Mathematics 0580/42 October/November 2022: question paper and mark scheme
Cambridge IGCSE Mathematics (0580), Paper 4 (Extended), October/November 2022. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
(a) (i) At a football club, season tickets are sold for seated areas and for standing areas.
The cost of season tickets are in the ratio seated : standing = 5 : 3.
The cost of a season ticket for the standing area is $45.
Find the cost of a season ticket for the seated area.
$ …… [2]
(ii) In 2021, the value of the team’s players was $2.65 million.
In 2022 this value has decreased by 12%.
Find the value in 2022.
$ …… million [2]
(iii) The number of people at a football match is 1455.
This is 6.25% of the total number of people allowed in the stadium.
Find the total number of people allowed in the stadium.
…… [2]
(iv) The average attendance increased exponentially by 4% each year for the three years from
2016 to 2019.
In 2019 the average attendance was 1631.
Find the average attendance for 2016.
…… [3]
(b) Another club sells season tickets for individuals and for families.
In 2018, the number of season tickets sold is in the ratio family : individual = 2 : 7.
(i) The number of family season tickets sold is x.
Write an expression, in terms of x, for the number of individual season tickets sold.
…… [1]
(ii) In 2019, the number of family season tickets sold increases by 12 and the number of individual season tickets sold decreases by 26.
Complete the table by writing expressions, in terms of x, for the number of tickets sold each year.
Year Family tickets Individual tickets
2018 x
2019
[2]
(iii) In 2019, the number of individual season tickets sold is 3 times the number of family season tickets sold.
Write an equation in x and solve it to find the number of family tickets sold in 2018.
x = …… [4]
Mark scheme
1(a)(i) 75 2 45
M1 for [× k] where k is 1, 5 or 8
3
1(a)(ii) 2.332 oe 2 12
M1 for 2.65 [million] × 1 − oe
100
or B1 for 0.318[million] seen
1(a)(iii) 23 280 cao 2
6.25
M1 for × x = 1455 or better
100
1(a)(iv) 1450 or 1449 to 1450 3 4
3
M2 for 1631 = k 1 + oe or better
100
3
4
or B1 for 1 + oe seen
100
n
4
or M1 for 1631 = k 1 + , n > 0 oe
100
1(b)(i) 7x 1
oe
2
1(b)(ii) 7x 2 FT their (b)(i)
x + 12 – 26 oe B1 for x + 12
2
final answer 7x
B1 for their – 26
2
1(b)(iii) 7x
− 26 = 3(x + 12) oe
4 7x
M1dep for their − 26 = 3 × their (x + 12) oe
2 2
leading to 124
M2dep for isolating x terms, dep on eqn with term in x and
constant on each side and with a bracket or fraction.
or M1dep for correctly removing brackets or dealing with
fractions, dep on eqn with term in x and constant on each side and
with a bracket or fraction.
All the lengths in this question are measured in centimetres.
NOT TO
9–x SCALE
x x
The diagram shows a solid cuboid with a square base.
(a) The volume, V cm 3 , of the cuboid is V = x 2 (9 - x) .
The table shows some values of V for 0 G x G 9 .
x 0 1 2 3 4 5 6 7 8 9
V 0 8 54 80 100 108 98 64 0
(i) Complete the table.
[1]
(ii) On the grid on the opposite page, draw the graph of V = x 2 (9 - x) for 0 G x G 9 . [4]
(iii) Find the values of x when the volume of the cuboid is 44 cm 3 .
x = …… or x = …… [2]
V
110
100
90
80
70
60
50
40
30
20
10
0
0 1 2 3 4 5 6 7 8 9 x
(b) (i) Show that the total surface area of the cuboid is (36x - 2x 2) cm 2 .
[2]
(ii) Find the surface area when the volume of the cuboid is a maximum.
…… cm 2 [3]
Mark scheme
2(a)(i) 28 1
2(a)(ii) Correct curve 4 B3FT for 9 or 10 correct points
or B2FT for 7 or 8 correct points
or B1FT for 5 or 6 correct points
2(a)(iii) 2.5 to 2.8 8.2 to 8.5 2 B1 for each value
2(b)(i) 2x2 + 4x(9 – x) oe M1 Accept the sum of individual areas if done in smaller parts
2x2 + 36x – 4x2 oe A1 With intermediate step shown and brackets removed with no
Leading to 36x – 2x2 errors or omissions
2(b)(ii) 144 3 B1 for x = 6 identified from graph or using calculus
M1 for 36 × their6 – 2 × (their 6)2
Kai and Ann carry out a survey on the distances travelled, in kilometres, by 200 cars.
Kai completes this frequency table for the data collected.
Distance (d km) 80 1 d G 100 100 1 d G 150 150 1 d G 200 200 1 d G 300 300 1 d G 400
Frequency 7 33 76 52 32
(a) (i) Calculate an estimate of the mean.
…… km [4]
(ii) Ann uses this frequency table for the same data.
There is a different interval for the final group.
Distance (d km) 80 1 d G 100 100 1 d G 150 150 1 d G 200 200 1 d G 300 300 1 d G 360
Frequency 7 33 76 52 32
Without calculating an estimate of the mean for this data, find the difference between Ann’s and Kai’s estimate of the mean.
You must show all your working.
…… km [2]
(iii) A histogram is drawn showing the information in Kai’s frequency table.
The height of the block for the interval 200 1 d G 300 is 2.6 cm.
Calculate the height of the block for each of the following intervals.
80 1 d G 100 …… cm
150 1 d G 200 …… cm
300 1 d G 400 …… cm [3]
(b) One car is picked at random.
Find the probability that the car has travelled more than 300 km.
…… [1]
(c) Two of the 200 cars are picked at random.
Find the probability that
(i) both cars have travelled 150 km or less,
…… [2]
(ii) one car has travelled more than 200 km and the other car has travelled 100 km or less.
…… [3]
Mark scheme
3(a)(i) 211.275 4 M1 for mid-points soi
(90, 125, 175, 250, 350)
M1 for use of Σfm with m in correct interval
including both boundaries
M1 for (dep on 2nd M1) for Σfm ÷ 200
3(a)(ii) 32 × 350 – 32 × 330 oe or better, or the reverse of this M1
3.2 or – 3.2 final answer B1
3(a)(iii) 1.75 3 B2 for two correct heights
or B1 for one correct height or 3 correct frequency densities
7.6
1.6 or M1 for scale factor of 5 or 0.2
3(b) 4 1
oe
25
3(c)(i) 39 2 40 39
oe M1 for × oe
995 200 199
3(c)(ii) 147 3
oe
4975 84 7
M2 for [2×] × oe
200 199
84 7 84 7
or B1 for and or and oe
200 199 199 200
147
If 0 scored, SC1 for answer oe
5000
y
8
7
6
A
5
4
C
3
2
1
D
–8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 x
–1
–2
B
–3
–4
–5
–6
(a) Describe fully the single transformation that maps
(i) shape A onto shape B,
…… [2]
(ii) shape A onto shape C,
…… [3]
(iii) shape A onto shape D.
…… [3]
(b) On the grid, draw the image of shape A after a reflection in the line y = x + 8 . [2]
Mark scheme
4(a)(i) Translation 2 B1 for each
7
oe
−8
4(a)(ii) Rotation 3 B1 for each
90º [anticlockwise] oe
(0, 8)
4(a)(iii) Enlargement 3 B1 for each
1
[sf] oe
2
[centre] (–1, –4)
4(b) Image at 2
(–4, 4) (–3, 4) (–2, 5) (–2, 3) (–4, 3) B1 for the line y = x + 8 drawn soi long enough to be fit for
purpose
or correct size and orientation but wrong position
(a) The diagram shows the speed–time graph for part of a journey for two vehicles, a car and a bus.
24 Car
Speed v Bus
(m/s) NOT TO
10 SCALE
0
0 18 40
Time (seconds)
(i) Calculate the acceleration of the car during the first 18 seconds.
…… m/s 2 [1]
(ii) In the first 40 seconds the car travelled 134 m more than the bus.
Calculate the constant speed, v, of the bus.
v = …… m/s [4]
(b) A train takes 10 minutes 30 seconds to travel 16 240 m.
Calculate the average speed of the train.
Give your answer in kilometres per hour.
…… km/h [3]
Mark scheme
5(a)(i) 14 1
oe
18
5(a)(ii) 17.5 4 1
M3 for (10 + 24)18 + 22 × 24 – 134 = 40v oe
2
1
or M2 for (10 + 24 )18 + 22 × 24 oe
2
or B2 for [distance covered by bus =] 700
or M1 for correct method for any partial area for the car
or for 40v
5(b) 4 3 figs162[4]
92.8 or 92 M1 for oe
5 their 10 min 30sec
60
M1 for correct conversion to km/h, e.g. ×
1000
(a) Solve.
4x + 15 = 9
x = …… [2]
(b) Factorise. a2 - 9
…… [1]
(c) Write as a single fraction in its simplest form.
4a 3ad
'
5 10c
…… [3]
(d) 5n + 5n + 5n + 5n + 5n = 5m
Find an expression for m in terms of n.
m = …… [2]
(e) Solve by factorisation.
4x 2 + 8x - 5 = 0
x = …… or x = …… [3]
(f) (i) y is directly proportional to (x + 3) 3 .
When x = 2 , y = 13.5 .
Find x when y = 108 .
x = …… [3]
(ii) g is inversely proportional to the square of d.
When d is halved, the value of g is multiplied by a factor n.
Find n.
n = …… [2]
(g) Expand and simplify.
(2x + 3) (x - 1) (x + 3)
…… [3]
dy
(h) Find the derivative, , of y = 3x 2 + 4x - 1. dx
…… [2]
Mark scheme
6(a) 1 3 2 15 9
–1.5 or –1 or – M1 for 4x = 9 – 15 or x + =
2 2 4 4
6(b) (a – 3)(a + 3) final answer 1
6(c) 8c 3 8ac 40c
final answer B2 for or
3d 3ad 15d
4 2c
or × seen
1 3d
or for correct answer seen then spoiled
4a 10c 8ac 3ad
or M1 for × or ÷ oe
5 3ad 10c 10c
6(d) n + 1 final answer 2 M1 for 5 × 5n or 5n+1 seen
6(e) (2x – 1)(2x + 5) [= 0] oe B2
M1 for 2x(2x + 5) – [1](2x + 5) [ = 0]
or 2x(2x – 1) + 5(2x – 1) [ = 0]
or for (2x + m)(2x + n) [ = 0] with and mn = –5 or n + m = 4
1 1 5 B1
or 0.5 and –2.5 or –2 or –
2 2 2
6(f)(i) 7 3 M1 for y = k(x + 3)3 or better
M1 for 108 = their k(x + 3)3
6(f)(ii) 4 2 1
2
M1 for oe
2
k
or oe seen or better
1 2
d
4
6(g) 2x3 + 7x2 – 9 final answer 3
B2 for correct expansion unsimplified
or for simplified 4 term expression of correct form with 3 terms
correct
or B1 for one pair of brackets expanded with at least 3 terms out
of 4 correct
6(h) 6x + 4 2
B1 for 6x or 4 or 6x + 4 with one extra term seen
(a)
R
39.4 cm 38.2 cm NOT TO
SCALE
P Q
46.5 cm
(i) Calculate angle QPR.
Angle QPR = …… [4]
(ii) Find the shortest distance from Q to PR.
…… cm [3]
(b) The diagram shows a cuboid.
H G
E 20 cm NOT TO
F
D C SCALE
21 cm
A 29 cm B
(i) Calculate the length AG.
AG = …… cm [3]
(ii) Calculate the angle between AG and the base ABCD.
…… [3]
(c)
North
K
NOT TO
SCALE
North
112 km
96°
M
L
The diagram shows the positions of a lighthouse, L, and two ships, K and M.
The bearing of L from K is 155° and KL = 112 km .
The bearing of K from M is 010° and angle KML = 96° .
Find the bearing and distance of ship M from the lighthouse, L.
Bearing ……
Distance …… km [5]
Mark scheme
7(a)(i) 52.[0] or 52.01… 4 39.42 + 46.52 − 38.22
M2 for [cosP = ] oe
2 × 39.4 × 46.5
or M1 for 38.22 = 39.42 + 46.52 − 2 × 39.4 × 46.5 × cos P oe
A1 for 0.616 or 0.6155…
7(a)(ii) 36.6 or 36.64 to 36.65 3 d
M2 for = sin(their 52.01) oe
46.5
or M1 for recognition that the line from Q is perpendicular to PR
7(b)(i) 41[.0] or 41.01… nfww 3
M2 for 292 + 212 + 202 oe or better
or M1 for 292 + 212 oe or 292 + 202 oe or 212 + 202 oe or better
7(b)(ii) 29.2 or 29.18 to 29.2 3 20
M2 for sin[GAC] = oe
their AG
or M1 for angle GAC identified
7(c) bearing 286 B2 B1 for angle MLK = 49 or for angle MKL = 35 correctly identified
or angle from North to ML = 106
distance 64.6 or 64.59… B3 112 × sin(their 35)
M2 for oe
sin(96)
or M1 for the implicit form
AB is a line with midpoint M.
A is the point (2, 3) and M is the point (12, 7).
(a) Find the coordinates of B.
( …… , …… ) [2]
(b) Show that the equation of the perpendicular bisector of AB is 2y + 5x = 74 .
[4]
(c) The perpendicular bisector of AB passes through the point N.
The point N has coordinates (2, n).
Find the value of n.
n = …… [1]
(d) Points A, M and N form a triangle.
Find the area of the triangle.
…… [2]
Mark scheme
8(a) (22, 11) 2 B1 for each value
8(b) their11 − 3 M1
oe or better
their 22 − 2
1 M1
−
their m
Substitution of (12, 7) into M1 Accept y – 7 = their m(x – 12) oe
y = (their m)x + c
leading to 2y + 5x = 74 final answer A1 Without error or omission
8(c) 32 1
8(d) 145 2 1
M1 for × (their 32 – 3) × 10 oe
2
or
1
× (7 − 3)2 + (12 − 2)2 × (their 32 − 7)2 + (2 − 12)2 oe
2
y
1
0
360° x
–1
(a) On the diagram, sketch the graph of y = sin x for 0° G x G 360° . [2]
(b) Solve the equation 5 sin x + 4 = 0 for 0° G x G 360° .
x = …… or x = …… [3]
Mark scheme
9(a) Correct sketch to go through (0, 0), and (360, 0) 2
y
M1 for correct sine curve shape through the origin
or for almost correct sketch fitting all tramlines but with an
0 omission at either end or incorrect curvature in one place only
360º x
9(b) 233.1 or 233.13… 3 B2 for one correct angle
and or M1 for sin x = –0.8 oe
306.9 or 306.86 to 306.87
If 0 scored SC1 for 2 reflex angles that add to 540 or two non-
reflex angles that add to 180
(a) The lengths of the sides of a triangle are 11.4 cm, 14.8 cm and 15.7 cm, all correct to 1 decimal place.
Calculate the upper bound of the perimeter of the triangle.
…… cm [2]
(b)
15.6 cm NOT TO
150° SCALE
The diagram shows a circle, radius 15.6 cm.
The angle of the minor sector is 150°.
Calculate the area of the minor sector.
…… cm 2 [2]
(c)
r cm
NOT TO x° SCALE
The diagram shows a circle, radius r cm and minor sector angle x°.
The perimeter of the major sector is three times the perimeter of the minor sector.
90 (r - 2)
Show that x = . r
[4]
9m
Mark scheme
10(a) 42.05 final answer 2 M1 for 11.4 + 0.05 oe or 14.8 + 0.05 oe
or 15.7 + 0.05 oe
10(b) 319 or 318.5 to 318.6 2 150
M1 for × π × 15.62 oe
360
10(c) 360 − x x M2
× 2πr + 2r = 3 × 2π r + 2r oe x
360 360 M1 for × 2πr oe seen
360
or
360 − x
× 2πr oe seen
360
4x M1 i.e. M mark for isolating and collecting terms in x
× 2π[r ] = 2π[r] – 4[r] oe
360
90 (π − 2 ) A1 With no errors or omissions
Leading to
π
(a) e o = 205
40m 2
Find the two possible values of m.
m = …… or …… [3]
(b)
A B
P a NOT TO
SCALE
O c C
OABC is a parallelogram.
OA = a and OC = c .
P is the point on CB such that CP : PB = 3 : 1.
(i) Find, in terms of a and/or c, in their simplest form,
(a) AC ,
AC = …… [1]
(b) CP ,
CP = …… [1]
(c) OP .
OP = …… [1]
(ii) OP and AB are extended to meet at Q.
Find the position vector of Q.
…… [2]
Mark scheme
11(a) 2.5 and – 2.5 oe 3 42025
M2 for 1681m2 = oe
4
or M1 for (9m)2 + (40m)2 oe
11(b)(i)(a) c – a final answer 1
11(b)(i)(b) 3 1
a final answer
4
11(b)(i)(c) 3 1 FT c + their (b)(i)(b), must be a vector in terms of a and/or c in its
c+ a final answer simplest form
4
11(b)(ii) 4 2 1 4
a+ c oe B1 for [ BQ = ] c or [ AQ = ] c
3 3 3
or
M1 for a correct route
or for answer a + kc oe, where k > 1
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Mathematics 0580/42 for October/November 2022 is out of 130 marks, with 2 hours 30 minutes to answer it.
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