Mathematics 0580/41 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)
5.7 cm NOT TO
SCALE
9.2 cm
19.4 cm
The diagram shows a brick in the shape of a cuboid.
(i) Calculate the total surface area of the brick.
…… cm 2 [3]
(ii) The density of the brick is 1.9 g/cm 3 .
Work out the mass of the brick.
Give your answer in kilograms.
[Density = mass ÷ volume]
…… kg [3]
(b) 9000 bricks are needed to build a house.
200 bricks cost $175.
Work out the cost of the bricks needed to build 5 houses.
$ …… [3]
(c) Saskia builds a wall using 1500 bricks.
She can build at the rate of 40 bricks each hour.
She works for 9 hours each day.
Saskia starts work on 6 July and works every day until the wall is completed.
Find the date when she completes the wall.
…… [3]
(d) Rafa has a cylindrical tank.
The cylinder has a height of 105 cm and a diameter of 45 cm.
Calculate the capacity of the tank in litres.
…… litres [3]
Mark scheme
1(a)(i) 683 3 M2 for [2]((19.4 × 9.2) + (5.7 × 9.2) +
(19.4 × 5.7)) oe
or M1 for one of 19.4 × 9.2 or 5.7 × 9.2
or 19.4 × 5.7
1(a)(ii) 1.93[0] or 1.932 to 1.933 3 M2 for 19.4 × 9.2 × 5.7 × 1.9
or M1 for 19.4 × 9.2 × 5.7
1(b) 39 375 3 M2 for 9000 ÷ 200 × 175 × 5
175
or M1 for 9000 ÷ 200 soi or for soi
200
1(c) 10th July 3 1
B2 for 4.1 to 4.2 or 4 or 4 days 1.5
6
hours
Or M2 for answer 9th July or 11th July
or M1 for 1500 ÷ (9 × 40)
1(d) 167 or 166.9 to 167.0… 3 B2 for answer with figs 167 or figs 1669 to
1670..
or M1 for π × 22.52 × 105 oe
If 0 scored SC1 for answer 668 or 667.9 to
668.1
Bob, Chao and Mei take part in a run for charity.
(a) Their times to complete the run are in the ratio Bob : Chao : Mei = 4 : 5 : 7.
(i) Find Chao’s time as a percentage of Mei’s time.
…… % [1]
(ii) Bob’s time for the run is 55 minutes 40 seconds.
Find Mei’s time for the run.
Give your answer in minutes and seconds.
…… min …… s [3]
(b) Chao collects $47.50 for charity.
(i) Bob collects 28% more than Chao.
Find the amount Bob collects.
$ …… [2]
(ii) Chao collects 60% less than Mei.
Find how much more money Mei collects than Chao.
$ …… [3]
(c) When running, Chao has a stride length of 70 cm, correct to the nearest 5 cm.
Chao runs a distance of 11.2 km, correct to the nearest 0.1 km.
Work out the minimum number of strides that Chao could take to complete this distance.
…… [4]
(d) In 2015, a charity raised a total of $1.6 million.
After 2015, this amount increased exponentially by 2.4% each year for the next 5 years.
Work out the amount raised by the charity in 2020.
$ …… million [2]
Mark scheme
2(a)(i) 71.4 or 71.42 to 71.43 1
2(a)(ii) 97 [min] 25 [s] 3 B2 for 13 min 55 sec seen or 97.4 or 97.41
to 97.42 seen or 5845 seen
OR
M2 for 55.66… ÷ 4 × 7 oe or 3340 ÷ 4 × 7
oe
or for 7/4 × 55 + 7/4 × 40 oe
or M1 for 55 min 40 sec ÷ 4 oe
or M1 for total time ÷ 16 soi
2(b)(i) 60.8[0] 2 28
M1 for 47.5 × 1 + oe
100
or B1 for 13.3[0]
2(b)(ii) 71.25 3 B2 for 118.75
60
Or M2 for 47.50 ÷ 1 − – 47.50
100
60
or M1 for x × 1 − = 47.50 oe or
100
better
2(c) 15 380 4 M3 for (1 120 000 – 5000) ÷ (70 + 2.5)
oe
or
B2 for answer figs 15 379 to figs 15 380
or
M2 for (1 120 000 ± 5000) ÷ (70 ± 2.5)
oe
or
M1 for one of figs 675, 725, 1115, 1125
seen
2(d) 1.8[0] or 1.801 to 1.802 [million] nfww 2 2.4
5
M1 for figs 16 × 1 + oe
100
The cumulative frequency diagram shows information about the mass, m kg, of each of 80 boys.
80
60
Cumulative frequency 40
20
0 m
30 40 50 60 70 80 90
Mass (kg)
(a)
30 40 50 60 70 80 90 m
Mass (kg)
On the grid, draw a box-and-whisker plot to show the information in the cumulative frequency diagram. [4]
(b) Use the cumulative frequency diagram to find an estimate of
(i) the 30th percentile,
…… kg [2]
(ii) the number of boys with a mass greater than 75 kg.
…… [2]
(c) (i) Use the cumulative frequency diagram to complete this frequency table.
Mass
30 1 m G 40 40 1 m G 50 50 1 m G 60 60 1 m G 70 70 1 m G 80 80 1 m G 90
(m kg)
Frequency 8 12 14 10
[1]
(ii) Calculate an estimate of the mean mass of the boys.
…… kg [4]
(iii) Two boys are chosen at random from those with a mass greater than 70 kg.
Find the probability that one of them has a mass greater than 80 kg and the other has a mass of 80 kg or less.
…… [3]
Mark scheme
3(a) Correct box-and-whisker plot 4 B1 for lowest value and highest value at
30 and 90
B1 for LQ and UQ at 50 and 72
B1 for median at 63
3(b)(i) 56 2 M1 for 24 soi
3(b)(ii) 16 2 B1 for 64 written
3(c)(i) 14, 22 1
3(c)(ii) 61.5 4 M1 for 35, 45, 55, 65, 75, 85 soi
M1 for Σfx
M1 dep for their Σfx ÷ (8 + 12 +their 14 +
their 22 + 14 +10) or Σfx ÷ 80
3(c)(iii) 35 3 10 14
oe M2 for [2] × oe
69 24 23
10 14
or M1 for or oe seen
24 24
35
If 0 scored, SC1 for answer oe
72
(a) Solve.
(i) 6 (7 - 2x) = 3x - 8
x = …… [3]
2x 2
(ii) = x-5 3
x = …… [3]
(b) Factorise completely.
(i) 2x 2 - 288y 2
…… [3]
(ii) 5x 2 + 17x - 40
…… [2]
(c) Solve x 3 + 4x 2 - 17x = x 3 - 9 .
You must show all your working and give your answers correct to 2 decimal places.
x = …… or x = …… [5]
Mark scheme
4(a)(i) 10 1 3 M1 for 42 – 12x = 3x – 8 oe
or 3 or 3.33[3…]
3 3 3x 8
or for 7 – 2x = − oe
6 6
M1 for reaching ax = b correctly FT their
first step
4(a)(ii) 1 5 3 M1 for 3 × 2x = 2(x – 5) oe
–2.5 or −2 or −
2 2
M1 for reaching ax = b correctly FT their
first step
4(b)(i) 2(x + 12y)(x – 12y) final answer 3 B2 for (2x + 24y)(x – 12y)
or (2x – 24y)(x + 12y)
or for 2(x + 12y)(x – 12y) seen
OR
M2 for k(x + 12y)(x – 12y)
or M1 for 2(x2 – 144y2)
4(b)(ii) (5x – 8) (x + 5) final answer 2 M1 for 5x(x + 5) – 8(x + 5)
or x (5x – 8)+ 5(5x – 8)
or for (5x + a)(x + b) where ab = – 40 or
a + 5b = 17
4(c) 4x2 – 17x + 9 [= 0] oe B1
B2 FT their 3 term quadratic
([−]17 ) − 4 ( 4 )( 9 )
2
[− −]17 ±
2× 4 B1FT for ([−]17 )2 − 4 ( 4) ( 9 ) ) or better
([ −]17 ) − 4 ( 4 )( 9 )
2 2
17
or x − oe or
8 4
or better
[− −]17 + q
and B1FT for or
2(4)
[− −]17 − q
or better
2(4)
17 145 17 145
or + oe or − oe or
8 64 8 64
[ − −]17 + q [ − −]17 − q
2 or 2
4 4
0.62 and 3.63 cao B2 B1 for each
SC1 for 0.6[0] or 0.619 to 0.620 and
3.6[0] or 3.6301 to 3.6302
or 0.62 and 3.63 seen in working
or –0.62 and–3.63 as final answers
(a)
D
A
NOT TO
SCALE
O 124°
B
35°
C
A, B, C and D are points on a circle, centre O.
Angle COD = 124° and angle BCO = 35°.
(i) Work out angle CBD.
Give a geometrical reason for your answer.
Angle CBD = …… because ……
…… [2]
(ii) Work out angle BAD.
Give a geometrical reason for each step of your working.
Angle BAD = …… because ……
…… [4]
(b)
R
42°
NOT TO
S SCALE
O
Q
5.9 cm
P
P, Q, R and S are points on a circle, centre O.
QS is a diameter.
Angle PRS = 42° and PQ = 5.9 cm.
Calculate the circumference of the circle.
…… cm [5]
3
Mark scheme
5(a)(i) 62 2 B1 for either
and
Angle at centre is twice angle at
circumference oe
5(a)(ii) 117 4 B2 for 117
and or B1 for [angle OCD =] 28
Isosceles [triangle] B1dep for isosceles [triangle]
and and
Opposite angles in a cyclic quadrilateral B1 for opposite angles in a cyclic
are supplementary quadrilateral are supplementary
5(b) 24.9 or 24.94 to 24.95 5 B1 for angle PQS = 42
M2 for QS = 5.9 ÷ cos 42 oe
5.9
or M1 for cos42= oe
QS
M1dep for their SQ × π oe
The table shows some values for y = x 2 - , x ! 0 , given correct to 1 decimal place.
2x
x -3 -2 -1 -0.5 -0.2 0.2 0.5 1 2 3 y 2.5 3.3 7.5 -7.5 -2.8 -0.5 3.3
(a) (i) Complete the table. [3]
3
(ii) On the grid, draw the graph of y = x 2 - for - 3 G x G - 0.2 and 0.2 G x G 3.
2x y
10
9
8
7
6
5
4
3
2
1
– 3 – 2.5 – 2 – 1.5 – 1 – 0.5 0 0.5 1 1.5 2 2.5 3 x
–1
–2
–3
–4
–5
–6
–7
–8
[5]
3 24
(b) By drawing a suitable straight line on the grid, solve the equation x 2 - = - 2x
2x 5 for - 3 G x G - 0.2 and 0.2 G x G 3.
x = …… or x = …… [4]
3 24
(c) The solutions to the equation x2 -
= - 2x are also the solutions to an equation of the
2x 5
3 2 form ax + bx + cx - 15 = 0 where a, b and c are integers.
Find the values of a, b and c.
a = ……
b = ……
c = …… [4]
Mark scheme
6(a)(i) 9.5, 4.8 and 8.5 3 B1 for each
6(a)(ii) correct curve 5 B4 for correct curve, but branches joined
or touching y axis
or B3FT for 9 or 10 correct plots
or B2FT for 7 or 8 correct plots
or B1FT for 5 or 6 correct plots
AND
B1 indep two separate branches not
touching or cutting y-axis
6(b) 24 4 B2 for correct ruled line crossing curve
y= − 2 x ruled twice
5
and
– 0.4 to – 0.2 and 1.45 to 1.7 or B1 for correct freehand or for short
ruled line or for line with negative gradient
through (0, 4.8)
or for line with gradient – 2
B1 for each value
6(c) [a =] 10 4 B3 for 10x3 – 15 = 48x – 20x2 oe or better
[b =] 20 or B2 for 2 correct values
[c =] – 48 or B1 for 1 correct value
15
or for 5 x 2 − = 24 − 10 x or better
2x
48
or for 2 x3 − 3 = x − 4 x 2 or better
5
3 24
or for x3 − = x − 2 x2
2 5
After 0 scored SC1 for correct elimination
of a denominator of 5, x or 2x from a four
term expression.
7(a)(i)(a) Shape at (–2, 1) ( –4, 1) ( –4, 7) (0, 7) 2 B1 for 3 correct points or for enlargement
SF2 from any centre
7(a)(i)(b) Shape at (2, –2) (2, –3) (5, –1) (5, –3) 3 B2 for correct orientation but wrong
position or for 3 correct points
or B1 for y = x – 1 drawn
(a) y
8
7
6
5
4
3
A
2
1
–5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 x
–1
B
–2
–3
–4
(i) On the grid, draw the image of
(a) shape A after an enlargement, scale factor 2, centre (0, 1), [2]
(b) shape A after a reflection in the line y = x - 1. [3]
(ii) Describe fully the single transformation that maps shape A onto shape B.
…… [3]
(b)
C B
NOT TO q M
SCALE
O p A
OABC is a trapezium and O is the origin.
M is the midpoint of AB.
OA = p , OC = q and OA = 2CB.
Find, in terms of p and q, the position vector of M.
Give your answer in its simplest form.
…… [3]
Mark scheme
7(a)(ii) Rotation 3 B1 for each
90 [anticlockwise] oe
(0, 0) oe
7(b) 3 1 1 3 p + 2q 3 ρ 1 1
p + q or ( 3p + 2q ) or M2 for AM = AM = - p + q + p oe
4 2 4 4 2 2
final answer ρ
or M1 for correct route for AB oe soi by
–½ p + q
ρ
or for OM soi
(a) f (x) = 3 - 5x
(i) Find x when f (x) =- 5.
x = …… [2]
(ii) Find f - 1 (x) .
f - 1 (x) = …… [2]
(b) g (x) = 18 - 3x - x 2
(i) Write g (x) in the form b - (a + x) 2 .
…… [3]
(ii) Sketch the graph of y = g (x) .
On your sketch, show the coordinates of the turning point.
y
O x
[3]
(iii) Find the equation of the tangent to the graph of y = 18 - 3x - x 2 at x = 4 .
Give your answer in the form y = mx + c .
y = …… [6]
Mark scheme
8(a)(i) 1.6 oe 2 M1 for 3 – 5x = – 5
8(a)(ii) 3− x 2 y 3
oe final answer M1 for x = 3 – 5y or = − x or better,
5 5 5
or y – 3 = – 5x oe
8(b)(i) 20.25 − (1.5 + x )
2 3 Method 1
B1 for ( ±1.5 ± x ) seen
2
B1 for [b =] 18 + their 1.52
OR
Method 2
B1 for b − a 2 − 2ax − x 2 or for b = 20.25
B1 for a = 1.5
8(b)(ii) Correct sketch with max in correct 3 FT their 20.25 − ( their1.5 + x )2 provided in
quadrant at ( −1.5, 20.25 )
that form
B1 for ∩ shape or for ∪ shape if in form
c + ( d + x ) in part (b)(i)
2
B1 for TP at ( −1.5, k ) or ( k , 20.25 ) FT
their 20.25 ± ( their 1.5 + x ) or for (–1.5,
2
20.25) seen
8(b)(iii) [y =] 34 – 11x 6 B2 for –3 – 2x
or B1 for either kx –3, k ≠ 0 or –2x + n or
for
18 – 3 – 2x
M1dep for gradient = their (–3 – 2(4) )
B1 for y-value at x = 4, is –10
M1dep for their –10 = (their –11)4 + c
oe
(a)
NOT TO x cm SCALE
(x + 3) cm
This rectangle has perimeter 20 cm.
Find the value of x.
x = …… [3]
(b)
M y°
NOT TO
SCALE
20°
This rhombus has perimeter 20 cm and angle y is obtuse.
M is the midpoint of one of the sides.
Find the value of y.
y = …… [5]
(c)
r cm NOT TO
SCALE z cm
40°
This sector of a circle has radius r and perimeter 20 cm.
Find the value of z.
z = …… [6]
Mark scheme
9(a) 3.5 oe 3 M1 for 2(x + x + 3) = 20 oe
M1 for correct ax = b for their linear
equation
9(b) 116.8 or 116.83 to 116.85 nfww 5 5sin 20
M2 for sin p =
2.5
2.5 5
or M1 for =
sin 20 sin p
A1 for 43.2 or 43.15 to 43.17
M1dep for 180 – (20 + their 43.2)
After 0 scored, SC1 for length of side = 5
9(c) 5.07 or 5.068 to 5.071 6 B3 for 7.41 or 7.412 to 7.413
40
or M2 for r + r + × 2 × π × r = 20 oe
360
40
or M1 for × 2 × π × r oe seen
360
M2 for 2 × 7.41 × sin 20 oe
or 7.412 + 7.412 – 2(7.412) cos 40 oe
7.41sin 40
or oe
sin 70
or M1 for implicit version
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On this page, free: the mark scheme for IGCSE Mathematics 0580/41, October/November 2021 (0580_w21_ms_41), and the question paper (0580_w21_qp_41), to read or download, with each question's marking points written out underneath.
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Mathematics 0580/41 for October/November 2021 is out of 130 marks, with 2 hours 30 minutes to answer it.
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