Mathematics 0580/41 October/November 2023: question paper and mark scheme
Cambridge IGCSE Mathematics (0580), Paper 4 (Extended), October/November 2023. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
y
9
8
7
6
5
4
C
3
2
1
0 x
-7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7
-1
-2
A
-3
B
-4
-5
(a) Describe fully the single transformation that maps
(i) shape A onto shape B
…… [2]
(ii) shape A onto shape C.
…… [3]
(b) On the grid, draw the image of
(i) shape A after a reflection in the line y = 2 [2]
(ii) shape A after an enlargement, scale factor - 2 , centre (0, 0). [2]
1
Mark scheme
1(a)(i) Translation 2 B1 for each
−7
oe
−1
1(a)(ii) Rotation 3 B1 for each
90º clockwise oe
(5, 1)
1(b)(i) Image at 2 B1 for reflection in y = k, k ≠ 2
(2, 6) (3, 6) (3, 8) or for reflection in x = 2
1(b)(ii) Image at 2 B1 for an enlargement, sf –2 in the wrong position
(–4, 4) (–6, 4) (–6, 8)
(a) s = at 2
2
Find the value of s when a = 9.8 and t = 20 .
s = …… [2]
(b) Solve.
5 (4y - 3) = 15
y = …… [3]
(c) Expand and simplify.
3 (5x - 8) - 2 (3x - 7)
…… [2]
(d) Rearrange A = 2b 2 - 3c 3 to make c the subject.
c = …… [3]
(e) Factorise completely.
6pq - 4q - 3p + 2
…… [2]
Mark scheme
2(a) 1960 2 1
M1 for × 9.8 × 202 oe
2
2(b) 3 3 M1 for a first correct step, e.g.
1.5 or 1½ or 20y – 15 = 15 or 4y – 3 = 3
2
M1FTdep for a second correct step, e.g.
20y = 30 or 4y = 6
15 15
or y – = oe
20 20
2(c) 9x – 10 final answer 2 B1 for kx – 10 or 9x + c
or M1 for 15x – 24 or –6x + 14
or B1 for correct answer seen and then spoiled
2(d) 2b2 − A 3
3 oe final answer
3 M1 for isolating 3c3, 3c3 = 2b2 – A oe or for
A 2b 2 A 2b 2
= − c3 or = + c3
3 3 −3 −3
M1FT for isolating c3, follow through their first
step dep on a 3-term expression with a kc3 term
M1FT taking the cube root to the final answer,
follow through their previous step
Maximum of two marks if answer incorrect
2(e) (2q – 1)(3p – 2) or (1 – 2q)(2 – 2 M1 for 2q(3p – 2) – [1](3p – 2)
3p) final answer or 3p(2q – 1) – 2(2q – 1)
or for correct answer seen then spoiled
(a) The table shows information about some of the planets in the solar system.
Planet Diameter (km) Average distance from the Sun (km)
Earth 12 800 1.496 # 108
Mars 6 800 2.279 # 108
Jupiter 143 000 7.786 # 108
Saturn 120 500 1.434 # 109
Neptune 49 500 4.495 # 109
(i) The average distance of Mars from the Sun is 2.279 # 10 8 km .
Write this distance as an ordinary number.
…… km [1]
(ii) The planet Uranus has a diameter that is 35.8% of the diameter of Jupiter.
Calculate the diameter of Uranus.
…… km [2]
(iii) The ratio diameter of Neptune : diameter of Saturn can be written in the form 1 : n.
Find the value of n.
n = …… [1]
(iv) Find the average distance of Neptune from the Sun as a percentage of the average distance of the Earth from the Sun.
…… % [2]
(v) Distances within the solar system are also measured in astronomical units (AU).
The average distance of Jupiter from the Sun is 5.20 AU.
Calculate the average distance of Mars from the Sun in astronomical units.
…… AU [2]
(vi) The diameter of Mars is 39.2% greater than the diameter of Mercury.
Calculate the diameter of Mercury.
…… km [2]
(b) One light year is the distance that light travels in a year of 365.25 days.
The speed of light is 2.9979 # 10 5 kilometres per second.
(i) Show that one light year is 9.461 # 10 12 km , correct to 4 significant figures.
[2]
(ii) The distance from the Andromeda Galaxy to Earth is 2.40 # 10 19 km .
Calculate the time taken for light to travel from this galaxy to Earth.
Give your answer in millions of years.
…… million years [2]
Mark scheme
3(a)(i) 227 900 000 1
3(a)(ii) 51 200 or 51 190 or 51 194 2 35.8
M1 for × 143 000
100
After 0 scored SC1 for answer figs 512 or figs
5119 or figs 51194
3(a)(iii) 2.43 or 2.434… 1
3(a)(iv) 3000 or 3004 to 3005 2 4.495 × 109
M1 for [× 100] oe
1.496 × 108
After 0 scored SC1 for answer figs 3 or figs
3004…. or figs 3005
3(a)(v) 1.52 or 1.522… 2 B1 for 1AU = 1.5[0] × 108 or 1.497… × 108 [km]
or
1km = 6.68 ×10−9 or 6.678×10−9 [AU]
OR
5.2 × 2.279 [×108 ]
M1 for oe
7.786 [×108 ]
After 0 scored SC1 for answer figs 152 or figs
1522……
3(a)(vi) 4890 or 4885… 2 39.2
M1 for d × 1 + = 6800 oe
100
3(b)(i) 2.9979 × 105 × 602 × 24 × M1
365.25 After M0
SC1 for 2.9979 × 105 × 31557600 oe
= 9.4606… × 1012 A1
3(b)(ii) 2.54 or 2.536 to 2.537 2 2.4 × 1019
M1 for oe
9.461 × 1012
(a) Lucia has two fair spinners.
Spinner A is five-sided and is numbered 1, 2, 3, 4, 5.
Spinner B is nine-sided and is numbered 3, 3, 3, 4, 4, 4, 4, 5, 5.
Lucia spins the two spinners and records whether they land on a prime number.
(i) Complete the tree diagram.
Spinner A Spinner B
prime
prime
3
5 not
…… prime
prime
…… not prime
not
…… prime
[2]
(ii) Find the probability that
(a) the two numbers are both prime
…… [2]
(b) the two numbers are not both prime.
…… [1]
(b) Lucia spins Spinner A 120 times.
Find the expected number of times the spinner lands on a prime number.
…… [1]
(c) Lucia spins Spinner B twice.
Find the probability that the two numbers it lands on add up to 9 or more.
…… [3]
(d) Lucia keeps spinning Spinner B until it lands on a 4.
Find an expression, in terms of n, for the probability that this happens on the nth spin.
…… [2]
Mark scheme
4(a)(i) 2 5 4 5 4 2 2
, , B1 for and a pair of probabilities for spinner B
5 9 9 9 9 5
that sum to 1
4a(ii)(a) 1 2 FT dep their tree diagram
oe
3 3 5
M1 for × their
5 9
4a(ii)(b) 2 1 1
oe FT dep 1 – their
3 3
4(b) 72 1
4(c) 20 3 2 4 2 2
oe M2 for × [× 2] + × oe
81 9 9 9 9
2 4 2 2
or M1 for × or × oe
9 9 9 9
4(d) 5
n−1
4 2 5
n−1
[×] oe final answer M1 for seen
9 9 9
(a)
D
83.2 m NOT TO
SCALE
38°
C B A
54.5 m
ACD is a right-angled triangle.
B is on AC and BC = 54.5 m.
AD = 83.2 m and angle ABD = 38° .
Calculate angle ACD.
Angle ACD = …… [5]
(b)
F
E G
EFG is a right-angled triangle.
A circle can be drawn that passes through the three vertices of the triangle.
On the diagram, mark the position of the centre of the circle with a cross.
Explain how you decide.
…… [2]
(c)
N R
NOT TO
5 cm SCALE
4 cm
Q
6 cm
L M P
In triangle LMN, the ratio angle L : angle M : angle N = 4 : 5 : 6.
In triangle PQR, PQ = 6 cm , PR = 4 cm and QR = 5 cm .
Calculate the difference between the largest angle in triangle PQR and the largest angle in triangle
LMN.
…… [7]
Mark scheme
5(a) 27.3 or 27.32 to 27.33 5 83.2
M4 for tan[ACD] = oe
83.2
+ 54.5
tan 38
or
83.2
M3 for [AC =] +54.5 oe
tan 38
or for
[CD =]
2
83.2 83.2
54.52 + − 2(54.5) cos(180 − 38)
sin 38 sin 38
oe
or
83.2 83.2
M2 for [AB =] oe or for [BD =] oe
tan 38 sin38
83.2 83.2
or M1 for tan38 = oe or sin38 = oe
AB BD
5(b) Centre marked at midpoint of B2 B1 for marking the centre at mid-point of FG
FG.
and
Angle in a semi-circle is 90
5(c) 10.8 or 10.81 to 10.82 7 B2 for 72
180
or M1 for [× 6]
4+5+6
and, for triangle PQR
B4 for [angle R=]82.8 or 82.81 to 82.83
5
or B3 for [cosR =] oe or better
40
4 2 + 52 − 6 2
or M2 for
2× 4×5
or M1 for 6 = 42 + 52 – 2×4×5×cosR
2
After 0 scored for triangle PQR, SC1 for
[P =] 55.8 or 55.77 to 55.78 or
[Q =] 41.4 or 41.40 to 41.41
(a)
Sequence 1st term 2nd term 3rd term 4th term 5th term nth term
A -7 -3 1 5
B 7 13 23 37
C 2 3 4 5
27 81 243 729
Complete the table for the three sequences.
[10]
(b) In a sequence, the sum of the first 49 terms is 7644.
The sum of the first 50 terms is 7975.
Find the 50th term of this sequence.
…… [1]
Mark scheme
6(a) A 9 B1
4n – 11 oe final answer B2 B1 for 4n – k or jn – 11 oe j ≠ 0
B 55 B1
2n2 + 5 oe final answer B2 B1 for any quadratic or second differences = 4
6 B1
C oe
2187
n +1 B3 B2 for 3n + 2 oe
oe final answer
3n + 2
OR
B1 for 3n + k seen oe
B1 for n + 1 as the numerator of a fraction
6(b) 331 cao 1
The frequency table shows the time of each of 42 athletes in a race.
Time (t seconds) Number of athletes
216 1 t G 219 9
219 1 t G 224 14
224 1 t G 234 14
234 1 t G 244 2
244 1 t G 264 3
(a) Calculate an estimate of the mean time.
…… seconds [4]
(b) Complete the histogram to show the information in the frequency table.
Two of the blocks have been drawn for you.
4
3
Frequency density 2
1
0
210 220 230 240 250 260 270 t
Time (seconds)
[3]
Mark scheme
7(a) 226 nfww or 226.2 to 226.3[0] 4 M1 for mid-points soi
nfww (217.5, 221.5, 229, 239, 254)
M1 for use of Σfm with m in correct interval
including both boundaries
M1 (dep on 2nd M1) for Σfm ÷
(9 + 14 + 14 + 2 + 3)
7(b) Blocks with heights 2.8, 1.4, 0.2 3 B1 for each correct block
and with correct widths
If 0 scored, SC1 for two correct frequency
densities soi
(a)
3.6 cm
6.5 cm
NOT TO
SCALE
5.4 cm
The diagram shows a solid formed by joining two hemispheres and a cylinder.
The radius of the large hemisphere is 5.4 cm.
The radius of the small hemisphere and the radius of the cylinder are both 3.6 cm.
The height of the cylinder is 6.5 cm.
(i) Show that the volume of the solid is 692 cm 3 , correct to the nearest cubic centimetre.
4
[The volume, V, of a sphere with radius r is V = rr 3 .]
3
[4]
(ii) A mathematically similar solid is made of silver.
In this solid, the cylinder has radius 0.6 cm.
1 cm 3 of silver has a mass of 10.49 grams.
Calculate the total mass of this silver solid.
…… g [4]
(b)
A
10 cm
O NOT TO
216° SCALE
B
AOB is a sector of a circle, centre O.
AO = 10 cm and the sector angle is 216°.
(i) Calculate the length of the arc of this sector.
Give your answer as a multiple of r.
……cm [2]
(ii) A cone is made from this sector by joining OA to OB.
Calculate the volume of the cone.
1
[The volume, V, of a cone with radius r and height h is V = rr 2 h .]
3
…… cm 3 [4]
Mark scheme
8(a)(i) 2 2 M3 2 2
π (3.6)3 + π (5.4)3 + M1 for either π (3.6)3 or π (5.4)3
3 3 3 3
π (3.6) × 6.5
2
M1 for π (3.6) × 6.5
2
692.1 to 692.2… A1
8(a)(ii) 33.6 or 33.60 to 33.62 4 0.6
3
M3 for × 692 × 10.49 oe
3.6
3
0.6
or M2 for × 692 oe
3.6
3 3
0.6 3.6
or M1 for or oe
3.6 0.6
If 0 scored, SC1 for their volume × 10.49
8(b)(i) 12π final answer 2 216
M1 for × 2π × 10 oe
360
After 0 scored
SC1 for final answer 8π or 12π + 20
8(b)(ii) 302 or 301.5 to 301.6… 4
M1 for 2πr = their (b)(i) oe or for
216
× π × 102 = π × r × 10 oe
360
and
M1 for [h =] 102 − their 62 oe
and
1
M1 for [V =] π (their 6)2 × (their 8)
3
f (x) = (3x + 1) (x + 5) (x - 4) g (x) = 2x - 3 h (x) = 4 2x - 1
(a) Find
(i) f (0)
…… [1]
(ii) g -1 (x)
g -1 (x) = …… [2]
(iii) gh(2).
…… [2]
(b) g (2x) = 7
Find the value of x.
x = …… [2]
(c) Simplify g (x 2) + gg (x) + 1.
…… [3]
(d) Find h -1 (16) .
…… [2]
(e) f (x) = (3x + 1) (x + 5) (x - 4)
This can be written in the form f (x) = ax 3 + bx 2 + cx + d .
Find the value of each of a, b, c and d.
a = …… b = …… c = …… d = …… [3]
Mark scheme
9(a)(i) –20 1
9(a)(ii) x+3 2 M1 for x = 2y – 3 or better or y + 3 = 2x or better
oe final answer y 3
2 or =x– or better
2 2
9(a)(iii) 125 2 M1 for g(64) or 2(42x – 1) – 3
9(b) 2.5 oe 2 M1 for 2(2x) – 3 = 7 or better
9(c) 2x2 +4x – 11 final answer 3 B2 for 2x2 and either +4x or – 11 in final 3 term
answer or for correct answer seen then spoiled
or M1 for 2x2 – 3 + 2(2x – 3) – 3 [+ 1]
9(d) 1.5 oe 2 M1 for 42x – 1 = 42 or better
9(e) a=3 3 B2 for 3 correct values
b=4 or for correct unsimplified expanded expression or
c = –59 for simplified four-term expression of correct
d = –20 form with 3 terms correct
or
B1 for 2 correct values
or for correct expansion of one pair of brackets
with at least 3 out of 4 terms correct.
(a) ABC is a triangle.
- 11
B is the point (1, - 10) , A is the point (4, 14) and CA = e o.
8
(i) Find the coordinates of C.
(…… , ……) [2]
(ii) Find BA.
BA = f p [1]
(iii) Find CA .
…… [2]
(b)
M
T
NOT TO a R
SCALE
O b N
OMN is a triangle.
OM = a and ON = b .
R is a point on MN such that MR : RN = 3 : 2.
ORT is a straight line.
2 3
(i) Show that OR = a + b .
5 5
[3]
(ii) (a) NT = 4a + kb and OT = c OR .
Find the value of k and the value of c.
k = …… c = …… [4]
(b) Find MT .
MT = …… [1]
Mark scheme
10(a)(i) (15, 6) 2 B1 for each
10(a)(ii) 3 1
24
10(a)(iii) 13.6 or 13.60… 2 M1 for (–11)2 + 82 oe
10(b)(i) 3 M3
a+ (b – a)
5
2
or b + (a – b)
5 3
2 3 M2 for [ MR =] (b – a) oe
leading to a + b with no 5
5 5 2
errors or [ NR = ] (a – b) oe
5
or M1 for MN = b – a or NM = a – b
or a correct route for OR
10(b)(ii)(a) k = 5, c = 10 4 B2 for c = 10
2 3
or M1 for c( a + b) = b + 4a + kb oe
5 5
2
or for c = 4
5
and
3
M1 for × their c = k + 1
5
10(b)(ii)(b) 3a + 6b final answer 1 FT 3a + ( their k + 1) b
(a) Differentiate x 3 - 4x 2 - 3x .
…… [2]
(b) A curve has equation y = x 3 - 4x 2 - 3x .
Work out the coordinates of the two stationary points.
Show all your working.
(…… , ……)
(…… , ……) [5]
(c) Determine whether each stationary point is a maximum or a minimum.
Show all your working.
[3]
Mark scheme
11(a) 3x2 – 8x – 3 2 B1 for two terms correct or correct answer seen
11(b) 3x2 – 8x – 3 = 0 M1 FT their part (a)
Correct method to solve their 3- M2
term quadratic
(3x + 1)(x – 3) [=0]
−(−8) ± (−8) 2 − 4(3)(−3) M1 for (3x + a)(x + b) [=0]
where ab = –3 or 3b + a = –8
2(3)
or for (−8)2 − 4(3)(−3)
p± q
or for where p = –(–8) and r = 2(3) seen
r
or for a correct method for solving a 2-term
quadratic
(3, –18) B2 B1 for one correct point or for two correct x-
1 14 values,
− , or M1 for substitution of their x-values into
3 27
y = x3 − 4 x 2 − 3x shown
11(c) (3, –18) minimum with reason 3 Reasons could be e.g.
1. A reasonable sketch of a positive cubic
1 14
− , maximum with 2. Correct use of 2nd derivative = 6x – 8 = 10,
3 27
10 > 0, so (3, –18) is a minimum oe.
reason
2nd derivative = 6x – 8 = –10, –10 < 0
1 14
so − , is a maximum oe.
3 27
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 a reason for that stationary
point
or for both x-values correct with correct
conclusions and reasonable sketch of a positive
cubic,
or for correct substitution of both of their
x-values into their second derivative shown,
or substitution shown for one x-value either side
of both of their stationary points to find the
gradients.
Or
M1 for showing [2nd derivative =] 6x – 8
or substitution shown for one x-value either side
of one of their stationary points to find the
gradients.
or for reasonable sketch of positive cubic.
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Mathematics 0580/41 for October/November 2023 is out of 130 marks, with 2 hours 30 minutes to answer it.
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