Mathematics 0580/42 October/November 2024: question paper and mark scheme
Cambridge IGCSE Mathematics (0580), Paper 4 (Extended), October/November 2024. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
(a) Anvi buys a new car.
(i) The price of the car is $28 240.
She is given a 7.5% discount.
Calculate the amount she pays.
$ …… [2]
(ii) The fuel tank in the new car has a capacity of 45 litres.
This is 72% of the capacity of the fuel tank in her old car.
Calculate the capacity of the fuel tank in her old car.
…… litres [2]
(b) Aadi buys a new car costing $28 000.
He pays for the car using a finance plan.
The finance plan is
• a deposit
• 47 equal monthly payments of $330
• a final payment of $11 490.
Using this finance plan, Aadi pays a total of $31 900 for the car.
Calculate the deposit paid as a percentage of $28 000.
…… % [4]
(c) A car travels 64 km and uses 2.5 litres of fuel.
It then travels 128 km and uses 6 litres of fuel.
Calculate the rate at which the car uses fuel during the whole journey.
Give your answer in litres per 100 km.
…… litres per 100 km [2]
(d) At the start of 2021 the value of a car was $46 500.
At the end of 2021 the value of the car was 20% less.
At the end of 2022 the value of the car was 15% less than its value at the end of 2021.
Calculate the value of the car at the end of 2022.
$ …… [2]
Mark scheme
1(a)(i) 26 122 cao 2 7.5
M1 for 28 240 × 1 − oe
100
or B1 for answer 2118
1(a)(ii) 62.5 2 72
M1 for C × = 45 oe or better
100
1(b) 17.5 4 31900 – 11490 – (47 × 330)
M3 for [× 100]
28000
or M2 for 31 900 – 11 490 – (47 × 330)
or M1 for 47 × 330 or for 31 900 – 11 490
1(c) 4.43 or 4.427… 2 2.5 + 6
M1 for [× 100] oe
64 + 128
1(d) 31 620 2 20 15
M1 for 46 500 × 1 − × 1 −
100 100
15 20
or 46 500 × 1 − × 1 −
100 100
20 15
or for 1 − × 1 −
100 100
The table shows some values for y = x 3 + 4x 2 - 4 .
x -4.5 -4 -3 -2 -1 0 1 1.5 y -14.1 5 4 -4 1 8.4
(a) Complete the table. [2]
(b) On the grid, draw the graph of y = x 3 + 4x 2 - 4 for - 4.5 G x G 1.5 .
y
10
5
x
–5 –4 –3 –2 –1 0 1 2
–5
–10
–15
[4]
(c) (i) Draw the tangent to the graph at the point (1, 1). [1]
(ii) Use your tangent to estimate the gradient of the curve at the point (1, 1).
…… [2]
(d) By drawing a suitable straight line on the grid, solve the equation x 3 + 4x 2 - x - 6 = 0 .
x = …… or x = …… or x = …… [4]
Mark scheme
2(a) –4, –1 2 B1 for each correct value
2(b) Correct graph 4 B3FT for 7 or 8 correct points
or B2FT for 5 or 6 correct points
or B1FT for 3 or 4 correct points
2(c)(i) Ruled tangent at x = 1 1
2(c)(ii) 6 to 14 nfww 2 dep on correct tangent or a close attempt at the
tangent at x = 1
M1 for rise/run for their tangent, or close
attempt at tangent at any point.
Must see correct or implied calculation from
a drawn tangent.
2(d) y = x + 2 ruled M2
M1 for [y =] x + 2 soi
or y = x + k ruled
or y = kx + 2 ruled, but not y = 2
x = –3.95 to –3.75 A2 A1 for any two values
x = –1.4 to –1.25
x = 1.1 to 1.25 If A0, SC1 for three correct values
(a) Simplify.
(i) 3m - 5n - 4m + 8n
…… [2]
(ii) (3a 2 c 3) 4
…… [2]
4x 3x 2x
(iii) - +
5 10 15
…… [2]
(b) This isosceles triangle has a perimeter of 35.5 cm.
NOT TO a cm SCALE
(3a + 2) cm
Find the value of a.
a = …… [3]
(c) Using the quadratic formula, solve 5x 2 - 4x - 3 = 0 .
You must show all your working.
x = …… or x = …… [3]
(d) Solve these simultaneous equations.
y = x 2 - 4x + 5
y = 2x - 3
You must show all your working.
x = …… y = ……
x = …… y = …… [5]
Mark scheme
3(a)(i) –m + 3n final answer 2
B1 for –m or [+] 3n in final answer or
for –m + 3n seen and then spoiled
3(a)(ii) 81a8 c12 final answer 2 B1 for final answer in correct form with any two
of 81, a8, c12 correct
or for 81a8 c12 seen and then spoiled
3(a)(iii) 19 x 2 6 × 4 x − 3 × 3x + 2 × 2 x
final answer M1 for oe
30 30
3(b) 4.5 oe 3 M1 for a + 2(3a + 2) = 35.5 oe
M1 for correct ka = b for their linear equation
3(c) −(−4) ± (−4)2 − 4 × 5 × (−3) M2 M1 for (−4)2 − 4 × 5 × (−3) or better
oe
2×5
−(−4) + q −(−4) − q
or for or or better
2×5 2×5
–0.472 or –0.4718 to –0.4717 B1
and
1.27 or 1.271 to 1.272
3(d) x2 – 6x + 8 [= 0] M2 M1 for x2 – 4x + 5 = 2x – 3
or or
y2 – 6y + 5 [= 0] y +3
2
y +3
y= − 4 +5
2 2
(x – 4)(x – 2) [= 0] M1 FT their 3-term quadratic but not if x2 – 4x + 5[=
or 0]
(y – 1)(y – 5) [=0]
OR
−(−6) ± (−6)2 − 4[×1] × 8
[x = ]
2[×1]
or
−(−6) ± (−6)2 − 4[×1] × 5
[y = ]
2[×1]
OR
[x = ] 3 ± −8 + 9
or
[y = ] 3 ± −5 + 9
[x = ] 2 , [y = ] 1 B2 B1 for one correct pair
or two correct x-values or two correct y-values
[x = ] 4 , [y = ] 5
If B0 scored and at least 2 method marks
scored, SC1 for correct substitution of both of
their x values or their y values into y = x2 – 4x +
5 or
y = 2x – 3
(a) The angles of a quadrilateral are w°, x°, y° and z°.
The ratio w : (x + y + z) = 3 : 5.
Find the value of w.
w = …… [2]
(b)
B
P 105° M
C NOT TO
N 49° SCALE
45°
A
D
Q
A, B, C and D are points on a circle.
PQ is the tangent to the circle at A.
BMND is a straight line.
Angle ACD = 49°, angle AMB = 105° and angle PAB = 45°.
(i) Find angle BAM.
Angle BAM = …… [2]
(ii) (a) Find angle BAD.
Angle BAD = …… [2]
(b) Give a geometrical reason why BD is not the diameter of the circle.
…… [1]
(c)
A
B O NOT TO
T D
SCALE
C
A, B, C and D are points on a circle, centre O.
TA and TC are tangents to the circle.
OA = 6.75 cm and OT = 11.5 cm.
(i) Show that angle AOC = 108.12°, correct to 2 decimal places.
[3]
(ii) Calculate the length of the minor arc ABC.
…… cm [2]
(iii) Calculate the area of the major sector OCDA.
…… cm 2 [3]
Mark scheme
4(a) 135 2 360
M1 for × k, where k = 1, 3 or 5 oe
5+3
4(b)(i) 26 2 B1 for ABD = 49
4(b)(ii)(a) 86 2 B1 for QAD = 49
or for BDA = 45
or for BCA = 45
4(b)(ii)(b) Angle in a semicircle = 90 1
4(c)(i) 6.75 M2 6.75
[2 × ] cos −1 oe M1 for cos(…) = oe
11.5 11.5
108.117… A1
4(c)(ii) 12.7 or 12.73 to 12.74 2 108.12
M1 for × 2 × π × 6.75
360
4(c)(iii) 100 or 100.1 to 100.2 3
360 − 108.12
M2 for × π × 6.752 oe
360
108.12
or M1 for × π × 6.752
360
360 − 108.12
If 0 scored, SC1 for ×π×k
360
5(a)(i)(a) 1480 1
5(a)(i)(b) 440 2 M1 for [UQ =] 1600 soi or [LQ =] 1160 soi
(a) The cumulative frequency diagram shows information about the distance travelled by each of
80 motorists in a month.
80
60
Cumulative
40 frequency
20
0
0 400 800 1200 1600 2000 2400
Distance (km)
(i) Use the cumulative frequency diagram to find an estimate for
(a) the median
…… km [1]
(b) the interquartile range
…… km [2]
(ii) One of these motorists is picked at random.
Find the probability that this motorist travels more than 1800 km.
…… [2]
(b) The distance around a racing track is 5.104 km.
The time taken by a car to complete one lap of the track is 1 min 18 s.
Calculate the average speed of the car.
Give your answer in km/h.
…… km/h [3]
(c) The top speed, v km/h, of each of 160 cars is recorded.
The histogram shows this information.
2.0
1.5
Frequency
1.0 density
0.5
0 v
100 140 180 220 260 300
Top speed (km / h)
(i) Show that there are 8 cars with a top speed in the interval 120 1 v G 160 .
[1]
(ii) Calculate an estimate of the mean top speed.
You must show all your working.
…… km/h [6]
3 2
Mark scheme
5(a)(ii) 8 2
oe M1 for 72 or 8 written
80
5(b) 236 or 235.5 to 235.6 3 5.104
M2 for × 60 oe
1.3
5.104
or M1 for
their time
5(c)(i) (160 – 120) × 0.2 [ = 8] 1 with no errors seen
5(c)(ii) 22, 36, 64, 30 seen B2 B1 for 2 or 3 correct frequencies
or
M1 for three of 1.1 × (180 – 160), 1.8 × (200 –
180), 1.6 × (240 – 200) and 0.5 × (300 – 240) oe
(8 × 140 + their22 × 170 M3
+ their36 × 190 + their64 × 220 M1 for midpoints soi
+ their30 × 270) ÷ 160
M1 for Σfx , x in interval or boundary of interval
M1 dep on second M1 for Σfx ÷ 160
211.75 B1
(a) Work out 2 e o - e o.
-5 -7
f p [2]
-6
(b) MN = e o.
4
(i) M is the point (2, -5).
Find the coordinates of N.
( …… , …… ) [1]
(ii) Find MN .
…… [2]
(c)
A Q C
NOT TO
SCALE
a
P
O 2c B
OACB is a trapezium with OB = 2AC.
OA = a and OB = 2c .
4
AP : PB = 4 : 1 and AQ = AC .
5
(i) Write each of the following in terms of a and c.
Give each answer in its simplest form.
(a) AB
…… [1]
(b) CB
…… [1]
(c) OP
…… [2]
(d) QP
…… [2]
(ii) Use your answers to make two statements about the relationship between lines QP and CB.
…… [2]
Mark scheme
6(a) 4 2 6 4 k
B1 for or answer or
−3 −10 k −3
6(b)(i) (–4, –1) 1
6(b)(ii) 7.21 or 7.211… 2 M1 for (–6)2 + 42
6(c)(i)(a) 2c – a 1
6(c)(i)(b) c–a 1
6(c)(i)(c) 1 2 4
(a + 8c) final answer M1 for [ AP =] × their(2c – a)
5 5
1
or [ BP = ] × – their (2c – a)
5
or for a correct vector route using the lines on
the diagram
6(c)(i)(d) 4 2 4 4
(– a + c) final answer M1 for [ QP = ] – c+ × their(2c – a)
5 5 5
or for a correct vector route
6(c)(ii) [QP is] parallel [to CB ] 2 Dep both statements consistent with
4 their (c)(i)(b) and their (c)(i)(d) and both vectors
QP = CB oe in terms of a and c
5
B1 for each dep on statement consistent with
their (c)(i)(b) and their (c)(i)(d) and both vectors
in terms of a and c
(a)
C
60°
North
B
NOT TO
SCALE
85 m 129 m
39°
A
72 m
D
The diagram shows a field, ABCD with B north of A.
BD is a path across the field.
AB = 85 m, AD = 72 m, BD = 129 m, angle BDC = 39° and angle BCD = 60°.
(i) Show that angle CBD = 81°.
[1]
(ii) Calculate CD.
…… m [3]
(iii) Show that angle ABD = 31.6°, correct to 1 decimal place.
[4]
(iv) Find the shortest distance from A to BD.
…… m [3]
(v) Find the bearing of B from C.
…… [2]
(vi) Trees are planted in the field.
The number of trees planted is 1100 per hectare.
Calculate the total number of trees planted in the field.
[1 hectare = 10 000 m 2 ]
…… [4]
(b) A rectangle has an area of 9400 cm 2 , correct to the nearest 100 cm 2 .
The length of the rectangle is 80 cm, correct to the nearest 10 cm.
Calculate the upper bound of the width of the rectangle.
…… cm [3]
Mark scheme
7(a)(i) 180 – 60 – 39 [ = 81] 1
7(a)(ii) 147 or 147.1… 3 129sin(81)
M2 for oe
sin 60
sin(81) sin 60
or M1 for = oe
CD 129
7(a)(iii) 852 + 1292 − 722 M2 M1 for 722 = 852 + 1292 − 2 × 85 ×129cos ABD
[cos = ]
2 × 85 × 129
31.58… A2 A1 for 0.851 to 0.852
9341
or or equivalent fraction
10965
7(a)(iv) 44.5 or 44.51 to 44.54 3 M2 for implicit correct method
d
e.g. = sin 31.6 oe
85
or M1 for recognition that the line from A is
perpendicular to BD
7(a)(v) 247 or 247.4… 2 M1 for 180 + (180 – 81 – 31.6) oe
or for NBC = 180 – 81 – 31.6 oe
or for NCB = 81 + 31.6 oe
7(a)(vi) 972 or 973 4
1
M1 for [ρABD] × 85 × 129sin31.6 oe
2
1
or × 129 × their 44.5 oe
2
1
M1 for [ρBCD ] × 129 × their147×sin39 oe
2
their total area
M1 for × 1100
10000
7(b) 126 nfww 3 9400 + 50 9400 to 9500
M2 for or
70 to 80 80 − 5
or M1 for 9350 or 9450 or 75 or 85 seen
(a) A bag contains 24 coloured beads.
Some are red, some are blue and 10 are yellow.
One bead is picked at random from the bag.
Find the probability that
(i) the bead is yellow
…… [1]
(ii) the bead is not yellow.
…… [1]
(b) Another bag contains 5 green marbles, 6 white marbles and 4 black marbles.
Meera picks 2 marbles at random from the bag, without replacement.
Find the probability that
(i) the first marble is black and the second marble is white
…… [2]
(ii) both marbles have different colours.
…… [4]
Mark scheme
8(a)(i) 5 1
oe
12
8(a)(ii) 7 1 FT 1 – their (a)(i)
oe
12
8(b)(i) 4 2 4 6
oe M1 for ×
35 15 14
8(b)(ii) 74 4 5 4 6 5 4 3
oe M3 for 1 – × + × + × oe
105 15 14 15 14 15 14
5 4 6 5 4 3
or M2 for × + × + × oe
15 14 15 14 15 14
k k −1
or M1 for × where k is 4, 5 or 6 oe
15 14
148
If 0 scored, SC1 for
225
ALTERNATIVE 1
5 10 6 9 4 11
M3 for × + × + × oe
15 14 15 14 15 14
or M2 for two of these products added oe
k 15 − k
or M1 for × where k is 4, 5 or 6 oe
15 14
148
If 0 scored, SC1 for
225
ALTERNATIVE 2
5 6 5 4 6 4
M3 for × × 2 + × × 2 + × × 2 oe
15 14 15 14 15 14
or M2 for at least two of these different products
added oe
or M1 for one correct product
148
If 0 scored, SC1 for
225
f (x) = 2x - 5 g (x) = x 2 - 2x
(a) Find
(i) f ( 7)
…… [1]
(ii) gf (7)
…… [1]
(iii) f -1 (x) .
f -1 (x) = …… [2]
(b) Find gf (x) - 3g (x) .
Give your answer in the form ax 2 + bx + c .
…… [4]
Mark scheme
9(a)(i) 9 1
9(a)(ii) 63 1 FT (their (a)(i))2 – 2 × their (a)(i)
9(a)(iii) x+5 2 y 5
oe final answer M1 for x = 2y – 5 or y + 5 = 2x or =x–
2 2 2
9(b) x2 – 18x + 35 final answer 4 M1 for (2x – 5)2 – 2(2x – 5) – 3(x2 – 2x)
B1 for 4x2 – 10x – 10x + 25
B1 for – 4x + 10 – 3x2 + 6x
A curve has the equation y = x 3 - 9x 2 - 48x .
(a) Differentiate x 3 - 9x 2 - 48x .
…… [2]
(b) Find the coordinates of the turning points of the graph of y = x 3 - 9x 2 - 48x .
You must show all your working.
( …… , …… ) and ( …… , …… )
[4]
(c) Determine whether each of the turning points is a maximum or a minimum.
Give reasons for your answers.
[3]
Mark scheme
10(a) 3x2 – 18x – 48 final answer 2 B1 for two correct terms
or for correct answer seen then spoiled
10(b) dy M1
their = 0 soi
dx
[3](x – 8)(x + 2) oe M1
−(−18) ± (−18)2 − 4(3)(−48)
or
2×3
oe
18 ± 900
oe
or 6
3 ± 16 + 32
(–2, 52) B2 B1 for one correct pair of coordinates or for two
(8, –448) correct values of x
10(c) (–2, 52) maximum with reason 3 Reasons could be e.g.
1. A reasonable sketch of a positive cubic
and 2. Correct evaluation and use of 2nd derivative
6x – 18 = –30, –30 < 0, so (–2, 52) is a
( 8, −448 ) minimum with reason maximum oe.
6x – 18 = 30, 30 > 0 , so ( 8, −448 ) is a
and no incorrect statement minimum 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.
Any incorrect statement MAX B2
B2 for 1 correct with correct reason for that
stationary point
or for both x-values correct and reasonable
sketch of a positive 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 =] 6x – 18
or correct FT their 2nd derivative from part (a)
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 positive cubic.
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 October/November 2024 mark scheme?+
On this page, free: the mark scheme for IGCSE Mathematics 0580/42, October/November 2024 (0580_w24_ms_42), and the question paper (0580_w24_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 October/November 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/O/N/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.