Mathematics 0580/42 October/November 2021: question paper and mark scheme
Cambridge IGCSE Mathematics (0580), Paper 4 (Extended), October/November 2021. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
(a) Malena has 450 fruit trees.
The fruit trees are in the ratio apple : pear : plum = 8 : 7 : 3.
(i) Show that Malena has 200 apple trees.
[2]
(ii) Find the number of plum trees.
…… [1]
(iii) Malena wants to increase the number of pear trees by 32%.
Calculate the number of extra pear trees she needs.
…… [2]
(iv) Each apple tree produces 48.5 kg of apples.
The apples have an average mass of 165 g each.
Calculate the total number of apples produced by the 200 trees.
Give your answer correct to the nearest 1000 apples.
…… [3]
(b) Malena’s land is valued at three million and seventy-five thousand dollars.
(i) Write this number in figures.
…… [1]
(ii) Write your answer to part (b)(i) in standard form.
…… [1]
(c) In 2020, each plum tree produced 37.7 kg of plums.
This was 16% more than in 2019.
Calculate the mass of plums produced by each plum tree in 2019.
…… kg [2]
(d) Malena invests $1800 at a rate of 2.1% per year compound interest.
Calculate the value of her investment at the end of 15 years.
$ …… [2]
Mark scheme
1(a)(i) 450 2 450
× 8 oe M1 for
8+7+3 8+7+3
1(a)(ii) 75 1
1(a)(iii) 56 2 32
M1 for × (450 – 200 – their 75) oe
100
32 450
or × × 7 oe
100 8 + 7 + 3
If 0 scored, SC1 for answer 231
1(a)(iv) 59 000 nfww 3 B2 for 58 600 to 58 800
or B1 for 293 to 294
figs485 × 200
or M1 for oe
165
If 0 scored, SC1 for their more accurate
answer seen and rounded to the nearest 1000
1(b)(i) 3 075 000 1
1(b)(ii) 3.075 × 106 1 FT their (b)(i)
1(c) 32.5 2 M1 for x × 16 = 37.7 or better
1 +
100
1(d) 2460 or 2458. … 2 15
2.1
M1 for 1800 1 + oe
100
(a) The cumulative frequency diagram shows information about the floor area, a m 2 , of each of
80 houses.
80
70
60
50
Cumulative frequency 40
30
20
10
0 a
40 60 80 100 120 140 160 180 200
Floor area (m2)
Use the diagram to find an estimate of
(i) the median,
…… m 2 [1]
(ii) the lower quartile,
…… m 2 [1]
(iii) the interquartile range,
…… m 2 [1]
(iv) the number of houses with a floor area greater than 120 m 2 .
…… [2]
(b) The information about the 80 floor areas is shown in this frequency table.
Floor area
40 1 a G 60 60 1 a G 80 80 1 a G 100 100 1 a G 130 130 1 a G 160 160 1 a G 200
(a m 2)
Frequency 14 17 18 15 9 7
(i) Calculate an estimate of the mean floor area.
…… m 2 [4]
(ii) Complete the histogram to show the information in the frequency table.
1.0
Frequency density 0.5
0 a
40 60 80 100 120 140 160 180 200
2
Floor area (m )
[4]
(iii) Two of the houses are picked at random.
Find the probability that one of the houses has a floor area greater than 130 m 2 and the other has a floor area 60 m 2 or less.
…… [3]
Mark scheme
2(a)(i) 90 1
2(a)(ii) 68 1
2(a)(iii) 52 1 FT 120 – their (a)(ii)
2(a)(iv) 20 2 B1 for 60 in working or as answer
2(b)(i) 97.5 4 M1 for mid-points soi
(50, 70, 90, 115, 145, 180)
M1 for use of Σfm with m in correct interval
including both boundaries
M1 for (dep on 2nd M1) for Σfm ÷ 80
2(b)(ii) Bars with heights 0.9, 0.5, 0.3, 0.175 4 B1 for each correct bar
and with correct widths If 0 scored, SC1 for 3 or 4 correct frequency
densities
2(b)(iii) 28 3 16 14
oe M2 for [2 ×] × oe
395 80 79
16 16 14
or M1 for or oe or oe or
80 79 80
14
oe seen
79
7
If 0 scored, SC1 for answer oe
100
(a)
C
38.6 m 56.5 m
D
B
94°
78.4 m NOT TO
SCALE
46.1 m
64°
E A
ABCDE is a pentagon.
(i) Calculate AD and show that it rounds to 94.5 m, correct to 1 decimal place.
[2]
(ii) Calculate angle BAC.
Angle BAC = …… [3]
(iii) Calculate the largest angle in triangle CAD.
…… [4]
(b)
Q
L 34.3 cm
P NOT TO
21.5 cm SCALE
111°
R
M N
27.6 cm
Triangle PQR has the same area as triangle LMN.
Calculate the shortest distance from R to the line PQ.
…… cm [3]
Mark scheme
3(a)(i) AD M1
= tan 64 oe or better
46.1
94.51 to 94.52 A1
3(a)(ii) 46[.0] or 45.96… nfww 3 sin 94
M2 for 56.5 × oe
78.4
56.5 78.4
or M1 for = oe
sin BAC sin 94
3(a)(iii) 102.3 or 102.4 or 102.34 to 102.38 4 38.6 2 + 78.4 2 − 94.5 2
M2 for [cosC = ]
2 × 38.6 × 78.4
or M1 for
94.5 2 = 38.6 2 + 78.4 2 − 2 × 38.6 × 78.4 × cos C
and
A1 for –0.214 or –0.2144 to –0.2137
If 0 scored, SC2 for [CAD =] 23.5 or 23.51 to
23.52 or for [CDA =] 54.1 or 54.14…
3(b) 16.2 or 16.15… 3 1 1
M2 for × 21.5 × 27.6sin111 = × 34.3 × d
2 2
oe
1
or M1 for × 21.5 × 27.6sin111 seen or
2
1
× 34.3 × d oe soi
2
y
9
8
7
6
5
4
3
A 2
B
1
–7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 x
–1
–2
–3
(a) On the grid, draw the image of triangle A after
-4
(i) a translation by the vector e o, [2]
5
(ii) a reflection in the line x = 1, [2]
(iii) an enlargement, scale factor 2 and centre ( - 5, - 2 ). [2]
(b) Describe fully the single transformation that maps triangle A onto triangle B.
…… [3]
Mark scheme
4(a)(i) Image at (–5, 6) (–5, 8) (–6, 7) 2 −4 k
B1 for translation by or
k 5
4(a)(ii) Image at (3, 1) (3, 3) (4, 2) 2 B1 for reflection in y = 1 or x = k
4(a)(iii) Image at (3, 4) (3, 8) (1, 6) 2 B1 for enlargement, sf 2, in wrong position
4(b) Rotation 3 B1 for each
90º [anticlockwise] oe
(–3, 0)
The table shows some values for y = x 3 - 3x 2 + 3.
x -1 - 0.5 0 0.5 1 1.5 2 2.5 3 y 2.125 3 2.375 1 -1 - 0.125
(a) Complete the table. [3]
(b) On the grid, draw the graph of y = x 3 - 3x 2 + 3 for - 1 G x G 3.
y
3.5
3
2.5
2
1.5
1
0.5
–1 – 0.5 0 0.5 1 1.5 2 2.5 3 x
– 0.5
–1
– 1.5
[4]
(c) By drawing a suitable straight line on the grid, solve the equation x 3 - 3x 2 + x + 1 = 0 .
x = …… or x = …… or x = …… [4]
Mark scheme
5(a) –1, –0.375, 3 3 B1 for each
5(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
5(c) y = 2 – x ruled correctly 4 B2 for y = 2 – x ruled
AND or B1 for [y = ] 2 – x soi
–0.45 to –0.35 or y = k – x ruled
1 or y = kx + 2 ruled, but not y = 2
2.35 to 2.45 B2 for all three values
or B1 for any two values
(a) Solve.
(i) 4 (2x - 3) = 24
x = …… [3]
(ii) 6x + 14 2 6
…… [2]
(b) Rearrange the formula V = 2x 3 - 3y 3 to make y the subject.
y = …… [3]
(c) Show that `2n - 5j - 13 is a multiple of 4 for all integer values of n.
2
[3]
(d) The expression 5 + 12x - 2x 2 can be written in the form q - 2 `x + pj .
2
(i) Find the value of p and the value of q.
p = …… , q = …… [3]
(ii) Write down the coordinates of the maximum point of the curve y = 5 + 12x - 2x 2 .
(…… , ……) [1]
(e) The energy of a moving object is directly proportional to the square of its speed.
The speed of the object is increased by 30%.
Calculate the percentage increase in the energy of the object.
…… % [2]
Mark scheme
6(a)(i) 1 9 3 M1 for 8x – 12 = 24 or 2x – 3 = 6
4.5, 4 or M1 for reaching ax = b correctly FT their
2 2
first step
6(a)(ii) 4 1 2 14
x> − or x > –1 final answer M1 for 6x > 6 – 14 or x + >1
3 3 6
6(b) 3 M1 for isolating term in y
2 x3 − V
[y =] 3 oe final answer M1 for division by 3 or FT their first step
3 M1 for cube root or FT their previous step to
the final answer
6(c) 4n2 – 20n + 12 M2 B1 for 4n2 – 10n – 10n + 25
4(n2 – 5n + 3) A1 with no errors seen
or e.g. 4, [–]20 and 12 are all multiples of 4 or
correct explanation linked to divides each term or each coefficient by 4
expression
6(d)(i) p = –3 and q = 23 3 B2 for 23 – 2(x –3)2
OR
M1 for [q] – 2x2 – 4px – 2p2 or –2(x – 3)2
seen
B1 for either p = –3 or q = 23
or FT q = 5 + 2(their p)2
6(d)(ii) (3, 23) 1 FT their (d)(i)
6(e) 69 2 M1 for figs 132 oe
(a) The diagram shows a container for storing grain.
The container is made from a hemisphere, a cylinder and a cone, each with radius 2 m. 2m
The height of the cylinder is 5.2 m and the height of the cone is h m.
5.2 m
NOT TO
SCALE
2m
hm
(i) Calculate the volume of the hemisphere.
Give your answer as a multiple of r .
4
[The volume, V, of a sphere with radius r is V = rr 3 .]
3
…… m 3 [2]
88r 3
(ii) The total volume of the container is m .
3
Calculate the value of h.
1
[The volume, V, of a cone with radius r and height h is V = rr 2 h .]
3
h = …… [4]
(iii) The container is full of grain.
Grain is removed from the container at a rate of 35 000 kg per hour.
1m 3 of grain has a mass of 620 kg.
Calculate the time taken to empty the container.
Give your answer in hours and minutes.
…… h …… min [3]
(b)
O
NOT TO r cm 140°
B SCALE
A
A and B are points on a circle, centre O, radius r cm.
The area of the shaded segment is 65cm 2 .
Calculate the value of r.
r = …… [4]
Mark scheme
7(a)(i) 16π 1 2 1 4
or 5 π final answer M1 for × π × 2 3 oe
3 3 2 3
7(a)(ii) 2.4[0] 4 B3 for answer in range 2.396… to 2.40…
OR
16π
M3 for their + π × 22 × 5.2 +
3
1 2 88π
π ×2 ×h = oe
3 3
88π 16π
or M2 for – their – π × 22 × 5.2
3 3
oe
or M1 for π × 22 × 5.2 oe
1
or π × 22 × h oe soi
3
7(a)(iii) 1 hour 38 min or 1 hour 37.8 min to 1 3 B2 for 1.63[2…] or 98 [mins] or 97.8 to
hour 37.9… min 97.9… ]
88π
× 620
or M1 for 3 [× 60] oe
35000
7(b) 8.5[0] or 8.496 to 8.497 4 65
M3 for [r= ] oe
140 1
π − sin140
360 2
140 1 2
or M2 for π × r2 – r × sin140 [=65]
360 2
oe
or M1 for either area expression seen
(a) Kaito runs along a 12 km path at an average speed of x km/h.
(i) Write down an expression, in terms of x, for the number of hours he takes.
…… hours [1]
(ii) Yuki takes 1.5 hours longer to walk along the same path as Kaito.
She walks at an average speed of (x - 4) km/h.
Write down an equation, in terms of x, and show that it simplifies to x 2 - 4x - 32 = 0 .
[4]
(iii) Solve by factorisation. x 2 - 4x - 32 = 0
x = …… or x = …… [3]
(iv) Find the number of hours it takes Yuki to walk along the 12 km path.
…… hours [2]
(b) A bus travels 440 km, correct to the nearest 10 km.
The time taken to complete the journey is 6 hours, correct to the nearest half hour.
Calculate the lower bound of the speed of the bus.
…… km/h [3]
-2
Mark scheme
8(a)(i) 12 1
or 12 ÷ x final answer
x
8(a)(ii) 12 12 M1 Accept 3 or more term equivalents
– their = 1.5oe
x−4 x
12x – 12(x – 4) = 1.5x(x – 4) M1 Correctly clearing fractions, or correctly
or collecting into a ‘single fraction’
12 x − 12( x − 4) [= 1.5 ] FT their expression dep on two fractions both
x ( x − 4) with algebraic denominators
12x – 12x + 48 = 1.5x2 – 6x M1 Correctly multiplying their two sets of
brackets
FT their expression dep on two fractions both
with algebraic denominators or first M1
given
[ 1.5x2 – 6x – 48 = 0 ] A1 One further step either 3 term equation or
division throughout by 1.5 leading to solution
x2 – 4x – 32=0 With no errors or omissions seen, dep on M3
8(a)(iii) (x + 4)(x – 8) M2 M1 for (x + a)(x + b)
where ab = –32 or a + b = –4
or for x(x + 4) – 8(x + 4)
or x(x – 8) + 4(x – 8)
–4 and 8 B1
8(a)(iv) 3 2 FT 12
their 8 − 4
M1 for 12 or 1 2 + 1.5 oe
their 8 − 4 th eir 8
or for answer 1 2
th eir 8
8(b) 69.6 3 430 to 440 440 − 5
M2 for or oe
6 + 0.25 6 to 6.5
or M1 for 440 + 5 oe or 440 – 5 oe
or 6 + 0.25 oe or 6 – 0.25 oe seen
(a) F is the point (5, - 2 ) and FG = e o.
3
Find
(i) the coordinates of point G,
(…… , ……) [1]
(ii) 5 FG ,
f p [1]
(iii) FG .
…… [2]
(b)
Q
A B
a P NOT TO
SCALE
O c C
OABC is a parallelogram.
P is a point on AC and Q is the midpoint of AB.
OA = a and OC = c .
(i) Find, in terms of a and/or c
(a) AQ,
AQ = …… [1]
(b) OQ .
OQ = …… [1]
2 1
(ii) OP = a + c
3 3
(a) Show that O, P and Q lie on a straight line.
[2]
(b) Write down the ratio OP : OQ.
Give your answer in the form 1 : n.
1 : …… [1]
Mark scheme
9(a)(i) (3, 1) 1
9(a)(ii) −10 1
15
9(a)(iii) 3.61 or 3.605 to 3.606 2 M1 for (–2)2 + 32 oe
9(b)(i)(a) 1 1
c
2
9(b)(i)(b) 1 1 FT a + their (b)(i)(a)
a+ c oe
2
9(b)(ii)(a) ρ 2 ρ ρ
1
O P = (2a + c) oe B1 for O P or PQ factorised
3
ρ 1
and OQ = (2a + c) oe or for correct multiplicative statement on
2 relationship without factorised vectors
OR ρ ρ 2 ρ ρ
ρ 2 1 e.g. OQ = 1.5 O P , OQ = O P ,
OP = (a + c) 3
3 2 ρ ρ
OR 2 PQ = O P ,
ρ 1
PQ = (a + 1 c) 1.5 2 a + 1 c = a + 1 c
3 2 3 3 2
and
correct comment
e.g. have the same base vector or that
they are multiples of one another and
they share a common point
OR
ρ ρ ρ ρ
e.g. OQ = 1.5 O P , 2 PQ = O P
9(b)(ii)(b) 1.5 oe 1
(a) Find the coordinates of the turning points of the graph of y = x 3 - 12x + 6 .
You must show all your working.
(…… , ……) and (…… , ……) [5]
(b) Determine whether each turning point is a maximum or a minimum.
Show how you decide.
[3]
Mark scheme
10(a) (2, –10) and (–2, 22) 5 B2 for 3x2 – 12 isw
or B1 for 3x2 + k or px2 – 12 (p ≠ 0)
or for 3x2 – 12 + 6 isw
M1 for setting their derivative = 0
dy
or =0
dx
B1 for x = ± 2 or for one correct coordinate
pair
10(b) (2, –10) minimum with correct reason 3 B2 for 1 correct with correct reasoning
or sketch or B2FT for correct evaluation with correct
2nd derivative for both of their different x
(–2, 22) maximum with correct reason values
or sketch
or M1 for showing [2nd derivative =] 6x
or gradients for one value on either side of
one correct stationary point
or for reasonable sketch of cubic
(a)
R
Z
NOT TO
20.4 cm 15.3 cm 13.6 cm
SCALE
X Y
P Q
Triangle PQR is mathematically similar to triangle XYZ.
(i) Find YZ.
YZ = …… cm [2]
(ii) The area of triangle XYZ is 63.6 cm 2 .
Calculate the area of triangle PQR.
…… cm 2 [3]
(b) Two containers are mathematically similar.
The larger container has a capacity of 64.8 litres and a surface area of 0.792 m 2 .
The smaller container has a capacity of 37.5 litres.
Calculate the surface area of the smaller container.
…… m 2 [3]
Mark scheme
11(a)(i) 10.2 2 YZ 15.3
M1 for = oe or better
13.6 20.4
11(a)(ii) 143.1 3 2
20.4
M2 for × 63.6 oe
13.6
2 2
20.4 13.6
or M1 for or oe
13.6 20.4
Alt method
M2 for ½ × 20.4 × 15.3 × sin R where
−1 63.6
R is sin
0.5 ×13.6 × their (a)(i)
−1 63.6
or M1 for R = sin
0.5 ×13.6 × their (a)(i)
11(b) 0.55 3 2
37.5
M2 for [ratio of areas] = 3 or
64.8
2
64.8
3 oe
37.5
or M1 for [ratio of lengths] = 3 37.5 or
64.8
64.8
3 oe
37.5
3 2
0.792 64.8
or for = oe
x 37.5
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Mathematics 0580/42 for October/November 2021 is out of 130 marks, with 2 hours 30 minutes to answer it.
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