Mathematics 0580/43 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) In 2023 a football club had 50 adult members and 70 child members.
The membership fee for an adult was $40 and the membership fee for a child was $15.
(i) Calculate the total of the membership fees received by the club in 2023.
$ …… [2]
(ii) The cost of running the club in 2023 was $2780.
Calculate $2780 as a percentage of the total of the membership fees received by the club.
…… % [1]
(iii) In 2023 there were 120 members.
This was a decrease by 4% of the number of members in 2022.
Calculate the number of members in 2022.
…… [2]
(iv) In 2024 the total number of members increased from the 120 members in 2023.
The number of adult members and the number of child members each increased by the same number.
The ratio number of adult members : number of child members changed to 14 : 19.
(a) Find the total number of members in 2024.
…… [2]
(b) Calculate the percentage increase in the total number of members from 2023 to 2024.
…… % [2]
(b) The population of a village is 2500.
The population is decreasing exponentially at a rate of 3% per year.
(i) Calculate the population at the end of 3 years.
…… [2]
(ii) Find the number of complete years it takes for the population to first fall below 2000.
…… years [2]
Mark scheme
1(a)(i) 3050 2 M1 for 50 × 40 + 70 × 15 or better
1(a)(ii) 91.1 or 91.14 to 91.15 1 2780
FT × 100
their 3050
1(a)(iii) 125 nfww 2 100 − 4
M1 for [...] × =120 oe
100
1(a)(iv)(a) 132 2 B1 for increase of 6 in adult or junior
or M1 for 56 : 76
or for multiples of 33 seen 33, 66, 99,
132, …
or 50 + x : 70 + x = 14 : 19 oe
19 + 14
or (70 – 50) × oe
19 − 14
14
or 50 + x = (120 + 2x) × oe
19 + 14
1(a)(iv)(b) 10 2 their (a) − 120
FT × 100
120
dep on their (a) > 120
their (a) − 120
M1 for [×100] or
120
their (a)
× 100 [−100]
120
1(b)(i) 2280 or 2281 to 2282 nfww 2 3
3
M1 for 2500 × 1 − oe
100
1(b)(ii) 8 2 3
n
M1 for 2500 × 1 − or 0.97
n
100
evaluated with n > 3
(a) The nth term of a sequence is 120 - n 3 .
(i) Find the 4th term of this sequence.
…… [1]
(ii) Find the value of n when the nth term is -1211.
n = …… [2]
(b) The nth term of a different sequence is 3 # (0.2) n - 1 .
Find the 5th term of this sequence.
…… [1]
(c) The table shows the first four terms of sequences A, B and C.
Sequence 1st term 2nd term 3rd term 4th term 5th term nth term
A 7 4 1 -2
B 1 2 3 4
4 5 6 7
C 0 2 6 12
Complete the table for each sequence.
[8]
Mark scheme
2(a)(i) 56 1
2(a)(ii) 11 2 M1 for 120 – n3 = – 1211 or 120 – 113 =
– 1211
2(b) 3 1
0.0048 or oe
625
2(c) 8
B1 for –5
A –5 10 – 3n B2 for 10 – 3n oe
or B1 for k – 3n or for 10 – kn
5
B1 for
8
5 n n
B B1 for oe
8 n+3 n+3
B1 for 20
2
B2 for n 2 − n oe
C 20 n −n or B1 for any quadratic or for at least two
second differences of 2
(a) Rahul rolls a dice 60 times.
The results are shown in the table.
Score 1 2 3 4 5 6
Frequency 10 6 11 13 14 6
Find the mode, the median and the mean.
mode = ……
median = ……
mean = …… [5]
(b) Sangita measures the speed of each of 100 cars.
The results are shown in the table.
Speed (v km/h) 20 1 v G 30 30 1 v G 50 50 1 v G 75
Frequency 10 72 18
(i) Calculate an estimate of the mean speed.
…… km/h [4]
(ii) Sangita draws a histogram to show the information in the table.
The height of the bar that represents 20 1 v G 30 is 3 cm .
Calculate the height of each of the other two bars on this histogram.
height of bar for 30 1 v G 50 …… cm
height of bar for 50 1 v G 75 …… cm [2]
Mark scheme
3(a) 5 B1
4 B1
3.55 3 M2 for (10 × 1 + 6 × 2 + 11 × 3 + 13 × 4
+
14 × 5 + 6 × 6) ÷ 60 oe
or M1 for 10 × 1 + 6 × 2 + 11 × 3 + 13 ×
4+
14 × 5 + 6 × 6 oe
3(b)(i) 42.55 or 42.6 4 M1 for 25, 40, 62.5 soi
M1 for Σfx with x values in correct
intervals, including boundaries
Σfx
M1 dep on second M1 for
100
3(b)(ii) 10.8 2 B1 for each or for frequency densities 3.6
2.16 and 0.72 seen
In this question all the measurements are in centimetres.
NOT TO r+2 SCALE
30° r+5 r+1
The area of the triangle is equal to the area of the square.
(a) Show that 3r 2 + r - 6 = 0 .
[4]
(b) Solve the equation 3r 2 + r - 6 = 0 .
Give your answer to 2 decimal places.
You must show all your working.
r = …… or r = …… [3]
(c) Find the perimeter of the square.
…… cm [2]
Mark scheme
4(a) 1 M2 1
(r + 5)(r + 2)sin 30 = (r + 1)2 M1 for (r + 5)(r + 2)sin 30 oe
2 2
r 2 + 5r + 2r + 10 or r 2 + r + r + 1 soi B1
Leading to 3r 2 + r − 6 = 0 with no errors A1 Dependent on both expansions seen
or omissions
4(b) −1 ± 12 − 4(3)(−6) B2 −1 + p
B1 for 12 − 4(3)(−6) or for
2(3) 2(3)
Or −1 − p
or
2(3)
2
or
1 1
− ± 2 + oe 2
6 6 1
or r +
6
1 1
2 or
1
− ± 18 + oe
3 2 2
1
2
3r +
2
–1.59 and 1.26 B1
4(c) 9.028 to 9.040 2 M1 for (their root (greater than –1) + 1)
×4
y
10
9
8
7
6
5
4
3
2
1
–3 –2 –1 0 1 2 x
3
–1
–2
–3
–4
–5
–6
–7
–8
The diagram shows the graph of y = f (x) for values of x from -3 to 3.
(a) (i) Use the graph to find f (2) .
…… [1]
(ii) Use the graph to solve the equation f (x) = 5.
x = …… or x = …… or x = …… [3]
(iii) The equation f (x) = k has exactly two solutions.
Write down the value of k.
k = …… [1]
(iv) tangent asymptote root perpendicular
Choose the correct word from the box to complete the statement.
The line x = 0 is the …… to the graph of y = f (x) . [1]
(b) (i) On the grid, draw the graph of y = x - 2 for values of x from -3 to 3. [2]
(ii) Find x when f (x) = x - 2 .
x = …… [1]
c
(c) f (x) = x 2 - ,x!0 x
Use the graph to show that c = 2 .
[2]
(d) The equation f (x) = x - 2 can be written as x 3 + px 2 + qx = 2 .
Find the value of p and the value of q.
p = ……
q = …… [2]
Mark scheme
5(a)(i) 3 cao 1
5(a)(ii) –2, –0.45 to –0.4, 2.40 to 2.45 3 B1 each
5(a)(iii) 3 cao 1
5(a)(iv) Asymptote 1
5(b)(i) Correct ruled line 2 B1 for ruled line through (0, –2)
but not y = –2
or for ruled line with gradient 1
5(b)(ii) 1 cao 1
5(c) Substituting values of x and y into M1
c
y = x 2 − for an exact point on graph
x
of y = f(x)
or
substituting their value of x from 5b(ii)
c
into x 2 − = x – 2
x
leading to c = 2 with no errors A1
5(d) [p = ] –1 and [q = ] 2 nfww 2 M1 for x3 – x2 + 2x = 2 seen
or B1 for each nfww
(a)
H
NOT TO
4m SCALE
G 1.5 m F
The diagram shows a ladder, GH, on horizontal ground, leaning against a vertical wall, HF.
GF = 1.5 m and HF = 4 m .
Calculate the length of the ladder, GH.
…… m [2]
(b)
W
NOT TO
120 m SCALE
V 50 m
W is 120 m north of V and 50 m east of V.
Calculate the bearing of V from W.
…… [3]
(c)
B
NOT TO
SCALE
D
A C
In the quadrilateral ABCD, AD = DC = 5 cm and AB = BC .
Angle ABD = 25° and angle BAD = 15° .
Calculate the perimeter of the quadrilateral ABCD.
…… cm [5]
(d)
S 8 cm R
110°
11 cm NOT TO
14 cm SCALE
P
10 cm
Q
PQRS is a quadrilateral.
Calculate angle PQR.
Angle PQR = …… [5]
Mark scheme
6(a) 4.27 or 4.272... 2 M1 for 42 + 1.52 oe
6(b) 203 or 202.6… 3 B2 for [angle at W = ] 22.6... or for
[angle at V =] 67.4 or 67.38…
5 12
or M1 for tan = or oe
12 5
6(c) 25.2 or 25.20 to 25.21[0] 5 B4 for [BC or AB = ] 7.6[0] or 7.604 to
7.605
OR
M3 for a complete explicit method
5sin140
leading to AB or BC, e.g.
sin 25
OR
M2 for a complete implicit method
leading to AB or BC, e.g.
sin 25 sin140
= oe
5 BC or AB
and
M1 (dep on AB from trig) for 2 × their
AB + 10
OR
B1 for any relevant angle
E.g. BDA or BDC = 140, DAE or
DCE = 50 or ADE or CDE = 40 or
ADC = 80
6(d) 79.5 or 79.6 or 79.54 to 79.55... 5 B2 for [PR2 =] 245 or 245.1 to 245.2
or [PR =] 15.65 to 15.66 or 15.7
or M1 for [PR2 = ] 112 + 82 – 2 × 11 × 8
× cos110
M2 for [cosPQR = ]
102 + 142 − (their PR)2
oe
2 × 10 × 14
or M1 for
(their PR)2 = 102 + 142 – 2 × 10 ×
14cosPQR oe
(a) (i) A car travels 50 km at an average speed of 75 km/h.
Find the time taken.
Give your answer in minutes.
…… min [2]
(ii) Another car travels 47 km, correct to the nearest kilometre.
The average speed of this car is 75 km/h, correct to the nearest 5 km/h.
Calculate the lower bound of the time taken.
Give your answer in minutes.
…… min [3]
(b) A train travels a total of 240 km.
The train travels for t minutes at an average speed of 100 km/h.
It then travels for (t + 60) minutes at an average speed of 110 km/h.
Find the average speed for the whole journey.
…… km/h [6]
Mark scheme
7(a)(i) 40 2 50
M1 for [× 60] oe
75
7(a)(ii) 36 nfww 3 47 − 0.5 46 to 47
M2 for [× 60] or [×
75 to 80 75 + 2.5
60]
or M1 for 47+0.5 or 47 – 0.5 or 75 + 2.5
or 75– 2.5
7(b) 107 or 107.2... 6 240
M5 for [speed = ] × 60 oe
( 2 × 260
7
+ 60 )
OR
B5 for [total time = ] 134 or 134.2 to
134.3
or 2.24 or 2.238...
or B4 for (t = ) 37.1 or 37.14...
OR
t t + 60
M2 for × 100 + × 110 = 240
60 60
oe
t t + 60
or M1 for × 100 or × 110 oe
60 60
M1 for correct equation of form at = b
from their equation containing two terms
in t and involving the speeds.
240
M1 for [× 60]
2 × their t + 60
(a)
NOT TO
SCALE
16 cm 1.5 cm
The diagram shows a solid made from a cylinder and a cone.
The height of the cylinder is 16 cm and the height of the cone is 1.5 cm.
The radius of the cylinder and the base radius of the cone are each 0.35 cm.
(i) Calculate the total surface area of the solid.
[The curved surface area, A, of a cone with radius r and slant height l is A = rrl .]
…… cm 2 [5]
(ii) Calculate the volume of the solid.
1
[The volume, V, of a cone with radius r and height h is V = rr 2 h .]
3
…… cm 3 [3]
(iii)
NOT TO
1.4 cm SCALE
3.5 cm
10 of the solids are placed in a box in the shape of a cuboid of length 17.5 cm.
The diagram shows one end of the box.
Calculate the volume of the empty space in the box.
…… cm 3 [3]
(b)
NOT TO
SCALE
The diagram shows two mathematically similar solids.
The surface area of the larger solid is 200 cm 2 and the surface area of the smaller solid is 98 cm 2 .
The volume of the larger solid is 450 cm 3 .
Calculate the volume of the smaller solid.
…… cm 3 [3]
Mark scheme
8(a)(i) 37.3 or 37.26 to 37.27 5
M2 for π × 0.35 × 0.352 + 1.52 oe
or M1 for 0.352 + 1.52 or better
M1 for π × 0.352
M1 for 2 × π × 0.35 × 16
8(a)(ii) 6.35 or 6.349 to 6.351 3 M1 for π × 0.352 × 16
1
M1 for × π × 0.352 × 1.5
3
8(a)(iii) 22.2 or 22.3 or 22.24 to 22.26 3 M2 for 17.5 × 3.5 × 1.4 – 10 ×
their(a)(ii)
or M1 for 17.5 × 3.5 × 1.4
8(b) 154 or 154.3 to 154.4 3 98
3
M2 for 450 × oe
200
3 3
98 200
or M1 for or oe
200 98
2 3
450 200
or for = oe
V 98
N A M I B I A
The diagram shows 7 cards.
(a) Amir picks a card at random.
Find the probability that the card shows
(i) the letter H
…… [1]
(ii) the letter B.
…… [1]
(b) Fumika picks one of the 7 cards at random.
She replaces it and picks a second card at random.
Find the probability that both cards show the letter I.
…… [2]
(c) Marcos picks two of the 7 cards at random, without replacement.
(i) Find the probability that one card shows the letter I and the other card shows the letter N.
…… [3]
(ii) Find the probability that the two cards show different letters.
…… [3]
(d) Nina picks one of the 7 cards at random without replacement.
She continues picking cards at random without replacement until she picks a card that shows the letter A.
4
The probability that this occurs when she picks the nth card is .
21
Find the value of n.
n = …… [2]
Mark scheme
9(a)(i) 0 1
9(a)(ii) 1 1
oe
7
9(b) 4 2 2 2
oe M1 for ×
49 7 7
9(c)(i) 2 3 2 1 1 2
oe M2 for × + × oe
21 7 6 7 6
2 1 1 2
or M1 for × or × oe seen
7 6 7 6
4
If 0 scored SC1 for
49
9(c)(ii) 19 3 2 1 2 1
oe M2 for 1 − × − × oe
21 7 6 7 6
2 1 2 1
or M1 for × + × oe
7 6 7 6
ALTERNATIVE
1 2 5
M2 for [× 1] × 3 + × × 2
7 7 6
2 5 1
or M1 for × or [× 1] × 3
7 6 7
38
If 0 scored SC1 for
49
9(d) 3 2 5 4 2or3
M1 for × ×
7 6 5
y = x 7 - 7x 6
(a) Find the derivative of y with respect to x.
…… [2]
(b) Find the equation of the tangent to the graph of y = x 7 - 7x 6 at the point where x =- 1.
Give your answer in the form y = mx + c .
y = …… [4]
(c) The graph of y = x 7 - 7x 6 has two turning points.
Find the coordinates of these points.
You must show all your working.
( …… , …… )
( …… , …… ) [5]
Mark scheme
10(a) 7 x6 − 42 x5 final answer 2 B1 for one correct term 7x6 or 42x5 or for
7 x6 − 42 x5 seen and spoiled
10(b) 49x + 41 4 M1 for substituting x = – 1 into [y = ] x7 –
7x6
M1 for x = – 1 substituted in their (a) or
the correct derivative to give their m
M1 for their –8 = (their m)(–1) + c oe
10(c) (0, 0) 5 B4 for (6, –46 656)
(6, –46 656) or B3 for x = 0 and 6
OR
dy dy
M1 for their = 0 or stating =0
dx dx
and
M1 for a correct method to solve their
7x6 – 42x5
(a)
Q
P
NOT TO x° SCALE
O
3
In the circle, centre O, the length of the minor arc PQ is of the length of the major arc PQ.
7
Show that x = 108 .
[3]
(b)
A
NOT TO
SCALE r y°
B
O
The diagram shows a sector, OAB, of a circle with centre O and radius r.
The area of triangle OAB is half the area of the sector.
Angle AOB = y° and is obtuse.
(i) Show that 360 sin y = ry .
[2]
(ii) Complete the table, giving your answers correct to two decimal places.
y 360 sin y ry
108.4 341.60 340.55
108.5 341.40 340.86
108.6 341.20
108.7
[3]
(iii) Complete the statement.
The value of y, correct to one decimal place, that satisfies
the equation 360 sin y = ry is …… . [1]
Mark scheme
11(a) 3 M2
× 360 oe 3 x
10 M1 for =
3 + 7 360
x 3 360 − x
or for [× 2πr] = × [×
360 7 360
2πr]
oe or better
or
10 360 − x
1 [× 2πr] = × [× 2πr] oe or
7 360
better
360
or × k (k = 1 or 7)
7+3
108 A1
11(b)(i) 1 2 1 y 2 y 1
r siny = × × πr2 M1 for × πr2 or for r2 siny
2 2 360 360 2
y 1
or × πr2 = [2 × ] r2 siny
360 2
and
one further step leading to 360siny = πy
with no errors
11(b)(ii) 341.18 or 341.22 3 B1 for each
341.00 341.49 or 341.54
11(b)(iii) 108.6 cao 1
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