Mathematics 0580/42 May/June 2023: question paper and mark scheme
Cambridge IGCSE Mathematics (0580), Paper 4 (Extended), May/June 2023. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
(a)
42°
NOT TO
SCALE
x°
The diagram shows an isosceles triangle with the base extended.
Find the value of x.
x = …… [3]
(b) The diagram shows three lines meeting at a point.
The ratio a : b : c = 3 : 4 : 5.
Find the value of c.
a° c° NOT TO b° SCALE
c = …… [3]
(c) A regular pentagon has an exterior angle, d.
A regular hexagon has an interior angle, h. d
Find the fraction . h
Give your answer in its simplest form.
…… [4]
(d)
S R x° (x + 20)°
NOT TO
SCALE
(3x – 40)°
Q
(2x – 5)°
P
Show that PQRS is a cyclic quadrilateral.
[5]
(e)
A B
50°
9 cm NOT TO
O SCALE
The diagram shows a circle of radius 9 cm, centre O.
The minor sector AOB, with sector angle 50°, is removed from the circle.
Calculate the length of the major arc AB.
…… cm [3]
Mark scheme
1(a) 111 3 180 − 42
M2 for 180 – oe or 42 +
2
180 − 42
oe
2
180 − 42
or M1 for oe
2
1(b) 150 3 M1 for k ÷ (3 + 4 + 5) [×p]
where p = 1, 3, 4 or 5
5
or oe
12
B1 for 360 used
1(c) 3 4 72
cao nfww B3 for
5 120
or B2 for [d = ] 72 or [h = ] 120
or M1 for 360 ÷ 5 oe isw
or 180 – (360 ÷ 6) isw
or for (6 – 2) × 180 [÷ 6]
1(d) x + 2x – 5 + x + 20 + 3x – 40 = 360 M1 Accept equivalent equation
e.g. 7x – 25 = 360
7x = 360 + 5 – 20 + 40 or better M1 FT their equation, accept e.g. 7x = 385
x = 55 B1
55 and 125 B1dep Dep on M1M1B1
or 105 and 75 Accept 55 + 3 × 55 – 40 = 180
or 2 × 55 – 5 + 55 + 20 = 180
If B0 scored, SC1 for 55, 75, 105 and 125
Opposite angles sum to 180 oe A1 Dep on M1M1B1B1
[so PQRS is a cyclic quadrilateral ]
1(e) 48.7 or 48.69 to 48.70… 3 360 − 50
M2 for × 2 × π × 9 oe
360
50
or M1 for × 2 × π × 9 oe
360
(a) Anil changes $830 into euros when the exchange rate is 1 euro = $1.16 .
He spends 500 euros.
He then changes the remaining money back into dollars at the same exchange rate.
Work out how much, in dollars, Anil receives.
$ …… [3]
(b) In 2021, Anil earns $37 000.
(i) He spends $12 400 on bills in 2021.
Calculate the percentage of his earnings he spends on bills.
…… % [2]
(ii) His earnings of $37 000 increase by 3.2% in 2022.
Calculate his earnings in 2022.
$ …… [2]
(c) Anil invests $3500 in an account that pays a rate of 2.4% per year compound interest.
(i) Calculate the total interest earned at the end of 5 years.
$ …… [3]
(ii) Find the number of complete years before Anil has at least $5000 in this account.
…… years [3]
Mark scheme
2(a) 249.98 to 250[.0…] 3 M2 for 830 – 500 × 1.16
or M1 for 500 × 1.16
OR
M1 for 830 ÷ 1.16
M1 for (their 715.5… – 500 ) × 1.16
2(b)(i) 33.5 or 33.51… 2 12 400
M1 for [×100] oe
37 000
If 0 scored, SC1 for answer 66.5 or 66.48
to 66.49
2(b)(ii) 38 184 cao 2 3.2
M1 for 37 000 × 1 + oe
100
or B1 for 1184
2(c)(i) 441 or 440.6 3 B2 for answer 3941 or 3940.6 or 3940.64
or 440.64 to 440.65 to 3940.65
5
2.4
or M2 for 3500 × 1 + – 3500
100
5
2.4
or M1 for 3500 × 1 + oe isw
100
2(c)(ii) 16 3 B2 for 15[.0] nfww to 15.1
15
2.4
or M2 for 3500 × 1 + oe seen
100
16
2.4
or 3500 × 1 + oe seen
100
or M1 for
n
2.4
(3500 or their 3941) × 1 +
100
associated with 5000 oe
C
NOT TO
SCALE
(x + 3) cm
A (2x + 5) cm B
The diagram shows a right-angled triangle ABC.
(a) (i) The area of the triangle is 60 cm 2 .
Show that 2x 2 + 11x - 105 = 0 .
[3]
(ii) Solve by factorisation.
2x 2 + 11x - 105 = 0
x = …… or x = …… [3]
(iii) Calculate angle ACB.
…… [3]
(b) Triangle ABC is similar to triangle DEF.
Triangle DEF has an area of 93.75 cm 2 .
(i) Find the size of the smallest angle of triangle DEF.
…… [1]
(ii) Find the length of the shortest side of triangle DEF.
…… cm [3]
Mark scheme
3(a)(i) ( x + 3) (2 x + 5) M1 Accept (x + 3)(2x + 5) = 2 × 60 or 120
= 60 Accept e.g. (x + 3) (x + 2.5) = 60 without
2
division by 2 shown for M1 (but not A1)
2x2 + 6x + 5x + 15 seen B1 Accept 2x2 + 11x + 15 seen
2x2 + 11x – 105 = 0 A1 Correct completion after M1B1 with the
fraction seen removed with no errors or
omissions seen
3(a)(ii) (2x + 21) (x – 5) [= 0] M2 M1 for partial factors
2x (x – 5) + 21(x – 5) [ = 0]
or x (2x + 21) – 5 (2x + 21) [ = 0]
OR
(2x + a)(x + b) [ = 0] where ab = – 105
or 2b + a = 11
–10.5 and 5 B1
3(a)(iii) 61.9 or 61.92 to 61.93 3 2 × their 5 + 5
M2 for tan = oe
their 5 + 3
or B1FT for 2 × their 5 + 5 and
their 5 + 3
3(b)(i) 28.1 or 28.07 to 28.08 1 FT their 90 – their (a)(iii) unless their
(a)(iii) < 45, in which case FT their
(a)(iii)
3(b)(ii) 10 3 93.75
M2 for (their 5 + 3) × oe
60
93.75 60
or M1 for or oe seen
60 93.75
2
their 5 + 3 60
or = oe
x 93.75
The table shows information about the heights of 80 children.
Height
1.2 1 h G 1.4 1.4 1 h G 1.5 1.5 1 h G 1.65 1.65 1 h G 1.8 1.8 1 h G 1.9
(h metres)
Frequency 2 13 24 32 9
(a) (i) Write down the interval containing the median.
…… 1 h G …… [1]
(ii) Calculate an estimate of the mean height.
…… m [4]
(b) (i) One of these children is chosen at random.
Calculate the probability that they have a height of 1.4 m or less.
…… [1]
(ii) Two of these children are chosen at random.
Calculate the probability that both children are taller than 1.5 m but only one of them is taller than 1.8 m.
…… [3]
(c) (i) Complete the cumulative frequency table for the heights.
Height h G 1.4 h G 1.5 h G 1.65 h G 1.8 h G 1.9
(h metres)
Cumulative
2 frequency
[2]
(ii) On the grid, draw the cumulative frequency diagram.
80
70
60
50
Cumulative frequency 40
30
20
10
0 h
1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9
Height (m)
[3]
(d) Use your diagram to find an estimate of
(i) the interquartile range
…… m [2]
(ii) the 60th percentile.
…… m [2]
Mark scheme
4(a)(i) 1.65 < h ≤ 1.8 1
4(a)(ii) 1.63875 4 M1 for midpoints soi
M1 for use of ∑fh with h in correct
interval including both boundaries
M1dep on 2nd M1 for ∑fh ÷ 80
4(b)(i) 1 1
oe
40
4(b)(ii) 63 3 56 9
oe M2 for × [×2] oe
395 80 79
56 9 9 56
or B1 for or or or oe seen
80 79 80 79
If 0 or B1 scored, instead award SC2 for
117
answer oe
632
63
or SC1 for answer oe
400
4(c)(i) 15, 39, 71, 80 2 B1 for 3 correct
or M1 for 1 error in addition with other
values then consistent
4(c)(ii) Correct curve 3 B1 for correct horizontal placement for 5
plots
B1FT for correct vertical placement for 5
plots
B1FT dep on at least B1 for reasonable
increasing curve or polygon through their
5 points
If 0 scored SC1 FT for 4 out of 5 points
correctly plotted
4(d)(i) Strict FT their UQ – their LQ 2dep B1dep for their UQ or their LQ seen
Dep on increasing curve/polygon for 2
marks or B1
4(d)(ii) Strict FT their reading at 48 2dep B1 for 48 written
(a)
NOT TO
15 cm
SCALE
8 cm
A cone has base diameter 8 cm and perpendicular height 15 cm.
(i) Calculate the volume of the cone.
1
[The volume, V, of a cone with radius r and height h is V = rr 2 h .]
3
…… cm 3 [2]
(ii) A label completely covers the curved surface area of the cone.
Calculate the area of the label as a percentage of the total surface area of the cone.
[The curved surface area, A, of a cone with radius r and slant height l is A = rrl .]
…… % [5]
(b)
NOT TO
SCALE
0.45 m
An empty cylindrical container has radius 0.45 m.
300 litres of water is poured into the container at a rate of 375 ml per second.
(i) Find the time taken, in minutes and seconds, for all the water to be poured into the container.
…… min …… s [3]
(ii) Calculate the height of the water in the container.
…… m [3]
n
Mark scheme
5(a)(i) 251 or 251.3 to 251.4 2 1
M1 for × π × 42 × 15 oe
3
5(a)(ii) 79.5 or 79.51… 5 M3 for π × 4 × 42 + 152 oe
or M2 for 152 + 42 oe
or M1 for [l2 = ] 42 + 152 oe
or π × 4 × their l
M1 for
their curved surfacearea
[×100]
their curved surfacearea + π × 42
oe
5(b)(i) 13 min 20 sec 3 40
B2 for 800 or oe seen
3
or M1 for figs 3 ÷ figs 375
or figs 3 ÷ 22 500
5(b)(ii) 0.472 or 0.4715 to 0.4716… 3 M2 for π × 0.452 × h = 0.3
or π × 452 × h = 300000 oe
or M1 for π × figs452 × h = figs3 oe
(a) A sequence has nth term .
2n + 3
(i) Find the first three terms of this sequence.
Give your answers as fractions.
…… , …… , …… [2]
12
(ii) The kth term of this sequence is .
25
Find the value of k.
k = …… [2]
(b) Find the nth term of each sequence.
(i) 6, 13, 32, 69, 130, …
…… [2]
(ii) 100, 50, 25, 12.5, 6.25, …
…… [2]
Mark scheme
6(a)(i) 1 2 3 2 B1 for 2 correct terms isw
, , final answer or for 0.2 and (0.286 or 0.2857…) and
5 7 9
0.333…
6(a)(ii) 36 2 12(2k + 3)
M1 for k = or better
25
6(b)(i) n3 + 5 oe final answer 2 B1 for any cubic
or common third differences of 6
(at least 2)
or for correct answer seen and spoilt
6(b)(ii) 100 × 21–n oe final answer 2 n[ + k ]
1
B1 for 2–n [+k] oe or oe in answer
2
or for correct answer seen and spoilt
North C NOT TO
SCALE
60 km 87 km
38°
A B
The diagram shows the straight roads between town A, town B and town C.
AC = 60 km , CB = 87 km and B is due east of A.
The bearing of C from A is 038°.
(a) Show that angle ACB = 95.1° , correct to 1 decimal place.
[5]
(b) Without stopping, a car travels from town A to town C then to town B, before returning directly to town A.
The total time taken for the journey is 3 hours 20 minutes.
Calculate the average speed of the car for this journey.
Give your answer in kilometres per hour.
…… km/h [6]
Mark scheme
7(a) Angle CAB = 52 B1
60sin their 52 M3 60sin their 52
180 – 52 – sin −1 M2 for [sin[...] =] oe
87 87
60 87
or M1 for = oe
sin B sin their 52
95.08… A1
7(b) 77.1 or 77.08 to 77.11 6 B4 for dist travelled = 256.9 to 257[.0…]
or B3 for [AB =] 109.9 to 110[.0…]
or M3 for
60 + 87 +
60 2 + 87 2 – 2 × 60 × 87 × cos 95.1 oe
or M2 for
60 2 + 87 2 – 2 × 60 × 87 × cos 95.1 oe
or AB2 = 12093. … to 12097. …
87sin95.1
or oe
sin their 52
or M1 for
AB2 = 602 + 872 – 2 × 60 × 87 × cos 95.1
oe
sin95.1 sin their 52
or = oe
AB 87
20
M1 for their total distance ÷ 3 oe
60
(a) (i) Show that the equation y = (x - 4) (x + 1) (x - 2) can be written as y = x 3 - 5x 2 + 2x + 8 .
[2]
(ii) On the diagram, sketch the graph of y = x 3 - 5x 2 + 2x + 8 , indicating the values where the graph crosses the axes.
y
O x
[4]
(b) The graph of y = x 3 - 5x 2 + 2x + 8 has two tangents with a gradient of 10.
Find the equations of these two tangents.
You must show all your working and give your answers in the form y = mx + c .
y = ……
y = …… [7]
Mark scheme
8(a)(i) Correct expansion of a pair of brackets M1 accept
x2 – 4x + [1]x – 4 x2 – 3x – 4
or x2 – 4x – 2x + 8 or x2 – 6x + 8
or x2 + [1]x – 2x – 2 or x2 – [1]x – 2
x3 – 4x2 + x2 – 4x – 2x2 + 8x – 2x + 8 A1 Accept
leading to and stating x3 – 3x2 – 4x – 2x2 + 6x + 8
[y = ] x3 – 5x2 + 2x + 8 or x3 – 6x2 +[1] x2 + 8x – 6x + 8
or x3 –[1] x2 – 2x – 4x2 + 4x + 8
leading to and stating
[y = ] x3 – 5x2 + 2x + 8
8(a)(ii) Correct labelled sketch 4
positive cubic
Crossing x-axis at –1, 2 and 4 only
Crossing y – axis at 8 only
B1 for positive cubic
B2 for three intercepts only with x -axis
labelled at – 1, 2 and 4
or B1 for 1 or 2 correctly labelled
x – intercepts
B1 for a single intercept on y-axis labelled
at 8 but not if line y = 8
8(b) 3x2 – 10x – 8 [= 0] M3 B2 for derivative = 3x2 – 10x + 2 isw
OR
B1 for derivative with 3x2 or –10x given
in expression isw
M1dep on B1 for their first derivative =
10
2 B1
x = 4 and x = −
3
2 112 B1
(4, 0) and − , oe
3 27
[y =] 10x – 40 B2 B1 for each
and or for two different equations of the form
292 [y = ] 10x + c (c must be numeric)
[y =] 10 x +
27 292
or for c = –40 and
27
(a) Simplify.
(i) (3x 2 y 4) 3
…… [2]
3
-
16 e 16 8 o
2
(ii) x y
…… [3]
(b) (i) Factorise. x2 - 9
…… [1]
(ii) Simplify. x2 - 9
2xy - 6y + 5x - 15
…… [3]
(c) Solve the simultaneous equations.
You must show all your working and give your answers correct to 2 decimal places.
2x + y = 7 y = 5x 2 + 2x - 13
x = …… , y = ……
x = …… , y = …… [6]
Mark scheme
9(a)(i) 27x6y12 final answer 2 B1 for two terms correct in answer e.g.
27x6yk or 27xky12 or kx6y12
or for correct answer seen then spoilt
9(a)(ii) x 24 y12 3 B2 for final answer with two correct
final answer elements
64
64 64−1
or final answer 24 12 or −24 −12 or
x y x y
better
or for correct answer seen
or B1 for 64 or x24 or y12 seen in final
answer
k
or final answer −24 −12
x y
or M1 for first correct step seen
3
[ −3]
x16 y8 2 4
eg or 8 4 or
16 x y
1
−
4096 2
48 24
x y
9(b)(i) (x + 3)(x – 3) final answer 1
9(b)(ii) x+3 3 M2 for (x – 3)(2y + 5)
final answer or M1 for 2y(x – 3) + 5(x – 3)
2y + 5
or x (2y + 5) – 3( 2y + 5)
9(c) 5x2 + 4x – 20 [= 0] oe M2 M1 for 7 – 2x = 5x2 + 2x – 13 oe seen
or 7− y
2
7− y
5y2 – 78y + 221 [= 0] oe or y = 5 + 2 − 13 oe seen
2 2
M2 FT their 3-term quadratic
−4 ± ( 4 )2 − 4(5)(−20)
oe
2(5)
or M1 for (4) 2 − 4(5)( −20) or better
or
2 −4 + q −4 − q
4 4 or for or
− ± 4 + oe 2×5 2×5
10 10 2
4
or for x + oe
10
x = 1.64 y = 3.72 B2 B1 for one correct pair or both x-values
and correct or both y – values correct
x = – 2.44 y = 11.88
(a)
H G
E F
NOT TO
SCALE
D C x cm
A x cm B
ABCDEFGH is a cuboid with a square base of side x cm.
CG = 20 cm and AG = 28 cm .
Calculate the value of x.
x = …… [4]
(b)
R Q
N P
NOT TO
SCALE
M L
J K
The diagram shows a different cuboid JKLMNPQR.
MR = 30 cm correct to the nearest centimetre.
KR = 37 cm correct to the nearest centimetre.
Calculate the lower bound of the angle between KR and the base JKLM of the cuboid.
…… [4]
Mark scheme
10(a) 13.9 or 13.85 to 13.86 4 M3 for 2x2 = 282 – 202 or better
or x = ( 28 − 20 ) sin45 oe
2 2
or M2 for x2 + x2 + 202 = 282 oe
x
or sin45 =
282 − 202 )
or M1 for any correct Pythag in 2D
or their AC × sin 45 oe dep on
trig/Pythagoras attempt for AC
10(b) 51.9 or 51.87 to 51.88 4 29 to 30 30 − 0.5
M3 for sin = or oe
37 + 0.5 37 to 38
or M2 for correct trig statement for
correct angle with values in range 29 to 31
and 36 to 38
or M1 for 30 + 0.5 or 30 – 0.5 or 37 + 0.5
or 37 – 0.5 seen
or for identifying correct angle RKM
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On this page, free: the mark scheme for IGCSE Mathematics 0580/42, May/June 2023 (0580_s23_ms_42), and the question paper (0580_s23_qp_42), to read or download, with each question's marking points written out underneath.
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Mathematics 0580/42 for May/June 2023 is out of 130 marks, with 2 hours 30 minutes to answer it.
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