Mathematics 0580/42 May/June 2024: question paper and mark scheme
Cambridge IGCSE Mathematics (0580), Paper 4 (Extended), May/June 2024. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
(a) A fruit drink is made using 1.5 litres of apple juice and 450 millilitres of mango juice.
Write the ratio apple juice : mango juice in its simplest form.
…… : …… [2]
(b) One litre of fruit drink is shared between three cups.
The amount in the cups is in the ratio 9 : 6 : 10.
Calculate the number of millilitres in each cup.
…… ml , …… ml , …… ml [3]
(c) A shop buys bottles of the fruit drink for $3.20 each.
It sells them at a profit of 15%.
Calculate the selling price of each bottle of fruit drink.
$ …… [2]
(d) The number of bottles of fruit drink sold has grown exponentially at a constant rate of 2.5% per year.
5 years ago, the shop sold 16 620 bottles.
Calculate the number of bottles sold this year.
…… [2]
(e)
d cm
NOT TO
23 cm SCALE
18.5 cm
The bottles of juice are 18.5 cm tall, correct to the nearest millimetre.
They are stored on shelves.
The distance between the shelves is 23 cm, correct to the nearest centimetre.
Calculate the lower bound for the distance, d cm, between the top of a bottle and the shelf above it.
…… cm [3]
Mark scheme
1(a) 10 : 3 final answer 2
M1 for 1500 : 450 oe in ratio form
If 0 scored SC1 for answer 3 : 10
1(b) 360 240 400 3
B2 for answer 0.36 0.24 0.4
or for answer two of 360 240 400
1000
or M1 for [×k ] where k = 1, 9,
9 + 6 + 10
6 or 10
If 0 scored, SC1 for answer with 3 values
in ratio
9 : 6 : 10 in that order
1(c) 3.68 cao 2 15
M1 for 1 + × 3.2 oe
100
or B1 for answer 0.48
1(d) 18 804[.0...] 2
5
2.5
1 for 16620 × 1 + oe
100
1(e) 3.95 3 M2 for 22.5 – (18.5 to 18.6)
or (22 to 23) −18.55
or M1 for 23 – 0.5 oe seen or 23 + 0.5 oe
seen
or 18.5– 0.05 oe seen
or 18.5 + 0.05 oe seen
(a)
38° NOT TO a° SCALE
b°
The diagram shows a straight line intersecting two parallel lines.
Find the value of a and the value of b.
a = ……
b = …… [2]
(b) Calculate the interior angle of a regular 12-sided polygon.
…… [2]
(c)
N NOT TO
SCALE f°
P
O g°
56°
A B
M
The diagram shows a circle, centre O.
The points M, N and P lie on the circumference of the circle.
AMB is a tangent to the circle at M.
Find the value of f and the value of g.
f = ……
g = …… [3]
(d)
NOT TO
SCALE
24°
k°
27°
The diagram shows a cyclic quadrilateral.
Find the value of k.
k = …… [2]
Mark scheme
2(a) 142 2 B1 for each
142 FT angle b = their angle a
2(b) 150 2 360
M1 for oe isw
12
or 180 × (12 − 2 ) oe isw
2(c) 56 B1
34 B2 M1 for angle at centre = 2 × their 56 oe
soi
or for angle OMB = 90 oe soi
2(d) 51 2 B1 for opp angle = 129 soi
(a) The table shows the time that each of 40 students takes to travel to school.
Time (m minutes) 0 1 m G 10 10 1 m G 25 25 1 m G 40 40 1 m G 60
Frequency 3 18 15 4
(i) Calculate an estimate of the mean.
…… min [4]
(ii) On the grid, draw a histogram to show the information in the table.
2
Frequency density 1
0 m
0 10 20 30 40 50 60
Time (minutes)
[3]
(iii) Two students are selected at random from the 40 students.
Calculate the probability that one student takes more than 25 minutes and the other student takes 10 minutes or less to travel to school.
…… [3]
(b) This is some information about the time that 200 people took to fill in a questionnaire:
• The longest time taken was 30 minutes.
• The median time was 22 minutes.
• The lower quartile was 8 minutes.
• The interquartile range was 19 minutes.
• The range was 25 minutes.
(i) Write down the shortest time taken.
…… minutes [1]
(ii) On the grid, draw a box-and-whisker plot to show this information.
0 10 20 30 40
Time (minutes)
[3]
(iii) George says that 101 of the 200 people took more than 22 minutes to fill in the questionnaire.
Explain why he is wrong.
…… [1]
Mark scheme
3(a)(i) 25.4375 4
M1 for mid-points soi
(5, 17.5, 32.5, 50)
M1 for use of Σfm with m in correct
interval including both boundaries
M1 for (dep on 2nd M1) for Σfm ÷ 40
3(a)(ii) correct histogram 3 B2 for 3 correct blocks
or B1 for 2 correct blocks
If 0 scored SC1 for 4 correct frequency
densities
0.3, 1.2, 1, 0.2 oe soi
3(a)(iii) 19 3
oe
260
19 3
M2 for × [×2] oe
40 39
19 3 19 3
or M1 for any of , , , oe
40 40 39 39
seen
57
If 0 scored, SC1 for oe
800
3(b)(i) 5 1
3(b)(ii) 3
5
30 B2 for with LQ at 8 and median at 22 and
8 22 27
UQ at 27 and boxed
or M1 for LQ at 8 and median at 22
Correct box plot or for UQ at 27
B1 for lowest = 5 and highest = 30
Max B1 if not box and whisker diagram
3(b)(iii) Correct explanation which states the 1
median is 22 and correct reference to 100
or 101
e.g.
Median is 22 which is 50%
of the people and 101 is more
than 50% oe
The median is 22 which is the
100th number (accept 100.5th
number)
(a)
NOT TO
12 cm SCALE
1m
The diagram shows a tank in the shape of a half-cylinder of radius 12 cm and length 1 metre.
The tank is fixed horizontally and is completely filled with water.
(i) Calculate the volume of water in the tank.
Give your answer correct to the nearest 10 cm 3 .
…… cm 3 [3]
(ii)
NOT TO
6 cm SCALE
Water is removed from the tank until the level of water is 6 cm below the top of the tank.
The diagram shows the cross-section of the tank.
Calculate the volume of water that is now in the tank.
…… cm 3 [5]
(b) A rectangular fish tank with length 42 cm and width 35 cm is full of water.
A stone lies at the bottom of the tank.
When the stone is removed from the tank, the depth of the water decreases by 0.2 cm.
The density of the stone is 2.2 g/cm 3 .
Calculate the mass of the stone in grams.
[Density = mass ' volume]
…… g [3]
(c)
H G
E
F
15 cm
NOT TO
SCALE
D
C
12 cm
A 8 cm B
The diagram shows a cuboid, ABCDEFGH.
Calculate the angle that AG makes with the base of the cuboid.
…… [4]
(a) Simplify `25x 6j .
3
Mark scheme
4(a)(i) 22 620 cao 3
B2 for 7200π or 22 608 to 22 629
1
or M1 for × π × 122 [× figs 1] oe
2
4(a)(ii) 8840 or 8850 or 8836 to 8850. 5 6
M1 for cos COM = oe
12
6
or sin AOC = oe
12
their COD
M1 for × π × 122 oe
M 360
1
M1 for × 122 × sin ( their COD ) oe
2
M1dep for (their area of sector COD–
their area of triangle COD) ×100 dep on
at least M1M1 oe
4(b) 647 or 646.8 3 m
M2 for 2.2 = oe
42 × 35 × 0.2
or M1 for [vol of stone =] 42×35×0.2 oe
If 0 scored SC1 for answer figs 647 or
figs 6468
4(c) 46.1 or 46.12 to 46.14 4 15
M3 for tan = oe
8 + 122
2
or M2 for 82 + 122 oe or 82 + 122 + 152 oe
or M1 for identifying the angle GAC
2
…… [2]
(b) These are the first five terms of a sequence.
1
1 6 36 216
6
Find the nth term of the sequence.
…… [2]
(c) Expand and simplify.
(x + 4) (x - 3) (3x - 1)
…… [3]
7
(d) (i) Show that (3x + 5) + = x simplifies to 2x 2 + x - 3 = 0 . x-2
[4]
(ii) Solve by factorisation 2x 2 + x - 3 = 0 .
x = …… or x = …… [3]
(e) A solid cylinder has base radius x and height 3x.
The total surface area of the cylinder is the same as the total surface area of a solid hemisphere of radius 5y.
2 75y 2
Show that x = .
8
[The surface area, A, of a sphere with radius r is A = 4rr 2 .]
[4]
Mark scheme
5(a) 125x9 final answer 2
B1 for answer 125 x k or m x9
or for correct answer seen then spoilt
5(b) 6n − 2 oe final answer 2
B1 for answer of form 6k oe
−k
1
or answer of the form oe
6
or for correct answer seen
5(c) 3x3 + 2x2 –37x + 12 final answer 3 B2 for correct expansion of three
brackets unsimplified or for simplified
four-term expression of correct form with
3 terms correct
or B1 for correct expansion of two
brackets with at least 3 terms out of 4
correct
5(d)(i) eliminates the fraction correctly M1
eg (3x + 5) (x – 2) + 7 = x (x – 2)
3x2 + 5x –6x – 10 + 7 = x2 – 2x oe B2
B1 for 3x2 + 5x –6x –10 [ + 7] oe seen
with at least 3 terms correct
leading to 2x2 + x – 3 = 0 A1 dep on M1 B2
with no errors or omissions
5(d)(ii) ( 2 x + 3)( x − 1) M2 or M1 for (2x + a)(x + b) where ab = −3
or 2b + a = [+]1
or for partial factors
2x(x – 1) + 3(x – 1)
or x(2x + 3) –[1](2x + 3)
−1.5 oe and +1 B1
5(e) [TSA cylinder =] 2π x 2 + 2π x × 3x M1
4π (5 y )2 M1
[TSA hemisphere=] π (5 y ) 2 +
2
Leading to M1 dep M1M1
2π x 2 + 6π x 2 = 50π y 2 + 25π y 2 oe
75 y 2 A1 dep on M1M1M1
x2 =
8
D
6.5 cm 26°
C NOT TO
A 64° SCALE
42°
10.4 cm
B
ABCD is a quadrilateral with AB = 10.4 cm and AD = 6.5 cm.
Angle DAB = 64° , angle BDC = 26° and angle DBC = 42° .
(a) Show that BD = 9.55 cm, correct to 2 decimal places.
[3]
(b) (i) Show that angle BCD = 112° .
[1]
(ii) Calculate CD.
CD = …… [3]
(c) Find the shortest distance from D to AB.
…… cm [3]
Mark scheme
6(a) 10.4 2 + 6.52 − 2 × 10.4 × 6.5 × cos 64 M2 M1 for 10.42 + 6.52 – 2× 10.4 × 6.5 ×
cos64
A1 for 91.1 to 91.2
9.546 to 9.547 A1
6(b)(i) 180 − ( 26 + 42 ) B1
6(b)(ii) 6.89 or 6.888 to 6.892... 3 9.55
M2 for × sin 42 oe
sin112
sin112 sin 42
or M1 for = oe
9.55 CD
6(c) 5.84[2…] 3 x
M2 for = sin 64 oe
6.5
or M1 for identifying shortest distance
from D is perpendicular to AB
(a) Solve 3x - 8 = 6 - 4x .
x = …… [2]
(b) Factorise fully 10a 2 + 5a .
…… [2]
(c) Factorise fully (2x - 3) 2 - 9 .
…… [2]
1 1
(d) f (x) = , x! g (x) = 3 x
4x - 1 4
(i) Find f (4) .
…… [1]
(ii) Find gg (2) .
…… [2]
(iii) Find k when g (k) = f (7) .
…… [2]
Mark scheme
7(a) 2 2 M1 for 3x + 4 x = 6 + 8 or better
7(b) 5a ( 2a + 1) final answer 2 B1 for a (10a + 5 ) or 5(2a2 +a)
or 5a ( 2a + 1) then spoilt
7(c) 4 x ( x − 3) final answer 2 M1 for ( (2 x − 3) − 3)( (2 x − 3) + 3) or
better
or for 4 x 2 − 6 x − 6 x + 9 [−9] oe or better
7(d)(i) 1 1
oe
15
7(d)(ii) 19 683 2 B1 for g(9), 39 or 33x seen
7(d)(iii) −3 2 1
M1 for 3k = or 3k = 3−3
27
or answer g(–3)
A baker decorates x small cakes and y large cakes.
In one day, he decorates:
• not more than 16 small cakes
• less than 10 large cakes
• more small cakes than large cakes
• a total of not more than 24 cakes.
One of the inequalities that shows this information is x G 16 .
(a) Write down the other three inequalities in x and/or y.
…… …… …… [3]
(b) On the grid, draw four straight lines and shade the unwanted regions to show these inequalities.
Label the region, R, which satisfies the four inequalities.
y
26
24
22
20
18
16
14
12
10
8
6
4
2
x
0 2 4 6 8 10 12 14 16 18 20 22 24 26
[6]
(c) The baker earns $8 for decorating a small cake and $12 for decorating a large cake.
Use your diagram to find the largest amount the baker can earn in one day by decorating cakes.
$ …… [2]
Mark scheme
8(a) y < 10 3 B1 for each
y < x oe
x + y ⩽ 24 oe If 0 scored, SC1 for y ⩽ 10 and y ⩽ x and
x + y < 24
8(b) Correct lines and region indicated 6 B1 for each correct line
and
c
c
c
R
B2 for R in correct region for all 4 correct
c
lines
or B1 for R in any one of the regions
marked c
or B1 for R that satisfies 3 of the correct
inequalities
8(c) 228 nfww 2 M1 for 8x + 12y for any (x, y) in their R,
x, y both integer
or x = 15, y = 9
(a)
O
NOT TO
10 cm 60° SCALE
17 cm
C D
A B
OAB is a sector of a circle, centre O, radius 17 cm.
OCD is a sector of a circle, centre O, radius 10 cm.
OCA and ODB are straight lines and angle AOB = 60° .
The perimeter of the shaded shape ABDC can be written in the form (ar + b) cm.
Find the value of a and the value of b.
a = ……
b = …… [3]
(b)
NOT TO
SCALE
The diagram shows a regular hexagon.
The area of the hexagon is 127.3 cm 2 .
(i) Show that the length of one side of the hexagon is 7.0 cm , correct to 1 decimal place.
[4]
(ii) The hexagon is the cross-section of a prism of length 10 cm.
127.3 cm2 NOT TO
SCALE
10 cm
7.0 cm
(a) Find the volume of the prism.
…… cm 3 [1]
(b) Calculate the surface area of the prism.
…… cm 2 [2]
Mark scheme
9(a) [a =] 9 3
B2 for a =9
[b =] 14 OR
M2 for
60 60
× 2 × π × 17 + × 2 × π × 10 + 7 + 7
360 360
oe
or M1 for
60 60
× 2 × π × 17 oe or × 2 × π × 10 oe
360 360
If 0 scored SC1 for b =14
9(b)(i) 60° at centre B1
or interior angle = 120°
[6×] 1 × d 2 × sin 60 oe M1
2
127.3 M1
[ d 2 =]
6 × 1 × sin 60
2
6.99[9…] to 7.00[…] A1 Dep on M1M1
9(b)(ii)(a) 1273 1
9(b)(ii)(b) 675 or 674.5 to 674.6 2 M1 for 2 ×127.3 oe or 6 × 7 × 10 oe
(a) A is the point (6, 2) and B is the point (3, - 4 ).
(i) Find the coordinates of the midpoint of AB.
( …… , …… ) [2]
(ii) Calculate the length AB.
…… [3]
(b) The equation of line l is 4x + 3y - 12 = 0 .
(i) Find the gradient of l.
…… [2]
(ii) Find the coordinates of the point where l crosses the y-axis.
( …… , …… ) [2]
(iii) Line p is perpendicular to l and passes through (6, 5).
Find the equation of p in the form y = mx + c .
y = …… [3]
Mark scheme
10(a)(i) (4.5, −1) 2 B1 for each
10(a)(ii) 6.71 or 6.708... 3
M2 for (6 – 3)2 + (2 – – 4)2 oe or better
or M1 for [–] ( 6 − 3) and [–] ( 2 − −4 ) oe
or for ([–]3)2 and ([–]6)2 oe
10(b)(i) 4 2 M1 for 3 y = −4 x + 12
−
3 4 12
or x + y − [= 0] or better seen
3 3
10(b)(ii) (0, 4) 2 B1 for each
or for y = 4 not in coordinate form
10(b)(iii) 3 1 3 3 −1
[ y =] x + final answer M1 for gradient or oe or
4 2 4 their(b)(i)
better
3
M1 for (6, 5) substituted into y = x+c
4
or y = their mx + c oe
(a) The point ( - 1, 6) lies on a curve. dy
This curve has the derived function =- 4x 3 - 9x 2 + 5. dx
Show that ( - 1, 6) is a stationary point of the curve.
[2]
(b) A different curve has equation y = 2x 3 - 6x + 8 .
(i) Calculate the gradient of the tangent to this curve at the point (- 2, 2) .
…… [3]
(ii) Find the x-coordinates of the stationary points of this curve.
x = …… and x = …… [2]
Mark scheme
11(a) –4 (–1)3 – 9 (–1)2 + 5 or better M1
= 0 [so stationary point] A1 with no errors
11(b)(i) 18 3 B2 for 6 x 2 − 6 isw
OR
B1 for 6 x 2 + k (any k) isw or px 2 − 6
isw (p ≠ 0)
or 6 x 2 − 6 + 8
M1dep on B1 for x = −2 substituted into
dy
their
dx
11(b)(ii) 1 and –1 2 M1 for 6 x 2 − 6 = 0 oe seen
dy
or for their = 0 if B1 scored in part
dx
(b)(i)
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 0580/42 May/June 2024 mark scheme?+
On this page, free: the mark scheme for IGCSE Mathematics 0580/42, May/June 2024 (0580_s24_ms_42), and the question paper (0580_s24_qp_42), to read or download, with each question's marking points written out underneath.
How many marks is 0580/42 out of?+
Mathematics 0580/42 for May/June 2024 is out of 130 marks, with 2 hours 30 minutes to answer it.
Can I get my 0580/42 answers marked?+
Yes. Upload your answers to 0580/42/M/J/24 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.