Mathematics 0580/42 February/March 2024: question paper and mark scheme
Cambridge IGCSE Mathematics (0580), Paper 4 (Extended), February/March 2024. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
A grocer sells potatoes, mushrooms and carrots.
(a) A customer buys 3 kg of mushrooms at $1.04 per kg and 4 kg of carrots at $1.28 per kg.
Calculate the total cost.
$ …… [2]
(b) In one week, the ratio of the masses of vegetables sold by the grocer is
potatoes : mushrooms : carrots = 11 : 8 : 6.
(i) Work out the mass of mushrooms sold as a percentage of the total mass.
…… % [2]
(ii) The total mass of potatoes, mushrooms and carrots sold is 1500 kg.
Find the mass of carrots the grocer sells this week.
…… kg [2]
(iii) The profit the grocer makes selling 1 kg of carrots is $0.75 .
Find the total profit the grocer makes selling carrots this week.
$ …… [1]
(iv) On the last day of the week, the grocer reduces the price of 1 kg of potatoes by 8% to $1.15 .
Calculate the original price of 1 kg of potatoes.
$ …… [2]
(c) The grocer buys 620 kg of onions, correct to the nearest 20 kg.
He packs them into bags each containing 5 kg of onions, correct to the nearest 1 kg.
Calculate the upper bound for the number of bags of onions that he packs.
…… [3]
Mark scheme
1(a) 8.24 cao 2 M1 for 3 ×1.04 + 4 ×1.28
1(b)(i) 32 2 8
M1 for [×100] oe
11 + 8 + 6
1(b)(ii) 360 2 1500
M1 for × k where k = 1 , 11, 8 or 6
11 + 8 + 6
1(b)(iii) 270 1 FT 0.75 × their 360
1(b)(iv) 1.25 cao 2 8
M1 for x × 1 − = 1.15 oe or better
100
1(c) 140 nfww 3 620 to 640 620 + 10
M2 for or oe
5 − 0.5 4 to 5
or M1 for 620 +10 oe or 620 – 10 oe
or 5 + 0.5 oe or 5 – 0.5 oe seen
X
D
A x°
NOT TO y° SCALE
C
B
A, B, C and D are points on a circle.
ADX and BCX are straight lines.
Angle BAD = x° and angle DCX = y°.
(a) Explain why x = y.
Give a geometrical reason for each statement you make.
[2]
(b) Show that triangle ABX is similar to triangle CDX.
[2]
(c) AD = 15 cm, DX = 9 cm and CX = 12 cm.
(i) Find BC.
BC = …… cm [3]
(ii) Complete the statement.
The ratio area of triangle ABX : area of triangle CDX = …… : 1. [1]
Mark scheme
2(a) y + angle BCD = 180 oe B2 B1 for angles on a straight line
AND
angles on a straight line
AND OR
x + angle BCD = 180 oe
AND
opposite angles of a cyclic quadrilateral opposite angles of a cyclic quadrilateral are
are supplementary supplementary
OR
OR
angles in opposite segments are
angles in opposite segments are
supplementary supplementary
leading to x = y with no errors
2(b) Allow any two statements from: M1
CXD is common angle
or angle AXB = angle CXD
x = y or angle BAX = angle DCX
angle ABX = angle CDX
States all three equal pairs of angles A1
OR
2/all angles equal so triangles similar
2(c)(i) 6 nfww 3 B2 for BX = 18 nfww
24 BC + 12
or M2 for = oe
12 9
24 BX
or M1 for = oe
12 9
If 0 scored, SC1 for answer 18
2(c)(ii) 4 1
(a) The table shows information about the marks gained by each of 10 students in a test.
Mark 15 16 17 18 19 20
Frequency 4 1 2 1 0 2
(i) Calculate the range.
…… [1]
(ii) Calculate the mean.
…… [3]
(iii) Find the median.
…… [1]
(iv) Write down the mode.
…… [1]
(b) Paulo’s mean mark for 7 homework tasks is 17.
After completing the 8th task, his mean mark is 17.5 .
Calculate Paulo’s mark for the 8th task.
…… [3]
(c) The table shows the percentage scored by each of 100 students in their final exam.
Percentage ( p) 0 1 p G 30 30 1 p G 50 50 1 p G 60 60 1 p G 70 70 1 p G 100
Frequency 12 18 35 20 15
On the grid, draw a histogram to show this information.
4
3
Frequency density 2
1
0
0 20 40 60 80 100 p
Percentage
[4]
Mark scheme
3(a)(i) 5 1
3(a)(ii) 16.8 3 M1 for 15 × 4 + 16 [× 1] + 17 × 2 + 18 [× 1 ]
[+ 19 × 0] + 20 × 2 oe
M1 dep on previous M1 for their Σfx ÷10
3(a)(iii) 16.5 1
3(a)(iv) 15 1
3(b) 21 3 M2 for 8 ×17.5 and 7 ×17 oe
or M1 for 7 ×17 or 8 ×17.5 oe seen
3(c) 5 correct blocks, with correct widths, 4 B3 for 4 correct blocks
heights 0.8cm, 1.8cm 7cm, 4cm, 1cm or B2 for 3 correct blocks
or B1 for 2 correct blocks
If 0 scored SC1 for correct frequency
densities
(0.4 0.9 3.5 2 0.5) soi
(a)
F
9 cm
NOT TO
SCALE
D C
12 cm
M
E 12 cm B
The diagram shows a pyramid with a square base BCDE.
The diagonals CE and BD intersect at M, and the vertex F is directly above M.
BE = 12 cm and FM = 9 cm.
(i) Calculate the volume of the pyramid.
1
[The volume, V, of a pyramid with base area A and height h is V = Ah .]
3
…… cm 3 [2]
(ii) Calculate the total surface area of the pyramid.
…… cm 2 [5]
(b)
NOT TO
SCALE
3r
r
The diagram shows a toy made from a cone and a hemisphere.
The base radius of the cone and the radius of the hemisphere are both r cm.
The slant height of the cone is 3r cm.
The total surface area of the toy is 304 cm 2 .
Calculate the value of r.
[The curved surface area, A, of a cone with radius r and slant height l is A = rrl .]
[The curved surface area, A, of a sphere with radius r is A = 4rr 2 .]
r = …… [4]
Mark scheme
4(a)(i) 432 2 M1 for 12 × 12 × 9 ÷ 3 oe
4(a)(ii) 404 or 403.5 to 403.7 5 1
M4 for 122 + 4 × × 12 × 62 + 92 oe
2
1
or M3 for ×12 × 62 + 92 oe
2
or M2 for explicit method to find height of
triangular face
e.g. 62 + 92 oe
or M1 for implicit method to find height of
triangular face
or for 62 + 92 oe seen
or B1 for slant height of triangle FC
153 or 3 17 or 12.4 or 12.36 to 12.37 soi
4(b) 4.4[0] or 4.398 to 4.399... nfww 4
304
M3 for oe
( 2 + 3) × π
4πr 2
or M2 for + πr × 3r = 304 oe
2
4πr 2
or M1 for oe seen or πr × 3r oe seen
2
(a) (i) Factorise. x 2 - x - 12
…… [2]
(ii) Simplify. x 2 - 16
2 x - x - 12
…… [2]
(b) Simplify.
`2x - 3j - `x + 1j
2 2
…… [3]
(c) Write as a single fraction in its simplest form.
2x + 4 x
- x+1 x-3
…… [4]
(d) Expand and simplify.
(x - 3) (x - 5) (2x + 1)
…… [3]
(e) Solve the simultaneous equations.
You must show all your working.
x - 3y = 13
2x 2 - 9y = 116
x = …… y = ……
x = …… y = …… [6]
Mark scheme
5(a)(i) ( x − 4 )( x + 3) final answer 2 M1 for ( x + a )( x + b ) where ab = −12
or a + b = −1
or for
x ( x + 3) − 4 ( x + 3) or x ( x − 4 ) + 3 ( x − 4 )
5(a)(ii) x+4 2 M1 for ( x − 4 )( x + 4 ) seen
final answer
x+3
5(b) 3x 2 − 14 x + 8 or ( x − 4 )( 3 x − 2 ) final 3 M2 for (( 2x − 3) − ( x + 1) ) (( 2x − 3) + ( x + 1))
answer
( ) (
or 4 x2 − 6 x − 6 x + 9 − x 2 + x + x + 1 or )
better
or correct answer seen
or M1 for ( x − 4 ) (ax + b) or ( 3 x − 2 ) ( x + c)
( ) (
or 4 x2 − 6 x − 6 x + 9 or ± x 2 + x + x + 1 oe )
5(c) x2 − 3x − 12 x − 3 x − 12
2 4
or 2 final B1 for common denominator ( x + 1)( x − 3)
( x + 1)( x − 3) x − 2x − 3
answer oe isw
B1 for ( 2 x + 4 )( x − 3) − x ( x + 1) or better
seen
B1 for 2 x 2 − 6 x + 4 x − 12 or − x 2 − x seen
5(d) 2 x3 − 15 x 2 + 22 x + 15 final answer 3 B2 for correct expansion of three brackets
unsimplified or for simplified four-term
expression of correct form with 3 terms
correct in final answer
or B1 for correct expansion of two brackets
with at least 3 terms out of 4 correct
5(e) 2 x 2 − 3 x − 77[ = 0] oe M2
( 6x − 9x − 231[= 0])
2
M1 for correct method to eliminate one
variable e.g. 2 (13 + 3 y ) − 9 y = 116
2
or
18 y 2 + 147 y + 222[ = 0] oe
or 2x2 − 3( x − 13) = 116 oe
( 6 y + 49 y + 74[= 0])
2
( 2 x + 11)( x − 7 ) [ = 0] M2
oe FT their 3-term quadratic in x or y , correct
factors, correct substitution into formula or
[−−]3 ± ([−]3)2 − 4 × 2 × −77 for correctly completing square
or oe
2× 2
or ( 6 y + 37 )( 3 y + 6 ) [ = 0]
−147 ± 147 2 − 4 × 18 × 222 M1 for a pair of factors giving 2 correct
or oe
2 × 18 terms when expanded their quadratic
or for e.g. ([−]3)2 − 4 × 2 × −77 oe
[− −]3 ± p
or oe
2× 2
x =7 and y = − 2 B2 B1 for both x-values or both y-values
or for 1 correct pair
1 1
x = − 5 oe and y = − 6 oe
2 6
A
NOT TO
SCALE
17.2 cm
54° 68°
B M C
12.8 cm
The diagram shows triangle ABC with AB = 17.2 cm.
Angle ABC = 54° and angle ACB = 68°.
(a) Calculate AC.
AC = …… cm [3]
(b) M lies on BC and MC = 12.8 cm.
Calculate AM.
AM = …… cm [3]
(c) Calculate the shortest distance from A to BC.
……cm [3]
8 -4
Mark scheme
6(a) 15[.0] or 15.00 to 15.01 3 17.2
M2 for × sin 54 oe
sin 68
sin 54 sin 68
or M1 for = oe
AC 17.2
6(b) 15.7 or 15.65 to 15.66 3 M2 for
their152 + 12.82 − 2 × their15 ×12.8 × cos68
OR
M1 for
their152 + 12.82 − 2 × their15 × 12.8 × cos68
A1 for 244.9 to 245.2
6(c) 13.9 or 13.90 to 13.92 3 x x
M2 for = sin 54 oe or = sin 68
17.2 their15
oe
or M1 for distance required is the
perpendicular from A to BC soi
(a) p =e o q =e o
-5 5
(i) Find 3q.
f p [1]
(ii) (a) Find p - q .
f p [1]
(b) Find p - q .
…… [2]
(b)
M
NOT TO
S SCALE a
O N b
In triangle OMN, O is the origin, OM = a and ON = b .
S is a point on MN such that MS : SN = 5: 3 .
Find, in terms of a and/or b, the position vector of S.
Give your answer in its simplest form.
…… [3]
Mark scheme
7(a)(i) −12 1
15
7(a)(ii)(a) 12 1
−10
7(a)(ii)(b) 15.6 or 15.62… 2
M1dep for their122 + (their [−]10) oe,
2
dep their 12 ≠ 0 and their –10 ≠ 0
7(b) 3 5 3
a + b final answer
8 8
B2 for an unsimplified correct answer
5
or MS = ( b − a ) soi
8
3
or NS = ( −b + a ) soi
8
or B1 for correct route for OS
or for MN = b – a or NM = a – b
(a) On the axes, sketch the graph of y = 4 - 3x .
y
O x
[2]
(b) On the axes, sketch the graph of y =- x 2 .
y
O x
[2]
(c) (i) Find the coordinates of the turning points of the graph of y = 10 + 9x 2 - 2x 3 .
You must show all your working.
( …… , …… ) and ( …… , …… ) [5]
(ii) Determine whether each turning point is a maximum or a minimum.
Show how you decide.
[3]
Mark scheme
8(a) Ruled line with negative gradient and 2
positive y-intercept B1 for ruled line with negative gradient
or for ruled line with positive y-intercept
or straight line with negative gradient and
positive
y-intercept
8(b) Negative quadratic, with vertex at origin 2 B1 for negative quadratic in other position
or for sketch in 3rd and 4th quadrants only
with single maximum at (0, 0) and no other
turning point
or for positive quadratic, with vertex at origin
8(c)(i) 18x – 6x2 isw B2 B1 for one correct term 18x or –6x2 seen
dy M1 Dep on at least B1 earned
setting their derivative = 0 or =0
dx or their derivative = ±18x ± 6x2
(0, 10) and (3, 37) B2 B1 for x = 0 and x = 3 or for (0, 10) or (3, 37)
8(c)(ii) (0, 10) minimum with correct reason 3 Reasons could be e.g.
1 A reasonable sketch of a negative cubic
AND 2 Correct use of 2nd derivative = –12(0) + 18
(3, 37) maximum with correct reason = 18,
18 > 0, so (0, 10) is a minimum oe.
2nd derivative = –12(3) + 18 = –18, –18 < 0
so (3, 37) is a maximum oe.
3 Evaluates correctly values of y on both
sides of both correct stationary points
4 Finds gradient on each side of both correct
stationary points.
B2 for 1 correct with correct reason for that
stationary point
or for both x-values correct and reasonable
sketch of a negative cubic,
or for correct substitution and evaluation of
both of their x-values into their second
derivative
or substitution and evaluation for one x-value
on both sides of both of their stationary
points to find the gradients soi
or M1 for showing [2nd derivative =] –12x +
18
or correct FT their 2nd derivative
or substitution and evaluation shown for one
x-value on both sides of one of their
stationary points to find the gradients soi
or for sketch of any negative cubic.
(a) Janna and Kamal each invest $8000.
At the end of 12 years, they each have $12 800.
(i) Janna invests in an account that pays simple interest at a rate of r% per year.
Calculate the value of r.
r = …… [3]
(ii) Kamal invests in an account that pays compound interest at a rate of R% per year.
Calculate the value of R.
R = …… [3]
(b) The population of a city is growing exponentially at a rate of 1.8% per year.
The population now is 260 000.
Find the number of complete years from now when the population will first be more than 300 000.
…… years [3]
Mark scheme
9(a)(i) 5 3 (12800 − 8000) × 100
M2 for
8000 × 12
8000 ×12 × r
or M1 for [12800 − 8000 =]
100
or 400 seen
If 0 scored, SC1 for answer 13.3 or 13.33…
9(a)(ii) 4[.0] or 3.99… 3 12800
M2 for 12
8000
or M1 for 12800 = 8000 × k12 for any k
9(b) 9 nfww 3 1.8
8
M2 for 260 000 × 1 + oe evaluated to 4
100
sf or better
9
1.8
or 260 000 × 1 + oe evaluated to 2 sf
100
or better
n
1.8
or M1 for [300 000 = ] 260 000 × 1 +
100
oe soi
(Accept any inequality sign in [300 000 = ])
The table shows some values for y = 2x 3 + 6x 2 - 2.5 .
x -3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 y 3.75 5.5 4.25 1.5 -2.5 -0.75
(a) Complete the table. [3]
(b) On the grid, draw the graph of y = 2x 3 + 6x 2 - 2.5 for - 3 G x G 1 .
y
6
5
4
3
2
1
–3 –2 –1 0 1 x
–1
–2
–3
–4
[4]
(c) By drawing a suitable line on the graph, solve the equation 2x 3 + 6x 2 = 4.5 .
x = …… or x = …… or x = …… [3]
(d) The equation 2x 3 + 6x 2 - 2.5 = k has exactly two solutions.
Write down the two possible values of k.
k = …… or k = …… [2]
1
Mark scheme
10(a) −2.5 −1.25 5.5 3 B1 for each
10(b) Correct graph 4 B3FT for 8 or 9 correct points
or B2FT for 6 or 7 correct points
or B1FT for 4 or 5 correct points
10(c) y = 2 drawn M1
−2.75 to –2.65 A2 A1 for 1 solution
–1.1 to −1.05
0.75 to 0.85
10(d) –2.5 5.5 2 B1 for each
f (x) = , x ! 0 g (x) = 3x - 5 h (x) = 2 x x
(a) Find.
(i) gf(2)
…… [2]
(ii) g -1 (x)
g -1 (x) = …… [2]
(b) Find in its simplest form g (x - 2) .
…… [2]
(c) Find the value of x when
(i) fg (x) = 0.1
x = …… [2]
(ii) h (x) - g (7) = 0 .
x = …… [2]
Mark scheme
11(a)(i) −3.5 oe 2 1 1
M1 for g seen or 3 − 5 or better
2 x
11(a)(ii) x+5 2 M1 for correct first step y + 5 = 3x ,
oe final answer
3 y 5
= x − or x = 3 y − 5
3 3
11(b) 3x −11 final answer 2 M1 for 3 ( x − 2 ) − 5
11(c)(i) 5 2 1
M1 for [ = 0.1]
3x − 5
11(c)(ii) 4 nfww 2
M1 for 2x − (3 × 7 − 5)[= 0] or better
(a)
NOT TO
SCALE
50°
12 cm
The diagram shows a circle of radius 12 cm, with a sector removed.
Calculate the perimeter of the remaining shaded shape.
…… cm [4]
(b) The diagram in part(a) shows the top of a cylindrical cake with a slice removed.
The volume of cake that remains is 3510 cm 3 .
Calculate the height of the cake.
…… cm [3]
Mark scheme
12(a) 88.9 or 88.92 to 88.93... 4
360 − 50
M3 for 2 × 12 + × 2 × π × 12 oe
360
or M2 for
( 360 − 50 ) × 2 × π × 12 oe isw
360
50
or M1 for × 2 × π × 12 oe isw
360
12(b) 9.01 or 9.009 to 9.010… 3
M2 for
( 360 − 50 ) × π ×122 × h = 3510
360
k
or M1 for × π × 122 × h oe seen
360
with k = 50 or 360 – 50
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 February/March 2024 mark scheme?+
On this page, free: the mark scheme for IGCSE Mathematics 0580/42, February/March 2024 (0580_m24_ms_42), and the question paper (0580_m24_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 February/March 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/F/M/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.