248 to 1249 oe 2 B1 for figs 125 or figs1248 to figs 1249
1(c)(i) 25.1 or 25.08… 2 7.53 ×109 − 6.02 ×109
M1 for oe
6.02 ×109
7.53 ×109
or × 100
6.02 ×109
1(c)(ii) 1.33 or 1.325… 3 7.53[×109 ] 17 their (c)(i)
M2 for 17 9
or 1 +
6.02[×10 ] 100
or M1 for 6.02[×10 ] × p = 7.53[×109]
9 17
their (c)(i)
or p17 = 1 +
100
y
8
7
6
5
4
A B
3
2
1
–6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 x
–1
–2
–3
–4
–5
–6
(a) On the grid, draw the image of
(i) triangle A after a rotation of 90° anticlockwise about (0, 0), [2]
3
(ii) triangle A after a translation by the vector e o. [2]
-5
(b) Describe fully the single transformation that maps triangle A onto triangle B.
…… [3]
Mark scheme
2(a)(i) Triangle at (–3, 2) (–3, 3) (–5, 2) 2 B1 for correct rotation about incorrect
point
or for rotation 90 clockwise around (0, 0)
2(a)(ii) Triangle at (5, –2) (6, –2) (5, 0) 2 3 k
B1 for translation by or
k −5
2(b) Enlargement 3 B1 for each
[SF] 3
[Centre] (1, 4)
(a) The average speeds, in km/h, of cars travelling along a road are recorded.
The box-and-whisker plot shows this information.
40 45 50 55 60 65 70 75 80 85
Speed (km/h)
Find
(i) the lowest speed recorded,
…… km/h [1]
(ii) the median,
…… km/h [1]
(iii) the interquartile range.
…… km/h [1]
(b) Another car takes 18 seconds to travel 400 m along this road.
Calculate the average speed of this car in km/h.
…… km/h [3]
Mark scheme
3(a)(i) 43 1
3(a)(ii) 65 1
3(a)(iii) 13 1
3(b) 80 3 400 60 × 60
M2 for × oe
18 1000
400
Or M1 for
18
60 × 60
or for their speed in m/s ×
1000
400 18
or for and soi
1000 60 × 60
P O S S I B I L I T Y
Morgan picks two of these letters, at random, without replacement.
(a) Find the probability that he picks
(i) the letter Y first,
…… [1]
(ii) the letter B then the letter Y,
…… [2]
(iii) two letters that are the same.
…… [3]
(b) Morgan now picks a third letter at random.
Find the probability that
(i) all three letters are the same,
…… [2]
(ii) exactly two of the three letters are the same,
…… [5]
(iii) all three letters are different.
…… [2]
Mark scheme
4(a)(i) 1 1
oe
11
4(a)(ii) 1 2 1 1
oe M1 for × oe
110 11 10
4(a)(iii) 4 3 2 1 3 2
oe M2 for × + × oe
55 11 10 11 10
2 1 3 2
or M1 for × or × seen
11 10 11 10
oe
4(b)(i) 1 2 3 2 1
oe M1 for × × oe
165 11 10 9
4(b)(ii) 1 5 2 1 9 3 2 8
oe M4 for 3 × × + 3 × ×
5 11 10 9 11 10 9
oe
3 2 8
or M3 for 3 × ×
11 10 9
2 1 9
or M2 for 3 × × or
11 10 9
3 2 8
× × oe
11 10 9
2 1 k
or M1 for × × where k is 3, 6
11 10 9
or 9
4(b)(iii) 131 2 M1 for 1 – (their (b)(i) + their (b)(ii)) oe
oe
165
(a)
B C z
A x NOT TO
O 162° SCALE
42° y
E D F
A, B, C and D are points on the circle, centre O.
EF is a tangent to the circle at D.
Angle ADE = 42° and angle COD = 162°.
Find the following angles, giving reasons for each of your answers.
(i) Angle x
x = …… because ……
…… [2]
(ii) Angle y
y = …… because ……
…… [2]
(iii) Angle z
z = …… because ……
…… [3]
(b)
P
NOT TO
T SCALE
U
R
Q
PQR is a triangle.
T is a point on PR and U is a point on PQ.
RQ is parallel to TU.
(i) Explain why triangle PQR is similar to triangle PUT.
Give a reason for each statement you make.
…… [3]
(ii) PT : TR = 4 : 3
(a) Find the ratio PU : PQ.
…… : …… [1]
(b) The area of triangle PUT is 20 cm2.
Find the area of the quadrilateral QRTU.
…… cm2 [3]
Mark scheme
5(a)(i) 81° 2 B1 for 81°
Angle at centre is twice angle at
circumference oe
5(a)(ii) 81° 2 FT their (a)(i)
Alternate segment [theorem] oe B1FT for 81°
5(a)(iii) 123° 3 FT their acute (a)(ii) + 42
Angles on a straight line [= 180] B1 for each element
Opposite angles in a cyclic quadrilateral
are supplementary oe
5(b)(i) Angle PTU = angle PRQ corresponding M2 Accept in any order
Angle PUT = angle PQR corresponding
Angle RPQ is common oe M1 for one correct pair with reason
If 0 scored, SC1 for two correct pairs of
equal angles identified with incorrect/no
reasons
Corresponding angles are equal oe A1
5(b)(ii)(a) 4 : 7 oe 1
5(b)(ii)(b) 41.25 oe 3 7
2
7 2 − 42
M2 for 20 × oe or 20 × oe
4 42
2 2
7 4 7 2 − 42
or M1 for or or or
4 7 42
42
7 2 − 42
H G
E F
11 cm NOT TO
SCALE
D
C
5 cm
A 8 cm B
ABCDEFGH is a cuboid.
AB = 8 cm, BC = 5 cm and CG = 11 cm.
(a) Work out the volume of the cuboid.
…… cm3 [2]
(b) Ivana has a pencil of length 13 cm.
Does this pencil fit completely inside the cuboid?
Show how you decide.
[4]
(c) (i) Calculate angle CAB.
Angle CAB = …… [2]
(ii) Calculate angle GAC.
Angle GAC = …… [2]
Mark scheme
6(a) 440 2 M1 for 8 × 5 × 11
6(b) 82 + 52 + 112 oe M3 M2 for 82 + 52 + 112 or 82 + 112 oe
or M1 for 82 + 52 or 52 + 112 oe
or
82 + 52 + 112 and 132
ALTERNATIVE
82 + 112 or 82 + 112 and 132
Yes and 14.5 or 14.4 or 14.49… A1 Accept equivalent conclusion
or
Yes and 13.6[0…]
6(c)(i) 32.0[…] 2 5
M1 for tan[..] = oe
8
6(c)(ii) 49.4 or 49.38 to 49.39 2 11
M1 for sin[..] = oe
their AG
(a) (i) Factorise 24 + 5x - x 2 .
…… [2]
(ii) The diagram shows a sketch of y = 24 + 5x - x 2 .
y
NOT TO c
SCALE
a O b x
Work out the values of a, b and c.
a = ……
b = ……
c = …… [3]
(iii) Calculate the gradient of y = 24 + 5x - x 2 at x =- 1.5 .
…… [3]
(b) (i) On the diagram, sketch the graph of y = (x + 1) (x - 3) 2 .
Label the values where the graph meets the x-axis and the y-axis.
y
O x
[4]
(ii) Write (x + 1) (x - 3) 2 in the form ax 3 + bx 2 + cx + d .
…… [3]
6 -2 2
Mark scheme
7(a)(i) (8 – x)(3 + x) 2 M1 for 8(3 + x) – x(3 + x)
or 3(8 – x) + x(8 – x)
or (a – x) (b + x) where ab = 24 or
a–b=5
7(a)(ii) [a = ] –3 3 FT their (a)(i) for a and b
[b = ] 8 B1FT for each of a and b or both correct
[c = ] 24 but reversed
B1 for [c =] 24
7(a)(iii) 8 3 M2 for 5 – 2x
or M1 for –2x or 5 – kx, k ≠ 0
7(b)(i) Correct sketch: 4 B1 for positive cubic shape with max on
positive cubic shape and max on the the y-axis or to the right of y-axis
y-axis or to the right of y-axis B1 for root at (–1, 0)
with one root at (–1, 0) B1 for turning point at (3, 0)
and B1 for y-intercept (0, 9)
turning point at (3, 0)
and If 0 score SC1 for all three intercepts on
y-intercept at (0, 9) all labelled axes identified
7(b)(ii) x3 – 5x2 + 3x + 9 final answer 3 B2 for correct expansion of three brackets
unsimplified
B1 for correct expansion of two brackets
with at least 3 terms correct
(a) AB = e o BC = e o DC = e o
-1 5 -3
Find
(i) AC ,
AC = f p [2]
(ii) BD,
BD = f p [2]
(iii) BC .
…… [2]
(b)
C
D NOT TO a SCALE
E b
O
A B
In the diagram, OAB and OED are straight lines.
O is the origin, A is the midpoint of OB and E is the midpoint of AC.
AC = a and CB = b .
Find, in terms of a and b, in its simplest form
(i) AB,
AB = …… [1]
(ii) OE ,
OE = …… [2]
(iii) the position vector of D.
…… [3]
Mark scheme
8(a)(i) 4 2 4 k
B1 for or
4 k 4
8(a)(ii) −4 2 −4 k
B1 for or
8 k 8
8(a)(iii) 5.39 or 5.385.. 2 M1 for (–2)2 + 52 oe
8(b)(i) a+b 1
ρ ρ
8(b)(ii) 3 2 M1 for a correct route, e.g. OA + AE
a+b
2
8(b)(iii) 4 3 ρ 4
2a + b M2 for unsimplified OD or for b
3 3
ρ
or M1 for OD attempted in terms of a
and b
ρ 1 ρ 2
or for CD = b or DB = b seen
3 3
(a) Find the integer values that satisfy the inequality 2 1 2x G 10 .
…… [2]
(b) Factorise completely.
(i) 6y 2 - 15xy
…… [2]
(ii) y 2 - 9x 2
…… [2]
(c) Simplify.
3 2
- x - 1 2x + 1
…… [3]
(d) The straight line y = 3x + 2 intersects the curve y = 2x 2 + 7x - 11 at two points.
Find the coordinates of these two points.
Give your answers correct to 2 decimal places.
( …… , …… )
( …… , …… ) [6]
Mark scheme
9(a) 2, 3, 4, 5 2 B1 for 3 correct and no extra or 4 correct
and one extra
or M1 for 1 < x ⩽ 5
9(b)(i) 3y (2y – 5x) 2 B1 for 3(2y2 – 5xy) or y (6y – 15x) or for
the correct answer seen and then spoiled
9(b)(ii) (y – 3x) (y + 3x) 2 B1 for (y + 3) (y – 3)
9(c) 4x + 5 3 M1 for 3(2x + 1) – 2(x – 1) oe isw
( x − 1)(2 x + 1)
M1 for (x – 1)(2x + 1) oe isw
4x + 5
or final answer
2 x2 − x − 1
9(d) (1.74 , 7.21 to 7.24) 6 For the y values accept any value rounded
and to 2 decimal places in the given range
(–3.74 , –9.20 to –9.22) cao
B5 for (1.74 , 7.21 to 7.24)
or (–3.74 , –9.20 to –9.22)
or x = 1.74 and x = – 3.74
OR
M2 for 2x2 + 4x – 13 = 0
or 2y2 + 4y – 133 = 0
or M1 for 2x2 + 7x – 11 = 3x + 2
2
y−2 y −2
or y = 2 + 7 – 11
3 3
AND
FT their quadratic expression (not
2x2 + 7x – 11)
−4 ± 42 − 4 × 2 × −13
M2FT for
2× 2
15
or −1 ± oe
2
or M1FT for 42 − 4 × 2 × −13 oe
−4 + k −4 − k
or for or
2× 2 2× 2
or (x + 1)2 [ – 13/2 – 1 = 0]
f (x) = 4 - 3x g (x) = x 2 + x h (x) = 3 x
(a) Find fh(2).
…… [2]
(b) Find f -1 (x) .
f -1 (x) = …… [2]
(c) Simplify.
(i) f (1 - 2x)
…… [2]
(ii) gf (x) - 9g (x)
…… [4]
1
(d) = 9 kx h (x)
Find the value of k.
k = …… [2]
11 The table shows the first four terms in sequences A, B, and C.
Sequence 1st term 2nd term 3rd term 4th term 5th term nth term
A 4 9 14 19
B 3 10 29 66
C 1 4 16 64
Complete the table.
[9]
Mark scheme
10(a) –23 2 M1 for 4 – 3(3x) oe soi
10(b) 4− x 2 M1 for x = 4 – 3y or y + 3x = 4
oe final answer or x + 3y = 4
3
y 4 x 4
or = + x oe or = + y oe
−3 −3 −3 −3
10(c)(i) 1 + 6x final answer 2 M1 for 4 – 3(1 – 2x)
10(c)(ii) 20 – 36x 4 B3 for 20 – 36x seen in working then
or 4(5 – 9x) final answer spoiled
OR
M1 for (4 – 3x)2 + 4 – 3x – 9(x2 + x) or
better
B1 for [(4 – 3x)2=] 16 – 12x – 12x + 9x2 or
better
B1 for answer 20 – kx or k – 36x oe
or answer 20 – 36x + kx2 k ≠ 0
10(d) 1 2 1
− x
− oe M1 for (32)kx or 9kx = 9 2 oe
2
11A 24 B1
5n – 1 oe B2 B1 for 5n – k or jn – 1 oe j ≠ 0
11B 127 B1
n3 + 2 oe B2 B1 for n3 oe
11C 256 B1
4(n–1) oe B2 B1 for 4k oe
Checked your answers against these? Let AI mark the whole paper, on your own working.
Examiners follow the generic marking principles at the front of the scheme, then each question's points: method (M), accuracy (A) and independent (B) marks, error carried forward where the scheme allows it, and positive marking, so you're credited for what you got right.
Check your answers against it
Done this paper? Upload your script and CIE Buddy's AI marker marks it question by question against this exact scheme, puts the ticks on your own working and says what each lost mark needed.
Where can I find the 0580/43 October/November 2020 mark scheme?+
On this page, free: the mark scheme for IGCSE Mathematics 0580/43, October/November 2020 (0580_w20_ms_43), and the question paper (0580_w20_qp_43), to read or download, with each question's marking points written out underneath.
How many marks is 0580/43 out of?+
Mathematics 0580/43 for October/November 2020 is out of 130 marks, with 2 hours 30 minutes to answer it.
Can I get my 0580/43 answers marked?+
Yes. Upload your answers to 0580/43/O/N/20 as a PDF or photos, and CIE Buddy marks every question against this paper's official mark scheme, using Cambridge's marking rules. You get your score, each lost mark explained on your own script, and the topics to revise. It's a guide, not an official Cambridge grade.