Mathematics 0580/42 February/March 2021: question paper and mark scheme
Cambridge IGCSE Mathematics (0580), Paper 4 (Extended), February/March 2021. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
Painter Plumber Electrician
$35 per hour Fixed charge $40 $48 per hour for the first 2 hours plus then
$26.50 per hour $32 per hour
These are the rates charged by a painter, a plumber and an electrician who do some work for Mr Sharma.
(a) The painter works for 7 hours.
Calculate the amount Mr Sharma pays the painter.
$ …… [1]
(b) Mr Sharma pays the plumber $252.
Calculate how many hours the plumber works.
…… hours [2]
(c) Mr Sharma pays the electrician $224.
Calculate how many hours the electrician works.
…… hours [2]
(d) Write down the ratio of the amount Mr Sharma pays to the painter, the plumber and the electrician.
Give your answer in its lowest terms.
painter : plumber : electrician = …… : …… : …… [2]
Mark scheme
1(a) 245 1
1(b) 8 2 M1 for 40 + 26.5x = 252 oe
or B1 for 212 seen
1(c) 6 2 M1 for (224 − 2 × 48 ) ÷ 32 oe
or 2 × 48 + 32 (x – 2) = 224 soi
1(d) 35 : 36 : 32 final answer 2 B1 for their (a) : 252 : 224 or equivalent ratio
y
14
12
10
B
8
A
6
4
2
– 14 – 12 – 10 –8 –6 –4 –2 0 2 4 6 8 10 x
–2
–4
C
–6
–8
– 10
(a) Describe fully the single transformation that maps
(i) triangle A onto triangle B,
…… [3]
(ii) triangle A onto triangle C.
…… [3]
-5
(b) Draw the image of triangle A after a translation by the vector e o. [2]
- 10
(c) Draw the image of triangle A after a reflection in the line y = 4 . [2]
Mark scheme
2(a)(i) rotation 3 B1 for each
90 anticlockwise oe
(–3, 2)
2(a)(ii) enlargement 3 B1 for each
1
−
2
( −2, − 1)
2(b) Image at (– 3, – 5) (1, − 5) (1, 3) 2 −5 k
B1 for translation by or
k −10
2(c) Image at (2, 3) (6, 3) (6, − 5) 2 B1 for reflection in y = k or x = 4
(a)
a°
NOT TO
126° SCALE c°
b° 63°
The diagram shows two straight lines intersecting two parallel lines.
Find the values of a, b and c.
a = ……
b = ……
c = …… [3]
(b)
Q
NOT TO
SCALE
S
R
58° x°
P
Points R and S lie on a circle with diameter PQ.
RQ is parallel to PS.
Angle RPQ = 58° .
Find the value of x, giving a geometrical reason for each stage of your working.
x = …… [3]
(c)
O NOT TO
SCALE
142° C
A y°
B
Points A, B and C lie on a circle, centre O.
Angle AOC = 142° .
Find the value of y.
y = …… [2]
Mark scheme
3(a) 126 3 B1 for each
54
117
3(b) angle [in a] semicircle is 90 B1 Do not accept triangle for angle
Allied, co-interior [add to 180] B1
or
Angles in triangle [ = 180] and
alternate oe
3(c) 109 2 B1 for 218 or 71 in correct places or correctly
labelled
(a) A shop gives each of 1000 people a voucher.
28 people use their voucher.
The shop now gives each of 16 500 people a voucher.
Calculate how many of these 16 500 people are expected to use their voucher.
…… [1]
(b) In a class activity, all the 15 students wear hats.
7 students wear red hats, 6 students wear green hats and 2 students wear white hats.
(i) One of these students is picked at random.
Find the probability that this student wears a red hat.
…… [1]
(ii) Two of the 15 students are picked at random.
37
Show that the probability that these two students wear hats of the same colour is .
105
[3]
(iii) Three of the 15 students are picked at random.
Find the probability that at least two of these three students wear red hats.
…… [4]
Mark scheme
4(a) 462 1
4(b)(i) 7 1
oe
15
4(b)(ii) 7 6 6 5 2 1 3 M2 for addition of two of
× + × + × 7 6 6 5 2 1
15 14 15 14 15 14 × + × + ×
37 15 14 15 14 15 14
= or M1 for one of the products seen
105
4(b)(iii) 29 4 M3 for
oe 7 6 5 7 6 6 7 6 2
65 × × + 3× × × + 3× × × oe
15 14 13 15 14 13 15 14 13
8 7 7 8 7 6
or 1 − 3 × × − × × oe
15 14 13 15 14 13
or M2 for the sum of at least
two of
7 6 5 7 6 6 7 6 2
× × , N× × × , N× × ×
15 14 13 15 14 13 15 14 13
seen
7 6 13
or for × ×
15 14 13
7 6 7 6 k
or × + N × × × seen
15 14 15 14 13
or M1 for
7 6 5 7 6 6 7 6 2
× × or N × × × or N × × ×
15 14 13 15 14 13 15 14 13
seen
1519
If 0 scored SC1 for oe
3375
C
B
65°
NOT TO
SCALE
4.4 cm
9.7 cm
A
8.6 cm 42°
D
(a) Calculate angle ADB.
Angle ADB = …… [3]
(b) Calculate DC.
DC = …… cm [4]
(c) Calculate the shortest distance from C to BD.
…… cm [3]
Mark scheme
5(a) 27[.0] or 26.97… nfww 3 8.62 + 9.7 2 − 4.42
M2 for [cos = ]
2 × 8.6 × 9.7
or M1 for implicit form
5(b) 9.19 or 9.192 to 9.193 4 B1 for [angle BCD =] 73 seen
9.7 × sin 65
M2 for oe
sin (180 − 65 − 42)
sin(180 − 65 − 42) sin 65
or M1 for = oe
9.7 DC
5(c) 6.15 or 6.149 to 6.151… 3 d
M2 for = sin42 oe
their 9.19
or M1 for right angle between line from C to BD
and BD soi
y
5
4
3
2
1
–2 –1 0 1 2 x
–1
–2
(a) The grid shows the graph of y = a + bx 2 .
The graph passes through the points with coordinates (0, 4) and (1, 1).
(i) Find the value of a and the value of b.
a = ……
b = …… [2]
(ii) Write down the equation of the tangent to the graph at (0, 4).
…… [1]
(iii) The equation of the tangent to the graph at x =- 1 is y = 6x + 7 .
Find the equation of the tangent to the graph at x = 1.
…… [2]
5
(b) The table shows some values for y = 1 + for - 2 G x G 1.5 .
3-x
x –2 – 1.5 –1 – 0.5 0 0.5 1 1.5
y 2 2.11 2.43 3 4.33
(i) Complete the table. [3]
5
(ii) On the grid, draw the graph of y = 1 + for - 2 G x G 1.5 . [4]
3-x
(c) (i) Write down the values of x where the two graphs intersect.
x = …… or x = …… [2]
(ii) The answers to part(c)(i) are two solutions of a cubic equation in terms of x.
Find this equation in the form ax 3 + bx 2 + cx + d = 0 , where a, b, c and d are integers.
…… [4]
Mark scheme
6(a)(i) [a = ] 4 2 B1 for [a = ] 4
B1 for [b = ] – 3 nfww
[b = ] – 3 nfww
6(a)(ii) y = 4 oe 1
6(a)(iii) y = − 6x + 7 oe final answer 2 B1 for answer − 6x + 7
or answers y = − 6x + c or y = kx + 7 (k < 0)
6(b)(i) 2.25 2.67 3.5 3 B1 for each
6(b)(ii) correct curve 4
B3 FT for 7 or 8 points
or B2 FT for 5 or 6 points
or B1 FT for 3 or 4 points
6(c)(i) –0.78 to –0.72 and 0.55 to 0.59 2 B1 for each
6(c)(ii) 3x3 − 9 x 2 − 3x + 4 [= 0] final answer 4 B3FT for 3 out of 4 correct terms
or for bx3 – 3bx2 + (a – 1)x + 8 – 3a [ = 0] oe
or B2FT for 2 out of 4 correct terms
or for 3 out of 4 terms from
bx3 – 3bx2 + (a – 1)x + 8 – 3a [ = 0]
5
or M1 for 1 + = their 4 + (their (−3)) x 2 oe
3− x
(a) The box-and-whisker plot shows information about the marks scored by some students in a test.
0 10 20 30 40 50 60 70 80 90 100
Mark
(i) Write down the median mark. …… [1]
(ii) Work out the range.
…… [1]
(iii) Jais scored a mark in the test that was higher than the marks scored by 75% of the students.
Write down a possible mark for Jais. …… [1]
(iv) This box-and-whisker plot shows information about the marks scored by the same students in a second test.
0 10 20 30 40 50 60 70 80 90 100
Mark
Make one comparison between the distributions of marks in the two tests.
…… [1]
(b) The table shows information about the height, h cm, of each of 50 plants.
Height (h cm) 0 1 h G 20 20 1 h G 30 30 1 h G 34 34 1 h G 40 40 1 h G 60
Frequency 4 9 20 15 2
Calculate an estimate of the mean.
…… cm [4]
(c) Some apples are weighed and the mass, m grams, of each apple is recorded.
The table shows the results.
Mass (m grams) 100 1 m G 110 110 1 m G 115 115 1 m G 125 125 1 m G 140
Frequency 50 x 44 51
The histogram shows some of the information from the table.
8
6
Frequency density 4
2
0 m
100 110 120 130 140
Mass (grams)
(i) Work out the value of x.
x = …… [1]
(ii) Complete the histogram.
[2]
Mark scheme
7(a)(i) 70 1
7(a)(ii) 78 1
7(a)(iii) Value in range 86 <V ≤ 90 1
7(a)(iv) One general comment interpreting 1
the median comparison nfww
e.g. Students did better on second test
oe
OR
One general comment interpreting
IQR/range comparison nfww
e.g. Students marks were more
consistent on the 2nd test oe
7(b) 31.2 4 M1 for mid-values soi
M1 for Σfm where m is any value in interval
including boundaries
M1 (dep on second M1) for their Σfm ÷ 50
7(c)(i) 38 1
7(c)(ii) Blocks of heights 4.4 and 3.4 with 2 B1 for each correct block
correct widths
If B0 scored, SC1 for both correct frequency
densities soi
(a)
O
53°
NOT TO
9.5 cm
A B SCALE
X Y
The diagram shows a sector OXY of a circle with centre O and radius 9.5 cm.
The sector angle is 53°.
A lies on OX, B lies on OY and OA = OB .
(i) Show that the area of the sector is 41.7 cm 2 , correct to 1 decimal place.
[2]
1
(ii) The area of triangle OAB is of the area of sector OXY.
3
Calculate OA.
OA = …… cm [4]
(b)
O
60°
NOT TO
24 cm SCALE
P Q
The diagram shows a sector OPQ of a circle with centre O and radius 24 cm.
The sector angle is 60°.
A cone is made from this sector by joining OP to OQ.
O
NOT TO
SCALE
P
Q
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 [6]
Mark scheme
8(a)(i) 53 M1
× π × 9.52
360
41.74 to 41.75 A1
8(a)(ii) 5.9[0] or 5.899 to 5.903.. 4 1
× 41.7
M3 for OA2 = 31 oe
2
sin 53
1 1
M2 for × OA2 × sin 53 = × 41.7 oe
2 3
1 1
M1 for × OA × OB × sin 53 = × 41.7 seen or
2 3
better
8(b) 396 or 397 or 396.4 to 396.6 6 60
M2 for [ r = ] × 2× π × 24 ÷ 2π oe or better
360
60
or M1 for 2π r = × 2 × π × 24 oe
360
M2 for 242 − a 2
or M1 for h2 + a2 = 242
1
M1 for π × their r 2 × their h
3
(a) Factorise.
(i) 5am + 10ap - bm - 2bp
…… [2]
(ii) 15 (k + g) 2 - 20 (k + g)
…… [2]
(iii) 4x 2 - y 4
…… [2]
(b) Expand and simplify.
(x - 3) (x + 1) (3x - 4)
…… [3]
(c) (x + a) 2 = x 2 + 22x + b
Find the value of a and the value of b.
a = ……
b = …… [2]
Mark scheme
9(a)(i) ( 5a − b )( m + 2 p ) final answer 2 M1 for 5a ( m + 2 p ) − b ( m + 2 p )
or m ( 5a − b ) + 2 p ( 5a − b )
or B1 for correct answer seen
9(a)(ii) 5 ( k + g )( 3k + 3g − 4 ) final answer 2 M1 for correct partial factorisation by 5 or (k + g)
isw
(
eg 5 3k 2 + 6kg + 3g 2 − 4k − 4 g )
or 5(3(k + g)2 – 4(k+ g)) or ( k + g ) (15 ( k + g ) − 20 )
or (5k + 5g) (3k + 3g – 4)
or B1 for correct answer seen
9(a)(iii)
( 2 x − y )( 2 x + y ) final answer
2 2 2 M1 for answer in form (a + b) (a – b)
or B1 for correct answer seen
9(b) 3 x 3 − 10 x 2 − x + 12 final answer 3 B2 for correct unsimplified expansion
or simplified expression with 3 terms correct in a
4-term expression of required form
or B1 for correct expansion of two of the brackets
with at least 3 terms correct
9(c) [a =] 11 2 B1 for each
[b =] 121
(a) A box is a cuboid with length 45 cm, width 30 cm and height 42 cm.
The box is completely filled with 90.72 kg of sand.
Calculate the density of this sand in kg/m 3 .
[Density = mass ÷ volume]
…… kg/m 3 [3]
(b) A bag contains 15000 cm 3 of sand.
Some of this sand is used to completely fill a hole in the shape of a cylinder.
The hole is 30 cm deep and has radius 10 cm.
Calculate the percentage of the sand from the bag that is used.
…… % [3]
(c) Sand costs $98.90 per tonne.
This cost includes a tax of 15%.
Calculate the amount of tax paid per tonne of sand.
$ …… [3]
(d) Raj buys some sand for 3540 rupees.
Calculate the cost in dollars when the exchange rate is $1 = 70.8 rupees.
$ …… [2]
Mark scheme
10(a) 1600 3 B2 for answer figs 16
or
M2 for 90.72 ÷ (figs45 × figs3 × figs42)
or M1 for volume = figs 45 × figs 3 × figs 42 isw
10(b) 62.8 or 62.83 to 62.84 3 π × 102 × 30
M2 for × 100
15000
or M1 for π × 10 2 × 30
10(c) 12.9[0] 3 B2 for 86
OR
98.9 98.9
M2 for × 0.15 oe or 98.9 − oe
15 15
1+ 1+
100 100
15
or M1 for 1 + a = 98.9 oe isw
100
10(d) 50 2 M1 for 3540 ÷ 70.8
Gaya spends $48 to buy books that cost $x each.
(a) Write down an expression, in terms of x, for the number of books Gaya buys.
…… [1]
(b) Myra spends $60 to buy books that cost $ (x + 2) each.
Gaya buys 4 more books than Myra.
Show that x 2 + 5x - 24 = 0 .
[4]
(c) Solve by factorisation. x 2 + 5x - 24 = 0
x = …… or x = …… [3]
(d) Find the number of books Myra buys.
…… [1]
Mark scheme
11(a) 48 1 Accept 48 ÷ x
final answer
x
11(b) 60 M1 FT their (a) provided expression in x
their (a) − = 4 oe
x+2
48 ( x + 2 ) − 60 x = 4 x ( x + 2 ) oe M2 FT their 3 term eqn with algebraic denominators, x
and
x + 2, for M2 or M1
M1 for common denominator x ( x + 2 ) oe seen
or any two terms in a 3 term equation from
± 48 (x + 2) , ± 60x , ± 4x(x + 2) oe seen
48x + 96 – 60x = 4x2 + 8x oe A1 With brackets expanded and no errors or omissions
leading to x 2 + 5 x − 24 = 0 seen
11(c) ( x − 3)( x + 8 ) B2 B1 for x(x + 8) – 3(x + 8)
or x(x – 3) + 8(x – 3)
or (x + a)(x + b) [= 0]
where ab = – 24 or a + b = 5 [a, b integers]
3 and − 8 B1
11(d) 12 1
(a) Find the gradient of the curve y = 2x 3 - 7x + 4 when x =- 2 .
…… [3]
(b) A is the point (7, 2) and B is the point (−5, 8).
(i) Calculate the length of AB.
…… [3]
(ii) Find the equation of the line that is perpendicular to AB and that passes through the point (−1, 3).
Give your answer in the form y = mx + c .
y = …… [4]
(iii) AB is one side of the parallelogram ABCD and
-a
• BC = e o where a 2 0 and b 2 0
-b
• the gradient of BC is 1
• BC = 8 .
Find the coordinates of D.
( …… , ……) [4]
Mark scheme
12(a) 17 3 M2 for 3 × 2 x 2 − 7 or better isw
or M1 for 3 × 2 x 2 oe or kx2 – 7 seen
12(b)(i) 13.4 or 13.41 to 13.42 3
M2 for ( −5 − 7 )2 + (8 − 2 )2 oe
or M1 for ( −5 − 7 ) + ( 8 − 2 ) oe
2 2
12(b)(ii) [ y =] 2 x + 5 final answer 4 8−2
M1 for [gradient of AB =] oe
−5 − 7
1
M1dep for gradient p = −1 ÷ their − oe
2
M1dep on previous M1 for substituting (−1, 3)
into
y = their px + c oe where their p ≠ 0
12(b)(iii) ( 5, 0) 4 ρ −2 ρ 2
B3 for AD = or DA =
−2 2
ρ 12
or coordinates of C (−7, 6) and CD = oe
−6
seen
or B2 for a = b = 2 soi or coordinates of C (−7, 6)
( 8 ) oe
2
or M1 for a = b oe soi or for a 2 + b 2 =
a
or cos 45 = oe
8
ρ −12 ρ 12
or for DC = or CD = seen
6 −6
y −8 y−2
or = 1 oe or =1
x − −5 x−7
Examiners follow the generic marking principles at the front of the scheme, then each question's points: method (M), accuracy (A) and independent (B) marks, error carried forward where the scheme allows it, and positive marking, so you're credited for what you got right.
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On this page, free: the mark scheme for IGCSE Mathematics 0580/42, February/March 2021 (0580_m21_ms_42), and the question paper (0580_m21_qp_42), to read or download, with each question's marking points written out underneath.
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Mathematics 0580/42 for February/March 2021 is out of 130 marks, with 2 hours 30 minutes to answer it.
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