Mathematics 0580/32 May/June 2024: question paper and mark scheme
Cambridge IGCSE Mathematics (0580), Paper 3 (Core), May/June 2024. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
(a) 6 7 10 12 18 32 49 63
From this list of numbers, write down
(i) a factor of 21
(ii) a square number
(iii) a prime number�
(b) Find the value of
(i) the cube root of 1728
(ii) 25
(iii) 50
1
(iv) 36 2 �
(c) Put one pair of brackets into this calculation to make it correct�
3#2-6-2'2= 4
[1]
(d) Find the lowest common multiple (LCM) of 30 and 68�
Mark scheme
1(a)(i) 7 1
1(a)(ii) 49 1
1(a)(iii) 7 1
1(b)(i) 12 1
1(b)(ii) 32 1
1(b)(iii) 1 1
1(b)(iv) 6 1
1(c) 3 × 2 – (6 – 2) ÷ 2 = 4 1
1(d) 1020 2 B1 for 1020k as final answer
or M1 for [30 =] 2 × 3 × 5 and [68 =] 2 ×
2 × 17
or for [30 =] 2 × 15 and [68 =] 2 × 34
or 2 correct factor trees or correct tables
or correct Venn diagram
or a list of multiples of both 30 and 68
with at least their first 3 correct
or 2 × 2 × 3 × 5 × 17 oe
(a) Simplify�
(i) 5a - 6a + 3a
(ii) 6x 2 - 6x - 4x 2 - x
(b) Find the value of c 2 + d 2 when c = 7 and d =- 5�
(c) The time, T minutes, to cook a chicken with a mass of m kg is T = 35m + 20 �
(i) Make m the subject of the formula�
(ii) Find the mass of a chicken that takes 83 minutes to cook�
(d) Solve these simultaneous equations�
You must show all your working�
5x - 6y = 24
15x + 8y = 33
Mark scheme
2(a)(i) 2a final answer 1
2(a)(ii) 2x2 – 7x final answer 2 B1 for 2x2 or −7x in final answer
or for 2x2 – 7x seen then spoilt.
2(b) 74 2 B1 for 49 or 25
2(c)(i) T − 20 2 T 20
[m =] final answer M1 for T – 20 = 35m or =m+
35 35 35
2(c)(ii) 1.8 2 FT their (c)(i) for 2 marks or 1 mark
M1 for (83 – 20) ÷ 35
2(d) Correctly eliminates one variable M1 Making the coefficients the same for one
of the variables and correct consistent use
of addition or subtraction using their
equations
alternative substitution method.
M1 for correct rearrangement of one
equation to make either x or y the subject
and correct substitution of their
rearrangement into 2nd equation.
[x =] 3 A1 If A0 scored SC1 for 2 values satisfying
one of the original equations.
[y =] −1.5 A1
The diagram shows four quadrilaterals, A, B, C and D, on a 1 cm 2 grid�
y
12
11
10
9
8
7
6
5
4
D
3
A
2
1
–8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 x
–1
–2
–3
–4
B
–5
–6
–7
–8
C
–9
– 10
– 11
(a) Write down the mathematical name of quadrilateral A�
(b) (i) Find the area of quadrilateral A�
(ii) Measure the perimeter of quadrilateral A�
(c) Describe fully the single transformation that maps
(i) quadrilateral A onto quadrilateral B
(ii) quadrilateral A onto quadrilateral C
(iii) quadrilateral A onto quadrilateral D�
(d) On the grid, enlarge quadrilateral A by scale factor 2, centre (- 3, - 3) � [2]
Mark scheme
3(a) Trapezium 1
3(b)(i) 7.5 1
3(b)(ii) 11 to 11.4 1
3(c)(i) Translation 2 B1 for each
9
−7
3(c)(ii) Reflection 2 B1 for each
y = −3 oe
3(c)(iii) Rotation 3 B1 for each
(0, 0)
90° clockwise
3(d) Trapezium drawn at 2 B1 for correct enlargement, scale factor
(−1, 5),(−7, 5),(−7, 11),(−3, 11) 2, but in the wrong position.
(a) Anton records the number of pets owned by each of 50 families�
The table shows some of his results�
Number of pets 0 1 2 3 4 5
Number of families 5 12 15 9
There are twice as many families with 4 pets than with 5 pets�
(i) Complete the table�
[3]
(ii) Complete the bar chart�
16
14
12
10
Number of families 8
6
4
2
0
0 1 2 3 4 5
Number of pets
[2]
(iii) Write down the mode�
(iv) Calculate the mean�
(b) 80 of the pets owned by the families are cats, rabbits or hamsters�
The table shows the number of each pet�
Type of pet Number of each pet Pie chart sector angle
Cat 38
Rabbit 28
Hamster 14
(i) Complete the table�
[2]
(ii) Complete the pie chart�
[2]
(iii) One of the pets is chosen at random�
Find the probability that a rabbit is chosen�
Mark scheme
4(a)(i) 6 3 3 M2 for (50 – (5 + 12 + 15 + 9)) ÷ 3
or M1 for 50 – (5 + 12 + 15 + 9)
4(a)(ii) Correct bar chart 2 FT their table entry for 4 and 5 pets
B1 for 3 or 4 correct heights
4(a)(iii) 2 1
4(a)(iv) 2.16 3 M1 for [0 × 5] + (1 × 12) + (2 × 15) + (3
× 9)
+ (4 × their6) + (5 × their3)
M1 dep for (their∑fx) ÷ 50
4(b)(i) 171 2 B1 for one correct angle
126 360
63 or M1 for × k where k = 1, 38, 28 or
80
14
4(b)(ii) Correct pie chart drawn 2 2FT provided the sum of their 3 angles is
360°
B1FT for one correct sector drawn, or
one correct FT sector drawn
4(b)(iii) 7 1
oe
20
(a)
NOT TO
SCALE
x°
125°
The diagram shows a pair of parallel lines and a straight line�
(i) Write down the mathematical name for the type of angle marked 125°�
(ii) Give the geometrical reason why the value of x is 125�
(b)
y°
NOT TO
SCALE
70°
58°
The diagram shows three straight lines�
Find the value of y�
Write down the geometrical properties needed to find the value of y�
(c)
NOT TO
C SCALE
D
E
74°
A B
O
The diagram shows a circle, centre O, with diameter AOB�
The line CDE touches the circle at D and angle DOB = 74° �
(i) Write down the mathematical name of the line CDE�
(ii) Work out angle ODB�
(iii) Work out angle BDE.
Give a geometrical reason for your answer�
(d) Find the interior angle of a regular 15-sided polygon�
Mark scheme
5(a)(i) Obtuse 1
5(a)(ii) Alternate angles 1
5(b) Opposite angles 2 B1 for each
angles in a triangle add to180
52 1
5(c)(i) Tangent 1
5(c)(ii) 53 2 M1 for (180 – 74) ÷ 2
5(c)(iii) 37 1 FT for 90 –their (c)(ii)
Angle between tangent and 1
radius = 90°
5(d) 156 2 360 (15 − 2) × 180
M1 for 180 − or oe
15 15
(a) (i) Complete the table of values for y =- x 2 + 8x + 1�
x 0 1 2 3 4 5 6 7 8
y 13 17 13 1
[3]
(ii) On the grid, draw the graph of y =- x 2 + 8x + 1 for 0 G x G 8 �
y
18
16
14
12
10
8
6
4
2
0
0 2 4 6 8 x
[4]
(iii) Write down the equation of the line of symmetry of the graph�
1
(b) A straight line has a gradient of and passes through the point (2, 7)�
2
(i) On the grid, draw this line for 0 G x G 8 �
[2]
(ii) Write down the equation of this line in the form y = mx + c �
(iii) Write down the coordinates of the points where this line intersects the graph of y =- x 2 + 8x + 1�
Mark scheme
6(a)(i) 1 8 16 16 8 3 B2 for 3 or 4 correct
or B1 for 1 or 2 correct
6(a)(ii) correct curve 4 B3FT for 8 or 9 correctly plotted points
or B2FT for 6 or 7 correctly plotted
points
or B1FT for 4 or 5 correctly plotted
points
6(a)(iii) x = 4 oe 1
6(b)(i) Correct line 2 1
B1 for a line with gradient
2
or a line through (2,7)
6(b)(ii) 1 2 B1 for [y =] ½ x + j oe or [y =] kx + 6 oe
y= x + 6 oe k≠0
2
6(b)(iii) (0.6 to 0.9 , 6.2 to 6.5) 2 B1FT for each
(6.6 to 6.9 , 9.2 to 9.5)
The area of some land is in the ratio park : gardens : playground = 11 : 2 : 3�
The park has an area of 4620 m 2�
(a) Work out the area of the gardens and the area of the playground�
(b) The park area of 4620 m 2 is made up of paths and grassland�
18% of the park area is paths�
(i) Show that the grassland area is 3788.4 m 2�
[1]
(ii) Seed for the grassland is sold in bags�
The seed in one bag covers an area of 280 m 2�
The bags cost $72 each for the first 5 bags and then $58 each for any extra bags�
Calculate the cost of the seed needed to cover the grassland�
(c) The owners of the land buy new equipment for the playground�
They borrow $8500 for 4 years at a rate of 6�5% per year compound interest�
Calculate the amount they repay at the end of the 4 years�
Give your answer correct to the nearest dollar�
(d) The café in the park sells water in bottles A, B and C�
NOT TO
SCALE
Bottle A Bottle B Bottle C
330 ml 500 ml 750 ml
$1.98 $3.20 $5.10
Work out which bottle is the best value�
You must show all your working�
Mark scheme
7(a) 840 3 B2 for 1 correct answer in correct place
1260 or answers reversed
or M1 for 4620 ÷ 11 × k (k = 1, 2, or 3)
7(b)(i) (100 − 18) M1 18
× 4620 accept 4620 – × 4620
100 100
18
or 4620 ( 1 – )
100
7(b)(ii) [$]882 4 B3 for 854.74, 853, 824
OR
M3 for 5 × 72 + (their 14 – 5) × 58 oe
or B2 for 14 [bags]
or M1 for 3788.4 ÷ 280
OR
M3 for 5 × 72 + (their 9 ) × 58 oe
or B2 for 9 [bags]
or M1 for (3788.4 – 5 × 280) ÷ 280 oe
OR
if 0 scored SC2 for 1056 or SC1 for 1027
or 998
7(c) [$]10935 cao 3 6.5
4
M1 for 8500 × 1 + oe
100
A1 for 10934.96……
If A0 scored SC1 for their answer correct
to at least 1 decimal place correctly
rounded to nearest whole number.
7(d) A 3
With correct comparisons made of the 3 M2 for 3 correct comparable values
bottles with suitable accuracy shown or for a correct method to compare 3
bottles shown but not evaluated to
enough accuracy.
or M1 for 2 correct comparable values
or for a correct method to compare 3
bottles but incorrect or not evaluated.
B 3.6 m A
NOT TO
SCALE
4.7 m
E
F
1.7 m
C D
5.5 m
The diagram shows a plan, ABCDE, of the floor of a room in Jo’s house�
F is a point inside the room�
(a) (i) Show that EF = 1.9 m �
[1]
(ii) Work out AF�
(b) Calculate the area of the floor�
(c) A cupboard in the room is in the shape of a cuboid�
The area of the base of the cupboard is 1�2 m 2 and the height of the cupboard is 2�3 m�
Calculate the volume of the cupboard�
Give the units of your answer�
(d) Jo buys 275 floor tiles which cost $1�64 each�
Calculate the total cost of the floor tiles�
(e) Jo builds a patio in the shape of a semicircle with radius 2�3 m�
Calculate the area of the patio�
Question 9 is printed on the next page.
Mark scheme
8(a)(i) 5.5 – 3.6 [= 1.9] 1
8(a)(ii) 3 1
8(b) 23 3 M2 for
4.7 × 3.6 + 1.9 × 1.7 + 0.5 × 1.9 × their
(a)(ii)
or
3.6 ×their(a)(ii) + 5.5 × 1.7 + 0.5 × 1.9 ×
their(a)(ii)
or 5.5 × 4.7 − 0.5 × 1.9 × their (a)(ii)
or 0.5 × (3.6 + 5.5) × their (a)(ii) + 1.7
× 5.5
or 0.5 × (1.7 + 4.7) × 1.9 + 3.6 × 4.7
or M1 for 4.7 × 3.6 + 1.9 × 1.7
or 3.6 × their (a)(ii) + 5.5 × 1.7
or 0.5 × 1.9 × their (a)(ii)
or 5.5 × 4.7
or 0.5 × (3.6 + 5.5) × their (a)(ii)
or 0.5 × (1.7 + 4.7) × 1.9
8(c) 2.76 1
m3 1
8(d) 451 1
8(e) 8.31 or 8.309 to 8.311 2 M1 for [0.5]π × 2.32 oe
(a) Sara rides her bicycle at a speed of 420 metres per minute�
Work out her speed in kilometres per hour�
(b) Jan cycles a distance of 51 km�
She starts at 11 55�
She has a rest stop for 25 minutes�
She finishes at 14 41�
Calculate her average speed, in km/h, for the time she is cycling�
Mark scheme
9(a) 25.2 2 M1 for 420 ÷ 1000 × 60
or B1 for figs 252 final answer
9(b) 21.7 or 21.70….. 4 B2 for 2 h 21 [min] or 2°21° or 2.35 [h]
or 141 [min]
or B1 for 2 h 46 [min] or 2°46° or 2.76…
[h]
or 2.77[h] or 166 [min]
or 1416 or 1220
M1 for 51 ÷ their time period
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Mathematics 0580/32 for May/June 2024 is out of 104 marks, with 2 hours to answer it.
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