Mathematics 0580/42 May/June 2021: question paper and mark scheme
Cambridge IGCSE Mathematics (0580), Paper 4 (Extended), May/June 2021. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
(a) A 2.5-litre tin of paint costs $13.50 .
In a sale, the cost is reduced by 14%.
(i) Work out the sale price of this tin of paint.
$ …… [2]
(ii) Work out the cost of buying 42.5 litres of paint at this sale price.
$ …… [2]
(b) Henri buys some paint in the ratio red paint : white paint : green paint = 2 : 8 : 5.
(i) Find the percentage of this paint that is white.
…… % [1]
(ii) Henri buys a total of 22.5 litres of paint.
Find the number of litres of green paint he buys.
…… litres [2]
(c) Maria paints a rectangular wall.
The length of the wall is 20.5 m and the height is 2.4 m, both correct to 1 decimal place.
One litre of paint covers an area of exactly 10 m2.
Calculate the smallest number of 2.5-litre tins of paint she will need to be sure all the wall is painted.
Show all your working.
…… [4]
Mark scheme
1(a)(i) 11.61 final answer 2 14
M1 for 13.5[0] × 1 − oe
100
or B1 for 1.89
1(a)(ii) 197.37 final answer 2 FT 17 × their (a)(i) exact or correct to
nearest cent
M1 for 42.5 ÷ 2.5
1(b)(i) 53.3 or 53.33… 1
1(b)(ii) 7.5 2 M1 for 22.5 ÷ (2 + 8 + 5) oe soi
1(c) 20.55 × 2.45 oe M2 M1 for 20.5 + 0.05 oe seen or 2.4 + 0.05
oe seen
If 0 scored, SC1 here for 20.45 × 2.35 oe
3 nfww A2 M1 for their area ÷ 10 ÷ 2.5 oe
The table shows some values for y = 2 # 0.5 x - 1.
x -1 -0.5 0 0.5 1 1.5 2 y 3 1.83 0.41 0 -0.29
(a) (i) Complete the table. [2]
(ii) On the grid, draw the graph of y = 2 # 0.5 x - 1 for - 1 G x G 2 .
y
3
2
1
–1 – 0.5 0 0.5 1 1.5 2 x
–1
–2
[4]
(b) By drawing a suitable straight line, solve the equation 2 # 0.5 x + 2x - 3.5 = 0 for - 1 G x G 2 .
x = …… [3]
(c) There are no solutions to the equation 2 # 0.5 x - 1 = k where k is an integer.
Complete the following statements.
The highest possible value of k is ……
The equation of the asymptote to the graph of y = 2 # 0.5 x - 1 is …… [2]
Mark scheme
2(a)(i) 1, –0.5 oe 2 B1 for each
2(a)(ii) Correct curve 4 B3FT for 6 or 7 correct plots
or B2FT for 4 or 5 correct plots
or B1FT for 2 or 3 correct plots
2(b) y = 2.5 − 2x ruled B2 B1 for y = k − 2 x or y = px + 2.5 ruled
(p ≠ 0)
or for [y =] 2.5 − 2x oe identified
1.3 to 1.4 B1
2(c) –1 B1
y = –1 B1 FT their k (must be negative)
(a) Simplify, giving your answer as a single power of 7.
(i) 75 # 76
…… [1]
(ii) 7 15 ' 7 5
…… [1]
(iii) 42 + 7
…… [1]
(b) Simplify.
(5x 2 # 2xy 4) 3
…… [3]
(c) P = 25 # 33 # 7 Q = 540
(i) Find the highest common factor (HCF) of P and Q.
…… [2]
(ii) Find the lowest common multiple (LCM) of P and Q.
…… [2]
(iii) P # R is a cube number, where R is an integer.
Find the smallest possible value of R.
…… [2]
(d) Factorise the following completely.
(i) x 2 - 3x - 28
…… [2]
(ii) 7 (a + 2b) 2 + 4a (a + 2b)
…… [2]
1
(e) 3 2x - 1 = #3
2y - x
9x
Find an expression for y in terms of x.
y = …… [4]
Mark scheme
3(a)(i) 711 cao 1
3(a)(ii) 710 cao 1
3(a)(iii) 72 cao 1 If answers 11, 10 and 2 in (a) then allow
SC1 in this part
3(b) 1000x9y12 final answer 3 B2 for correct answer seen or answer of
the form 1000x9yk or 1000xky12 or kx9y12
or B1 for answer with one correct element
in product
or (10x3y4)[3] seen
3(c)(i) 108 2 M1 for [540 =] 22 [×] 33 [×] 5
or B1 for 108 oe not in prime factor form
e.g. 22 × 3 × 9
3(c)(ii) 30 240 2 M1 for (540 × 25 × 33 × 7) ÷ their (c)(i) oe
or B1 for answer 30 240 oe not in prime
factor form
e.g. 25 × 33 × 35
3(c)(iii) 98 2 B1 for 592 704 seen or 26 × 33 × 73 seen
or 2 × 72 oe seen
3(d)(i) (x – 7) (x + 4) final answer 2 M1 for x(x – 7) + 4(x – 7) or
x(x + 4) – 7 (x + 4) or better
or for (x + a)(x + b) where ab = – 28
or a + b = – 3
3(d)(ii) (a + 2b)(11a + 14b) final answer 2 M1 for (a + 2b) (7(a + 2b) + 4a)
or (a + pb)(11a + qb) where pq = 28
or 11p + q = 36
If 0 scored, SC1 for a + 2b (11a + 14b)
3(e) 5x − 1 4 B2 for 2x – 1 = –2x + 2y – x oe
[ y =] oe final answer or B1 for 9x = 32x or better
2
M1dep for correct rearrangement of their
5 term ‘linear’ equation in y and x to make
y the subject
(a) The mass, m kg, of each of 40 parcels in a warehouse is recorded.
The table shows information about the masses of these parcels.
Mass (m kg) 0.5 1 m G 1 11mG2 21mG4 41mG7 7 1 m G 12
Frequency 4 7 15 10 4
(i) Complete the histogram to show this information.
9
8
7
6
5
Frequency density
4
3
2
1
0
0 1 2 3 4 5 6 7 8 9 10 11 12 m
Mass (kg)
[3]
(ii) Calculate an estimate of the mean mass of the parcels.
…… kg [4]
(iii) A parcel is picked at random from the 40 parcels.
Find the probability that this parcel has a mass of 2 kg or less.
…… [1]
(iv) Two parcels are picked at random without replacement from those with a mass greater than 2 kg.
Work out the probability that one of them has a mass greater than 7 kg and the other has a mass of 4 kg or less.
…… [3]
(b) A van delivers parcels from a different warehouse.
The box-and-whisker plot shows information about the masses of the parcels in the van.
0 1 2 3 4 5 6 7 8 9
Mass (kg)
(i) Find the median.
…… kg [1]
(ii) Find the interquartile range.
…… kg [1]
(iii) Two parcels are removed from the van at the first delivery.
The masses of these parcels are 2.4 kg and 5.8 kg.
Describe the effect that removing these parcels has on the median mass of the remaining parcels.
Give a reason for your answer.
…… [2]
-3 2
Mark scheme
4(a)(i) Correct histogram 3 B1 for each correct block
If 0 scored, SC1 for any two of fds 7.5,
3.33…, 0.8 oe soi
4(a)(ii) 3.7875 or 3.79 or 3.787 or 3.788 4 M1 for 0.75, 1.5, 3, 5.5, 9.5 soi
M1 for Σfx
M1 dep for their Σfx ÷ 40
4(a)(iii) 11 1
oe
40
4(a)(iv) 30 3 4 15
oe M2 for [2 ×] × oe
203 29 28
4 15
or M1 for or oe seen
29 29
4 26
After 0 scored, SC1 for [ 2 ×] ×
40 39
oe
120
or for answer oe
841
4(b)(i) 4.6 1
4(b)(ii) 3.2 1
4(b)(iii) [median] remains the same oe 2 B1 for each statement
and
one is below [the median/middle] and one
is above oe
5(a)(i)(a) 5 1
final answer
−13
5(a)(i)(b) −4 2 −4 k −6
final answer B1 for answer or or
11 k 11 16
seen
5(a)(i)(c) 5.39 or 5.385… 2 M1 for 22 + ([–]5)2
(a) a=e o b =e o
8 -5
(i) Find
(a) b - a ,
f p [1]
(b) 2a + b ,
f p [2]
(c) b.
…… [2]
13
(ii) a + kb = e o , where k and m are integers. m
Find the value of k and the value of m.
k = ……
m = …… [3]
(b)
C B
M N NOT TO q SCALE
O p A
OABC is a parallelogram and O is the origin.
M is the midpoint of OB.
N is the point on AB such that AN : NB = 3 : 2.
OA = p and OC = q.
(i) Find, in terms of p and q, in its simplest form.
(a) OB
OB = …… [1]
(b) CM
CM = …… [2]
(c) MN
MN = …… [2]
(ii) CB and ON are extended to meet at D.
Find the position vector of D in terms of p and q.
Give your answer in its simplest form.
…… [3]
Mark scheme
5(a)(ii) [k =] 8 3 B2 for k = 8 or m = –32
[m =] – 32 or M1 for – 3 + 2k = 13 oe
or for m = –5 × their k + 8 correctly
evaluated
5(b)(i)(a) p + q final answer 1
5(b)(i)(b) 1 1 1 p–q 2 M1 for unsimplified answer or any correct
p – q or (p – q) or final ρ
2 2 2 2 vector route for CM , e.g.
answer 1
–q + their (b)(i)(a)
2
5(b)(i)(c) 1 1 5p + q 2 M1 for unsimplified answer or any correct
p+ q or final answer ρ
2 10 10 vector route for MN
5(b)(ii) 5 5p + 3q 3 B2 for unsimplified correct answer
p + q or final answer OR
3 3
3
M1 for p + q seen
5
B1 for final answer of form kp + q (k > 1)
5
or final answer p + jq oe (any j)
3
B
16 m
NOT TO
A 57° 32 m SCALE
19 m
C
75°
D
The diagram shows a quadrilateral ABCD made from two triangles, ABD and BCD.
(a) Show that BD = 16.9 m, correct to 1 decimal place.
[3]
(b) Calculate angle CBD.
Angle CBD = …… [4]
(c) Find the area of the quadrilateral ABCD.
…… m2 [3]
(d) Find the shortest distance from B to AD.
…… m [3]
Mark scheme
6(a) 16 2 + 19 2 – 2 × 16 × 19cos57 oe M2 or M1 for 162 + 192 – 2 × 16 × 19 cos 57
A1 for 285.8 to 285.9
16.90 to 16.91 A1
6(b) 74.3 or 74.30 to 74.33 4 16.9 × sin 75
M2 for [sin ... =] oe
32
16.9 32
or M1 for = oe
sin C sin 75
B1 for [angle BCD =] 30.7 or 30.67 to
30.69…
or M1dep for 105 – their angle BCD
6(c) 388 or 387.7 to 387.9… nfww 3 1
M1 for × 16 × 19 × sin 57 oe
2
1
M1 for × 16.9 × 32 × sin their (b) oe
2
6(d) 13.4 or 13.41 to 13.42 nfww 3 x
M2 for = sin 57 oe
16
or M1 for distance required is
perpendicular to AD soi
y
6
5
4
T
3
2
1
–6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 x
–1
–2
A –3
–4
–5
–6
(a) On the grid, draw the image of
2
(i) triangle T after a translation by the vector e o, [2]
-1
(ii) triangle T after a rotation, 90° clockwise, about the origin, [2]
1
(iii) triangle T after an enlargement, scale factor - , centre (-2, 3). [2]
2
(b) Describe fully the single transformation that maps triangle T onto triangle A.
…… [2]
Mark scheme
7(a)(i) Triangle at (4, 0) (4, 3) (6, 3) 2 2 k
B1 for translation by or
k −1
If 0 scored SC1 for triangle at (3, 0.5)
(3, 3.5) (5, 3.5)
7(a)(ii) Triangle at (1, –2) (4, –4) (4, –2) 2 B1 for rotation 90 clockwise wrong centre
or for rotation 90 anticlockwise about the
origin
7(a)(iii) Triangle at (–4, 4) (–4, 2.5) 2 1
(–5, 2.5) B1 for enlargement SF − with wrong
2
centre
1
or for enlargement SF with centre
2
(–2, 3)
7(b) Reflection 2 B1 for each
y = – x oe
(a) A cuboid has length L cm, width W cm and height H cm.
L cm
H cm
NOT TO
SCALE
20.1 cm W cm
37.8 cm
The diagram shows the net of this cuboid.
The ratio W : L = 1 : 2.
Find the value of L, the value of W and the value of H.
L = ……
W = ……
H = …… [5]
(b)
E
NOT TO
SCALE
24 cm
D C
15 cm
A 18 cm B
The diagram shows a solid pyramid with a rectangular base ABCD.
E is vertically above D.
Angle EDC = angle EDA = 90°.
AB = 18 cm, BC = 15 cm and EC = 24 cm.
(i) The pyramid is made of wood and has a mass of 800 g.
Calculate the density of the wood.
Give the units of your answer.
1
[The volume, V, of a pyramid is V = # area of base # height.]
3
[Density = mass ' volume]
…… …… [5]
(ii) Calculate the angle between BE and the base of the pyramid.
…… [4]
Mark scheme
8(a) [L =] 11.8 5 M1 for L = 2W oe soi
[W =] 5.9 M1 for W + 2H = 20.1 oe
[H =] 7.1 M1 for 2L + 2H = 37.8 oe
B1 for at least one correct answer
8(b)(i) 0.559 to 0.56[0…] B4 1
M2 for × 18 × 15 × 242 − 182 isw
3
conversion
or M1 for h2 + 182 = 242 oe or better
M1 for figs 800 ÷ figs their volume isw
g/cm3 or g cm–3 final answer B1
8(b)(ii) 34.1 or 34.11 to 34.12 4 242 − 182
M3 for tan [ ] = oe
182 + 152
or M2 for 182 + 152 isw
or 24 2 + 152 isw
or M1 for 182 + 152 isw or 242 + 152 isw
or M1 for indicating required angle is EBD
(a) (i) The equation y = x 3 - 4x 2 + 4x can be written as y = x (x - a) 2 .
Find the value of a.
a = …… [2]
(ii) On the axes, sketch the graph of y = x 3 - 4x 2 + 4x , indicating the values where the graph meets the axes.
y
O x
[4]
(b) Find the equation of the tangent to the graph of y = x 3 - 4x 2 + 4x at x = 4.
Give your answer in the form y = mx + c .
y = …… [7]
Question 10 is printed on the next page.
Mark scheme
9(a)(i) 2 2 M1 for x(x2 – 4x + 4) or x (x – 2)2
or (x2 – 2x) (x – 2) or x3 – 2ax2 + a2x
9(a)(ii) Correct sketch with curve passing through 4 B1 for any positive cubic
O and touching (2, 0) B1 for sketch through or touching O
B1 for sketch with min or max touching
x-axis once only but not at (0, 0)
B1FT their (a)(i) for sketch with min or
max touching x-axis at (their 2, 0) and
their 2 is labelled or clearly indicated
9(b) y = 20x – 64 final answer nfww 7 B6 for equivalent correct equation
OR
B2 for 3x2 – 8x + 4 isw
or B1 for 3x2 or –8x seen
M2dep for [grad =] 20 soi nfww
or M1dep for substituting 4 into their
derivative isw
B1 for (4, 16) soi
M1dep for 16 = their 20 × 4 + c oe
The table shows four sequences A, B, C and D.
Sequence 1st term 2nd term 3rd term 4th term 5th term nth term
A 1 8 27 64
B 5 11 17 23
C 0.25 0.5 1 2 4
D 4.75 10.5 16 21
Complete the table.
[9]
Mark scheme
10 125 n3 oe final ans B2 B1 for 125
B1 for n3
29 6n – 1 oe final ans B3 B1 for 29
B2 for 6n – 1 oe
or B1 for 6n + k or an – 1 (a ≠ 0)
2n – 3 oe final ans B2 B1 for 2n [+ k] oe
25 6n – 1 – 2n – 3 oe final ans B2 FT their 29 – 4 and
their 6n – 1 – their 2n – 3
OR B1FT for each
1 3 1 2 17 OR
25.25 − n + n + n − 1 oe B1 for each
24 8 3
final ans
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