Mathematics 0580/41 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) The total cost of a taxi journey is calculated as
• $0.50 per kilometre plus
• $0.40 per minute.
(i) Calculate the total cost of a journey of 32 km that takes 30 minutes.
$ …… [2]
(ii) The total cost of a journey of 100 km is $98.
Show that the time taken is 2 hours.
[3]
(b) Three taxi drivers travel a total of 8190 km in the ratio 5 : 2 : 7.
Calculate the distance each driver travels.
Driver 1 …… km
Driver 2 …… km
Driver 3 …… km [3]
(c) After midnight, the cost of any taxi journey increases by 45%.
One journey costs $84.10 after midnight.
Calculate the cost of the same journey before midnight.
$ …… [2]
Mark scheme
1(a)(i) 28 2 M1 for 32 × 0.50 + 30 × 0.40
1(a)(ii) 98 – 100 × 0.5 M3 M1 for 100 × 0.50 + x × 0.40 = 98
48 ÷ 0.4 = 120 [minutes] = 2 [hrs] M1 for 50 + 0.4 x = 98 or 0.4 x = 48
48
M1 for x = x = 120 [min] = 2 [hr]
0.4
OR
M1 for 100 × 0.5 [ = 50]
M1 for 98 − 50 [ = 48]
M1 for 48 ÷ 0.4 = 120 [ min ] = 2 [ hr ]
1(b) 2925 3 B2 for one correct answer
1170 or M1 for 8190 ÷ (5 + 2 + 7)
4095
1(c) 58 2 45
M1 for 1 + k = 84.1 oe
100
The diagram shows the speed−time graph for the first 180 seconds of a train journey.
9
Speed
(m/s)
0
0 50 100 150 200 250 300
Time (s)
(a) Find the acceleration, in m/s2, of the train during the first 50 seconds.
…… m/s2 [1]
(b) After 180 seconds, the train decelerates at a constant rate of 1944 km/h2.
Show that the train decelerates for 60 seconds until it stops.
[2]
(c) Complete the speed−time graph. [1]
(d) Calculate the average speed of the train for the whole journey.
…… m/s [4]
Mark scheme
2(a) 9 1
0.18 or
50
2(b) 1000 M1
1944 ×
3600 × 3600
9 ÷ 0.15 = 60 M1
2(c) 1 ruled line to axis with point of contact at 240
240
2(d) 6.9375 4 1
M2 for area = × (130 + 240 ) × 9 oe
2
or M1 for one correct partial area
M1dep for their total area ÷ 240
(a)
NOT TO
SCALE
6.3 cm
R cm
2.4 cm
The diagram shows a solid cone and a solid hemisphere.
The cone has radius 2.4 cm and slant height 6.3 cm.
The hemisphere has radius R cm.
The total surface area of the cone is equal to the total surface area of the hemisphere.
Calculate the value of R.
[The curved surface area, A, of a cone with radius r and slant height l is A = rrl .]
[The curved surface area, A, of a sphere with radius r is A = 4rr 2 .]
R = …… [4]
(b)
NOT TO
SCALE
16 cm
12 cm
7.6 cm 7.6 cm
The diagram shows a solid cone with radius 7.6 cm and height 16 cm.
A cut is made parallel to the base of the cone and the top section is removed.
The remaining solid has height 12 cm, as shown in the diagram.
Calculate the volume of the remaining solid.
[The volume, V, of a cone with radius r and height h is V = 13 rr 2 h .]
…… cm3 [4]
Mark scheme
3(a) 2.64 or 2.638… 4 π × 2.42 + π × 2.4 × 6.3
M3 for [ R 2 =] oe
π + 2π
or M2 for
1
π × 2.42 + π × 2.4 × 6.3 = π R 2 + × 4π R 2
2
2
or M1 for [ π × 2.4 ] + π × 2.4 × 6.3 oe
1
or [ π R 2 ] + × 4π R 2 oe
2
3(b) 953 or 952.6 to 952.8 4 1 16 − 12 3
M3 for × π × 7.62 × 16 × 1 −
3 16
1 1
or × π × 7.62 × 16 − × π × 1.92 × (16 − 12 )
3 3
OR
3
16 − 12
B1 for top radius = 1.9 or oe
16
M2 for
1 1
× π × 7.62 × 16 − × π × (their 1.9) 2 × (16 − 12 )
3 3
1 2
16 − 12
3
or × π × 7.6 × 16 × 1 − their
3 16
1
or M1 for × π × 7.62 × 16
3
1
or for × π × (their 1.9)2 × (16 − 12 )
3
(a) The exchange rate is 1 euro = $1.142 .
(i) Johann changes $500 into euros.
Calculate the number of euros Johann receives.
Give your answer correct to the nearest euro.
…… euros [2]
(ii) Johann buys a computer for $329.
The same computer costs 275 euros.
Calculate the difference in cost in dollars.
$ …… [2]
3
(b) Lucy spends of the money she has saved this month on a book that costs $5.25 .
8
Calculate how much money Lucy has saved this month.
$ …… [2]
(c) Kamal invests $6130 at a rate of r% per year compound interest.
The value of his investment at the end of 5 years is $6669.
Calculate the value of r.
r = …… [3]
Mark scheme
4(a)(i) 438 cao 2 500
M1 for
1.142
4(a)(ii) 14.95 2 M1 for [329 –] 275 × 1.142 oe
4(b) 14 2 8
M1 for 5.25 × oe
3
4(c) 1.7[0] or 1.699… 3 6669
M2 for 5
6130
or M1 for 6669 = 6130 ( k )
5
A
NOT TO
SCALE
10.6 cm
58° 78°
B C
X 6.4 cm
The diagram shows triangle ABC.
X is a point on BC.
AX = 10.6 cm, XC = 6.4 cm, angle ABC = 58° and angle AXB = 78°.
(a) Calculate AC.
AC = …… cm [4]
(b) Calculate BX.
BX = …… cm [4]
(c) Calculate the area of triangle ABC.
…… cm2 [3]
Mark scheme
5(a) 13.5 or 13.47… 4 B1 for angle 102 seen
M2 for
10.62 + 6.42 − 2 × 10.6 × 6.4 × cos (180 − 78 )
OR
M1 for
10.6 2 + 6.4 2 − 2 × 10.6 × 6.4 × cos (180 − 78 )
A1 for 181.5…
5(b) 8.68 or 8.682 to 8.683 nfww 4 B1 for angle = 44
10.6
M2 for sin(180 – 58 – 78) × oe
sin 58
sin(180 − 58 − 78) sin 58
or M1 for = oe
x 10.6
5(c) 78.2 or 78.17 to 78.19… 3 1
M2 for × 10.6 × ( 6.4 + their 8.68) × sin ( 78)
2
oe
OR
1
M1 for × 10.6 × 6.4 × sin(180 – 78) oe
2
1
M1 for × 10.6 × their 8.68 × sin 78 oe
2
(a) In the Venn diagram, shade the region P l , Q .
P Q
[1]
(b) There are 50 students in a group.
34 have a mobile phone (M ).
39 have a computer (C ).
5 have no mobile phone and no computer.
Complete the Venn diagram to show this information.
M C
[2]
(c) The Venn diagram shows the number of students in a group of 30 who have brothers (B ), sisters (S ) or cousins (C ).
B S
2 2 5
1
3 7
8 2
C
(i) Write down the number of students who have brothers.
…… [1]
(ii) Write down the number of students who have cousins but do not have sisters.
…… [1]
(iii) Find n (B , S , C )l .
…… [1]
(iv) Use set notation to describe the set of students who have both cousins and sisters but do not have brothers.
…… [1]
(v) One student is picked at random from the 30 students.
Find the probability that this student has cousins.
…… [1]
(vi) Two students are picked at random from the students who have cousins.
Calculate the probability that both these students have brothers.
…… [3]
(vii) One student is picked at random from the 30 students.
Event A This student has sisters.
Event B This student has cousins but does not have brothers.
Explain why event A and event B are equally likely.
…… [1]
Mark scheme
6(a) 1
6(b) 2 B1 for 2 or 3 correct elements
or M1 for 34 – x , x and 39 – x correctly placed
28 11
6 on diagram and x = 28
5
6(c)(i) 8 1
6(c)(ii) 11 1
6(c)(iii) 2 1
6(c)(iv) C ∩ S ∩ B′ oe 1
6(c)(v) 19 1
oe
30
6(c)(vi) 2 3 4 3
oe M2 for ×
57 19 18
4
or M1 for seen
19
6(c)(vii) Equal numbers 15 1
15
or equal probability oe
30
(a) Simplify. x 2 - 25 x 2 - x - 20
…… [3]
(b) Write as a single fraction in its simplest form.
x+5 x+8
+ x x-1
…… [3]
(c) A curve has equation y = 2x 3 - 4x 2 + 6 . dy
(i) Find , the derived function of y. dx
…… [2]
(ii) Calculate the gradient of the curve y = 2x 3 - 4x 2 + 6 at x = 4.
…… [2]
(iii) Find the coordinates of the two stationary points on the curve.
(…… , ……) and (…… , ……) [4]
Mark scheme
7(a) x+5 3 B1 for ( x − 5 )( x + 5 )
final answer
x+4 B1 for ( x − 5 )( x + 4 )
7(b) 2 x 2 + 12 x − 5 2 x 2 + 12 x − 5 3 B1 for common denominator x ( x − 1) oe
or
x ( x − 1) x2 − x B1 for ( x − 1)( x + 5 ) + x ( x + 8 ) or better
final answer
7(c)(i) 6 x 2 − 8 x final answer 2 B1 for each term in final answer
or M1 for correct answer seen and spoilt
7(c)(ii) 64 2 FT their (c)(i) correctly evaluated provided at
least 2 terms but not the original equation
M1 for substituting x = 4 into their (c)(i)
7(c)(iii) (0, 6) 4 dy
M1 for their derivative = 0 or = 0 soi
4 98 dx
, oe
3 27 4
B1 for x = 0 and x =
3
M1dep for substituting one of their x values
into y = 2 x3 − 4 x 2 + 6 soi
(a) The table shows information about the mass, in kilograms, of each of 50 children.
Mass (k kg) 0 1 k G 10 10 1 k G 25 25 1 k G 35 35 1 k G 40 40 1 k G 50
Frequency 3 19 21 5 2
(i) Complete the cumulative frequency table.
Mass (k kg) k G 10 k G 25 k G 35 k G 40 k G 50
Cumulative frequency
[2]
(ii) On the grid, draw a cumulative frequency diagram to show this information.
50
40
30
Cumulative frequency
20
10
0
0 10 20 30 40 50 k
Mass (kg)
[3]
(iii) Use your diagram to find an estimate of the number of children with a mass of 32 kg or less.
…… [1]
(b) This cumulative frequency diagram shows information about the height, in metres, of each of
100 students.
100
80
60
Cumulative frequency
40
20
0
1.4 1.5 1.6 1.7 1.8 1.9
Height (m)
The height of the tallest student is 1.83 metres.
The height of the shortest student is 1.45 metres.
1.4 1.5 1.6 1.7 1.8 1.9
Height (m)
On this grid, draw a box-and-whisker plot for the heights of the 100 students. [4]
D
Mark scheme
8(a)(i) 3 22 43 48 50 2 B1 for 4 correct
or M1 for one error in adding.
8(a)(ii) correct diagram 3 B1FT their (a)(i) for 5 correct heights
B1 for 5 points at upper ends of intervals on
correct vertical line
B1FT dep on at least B1 for increasing curve
through their 5 points
After 0 scored, SC1 for 4 of their points
correctly plotted
8(a)(iii) 35 to 38 1
8(b) Correct box-and-whisker diagram 4 B1 for median 1.64 drawn
B1 for LQ 1.57 drawn
B1 for UQ 1.71 drawn
If 0 scored SC1 for 1.64, 1,57 or 1.71 seen
1.45 1.57 1.64 1.71 1.83
NOT TO
A SCALE
13 cm 20 cm E F
24 cm
B
C
The diagram shows a prism, ABCDEF.
AB = 13 cm, AC = 20 cm, CF = 24 cm and angle ABC = 90°.
(a) Calculate the total surface area of the prism.
…… . cm2 [6]
(b) Calculate the volume of the prism.
…… cm3 [1]
(c) Calculate the angle that AF makes with the base BCFE.
…… [4]
Mark scheme
9(a) 1350 or 1354…. 6 M2 for 202 − 132
or M1 for BC2 + 132 = 202
A1 for 231 or 15.2 or 15.19 to 15.20
M1 for 20 × 24 and 13 × 24 and their
15.2 × 24
M1 for [½ ×] their 15.2 × 13
9(b) 2370 or 2369 to 2371… cao 1
9(c) 24.6 or 24.58 to 24.59 4 13
M3 for sin [...] = oe
20 + 242
2
or M2 for 202 + 242 or 242 + 202 − 132
or M1 for AF2 = 202 + 242 or 242 + 202 - 132
or M1 for correct angle identified
The table shows some values of y = 3 + 4x - x 2 for - 1 G x G 5.
x -1 -0.5 0 1 2 3 4 4.5 5 y -2 6 6 -2
(a) Complete the table. [3]
(b) On the grid, draw the graph of y = 3 + 4x - x 2 for - 1 G x G 5.
y
8
6
4
2
–1 0 1 2 3 4 5 x
–2
[4]
(c) Write down an integer value of k for which the equation 3 + 4x - x 2 = k has no solutions.
…… [1]
9
(d) By drawing a suitable straight line on the grid, solve the equation - 1 + x - x 2 = 0 .
2
x = …… or x = …… [4]
Mark scheme
10(a) 0.75 3 7 3 0.75 3 B2 for 4 or 3 correct
or B1 for 2 correct
10(b) correct curve 4 B3FT for 8 or 9 correct plots
B2FT for 6 or 7 correct plots
B1FT for 4or 5 correct plots
10(c) Accept any integer ≥ 8 1
10(d) 1 B3 1
line y = 4 − x ruled B2 for [ y =] 4 − x identified
2 2
1
or B1 for ruled line with gradient −
2
or B1 for ruled line through (0, 4) but not y = 4
0.2 to 0.3 4.2 to 4.3 B1
(a) Find the size of an exterior angle of a regular polygon with 18 sides.
…… [2]
(b)
A
5.2 cm
NOT TO
SCALE
B E
2.6 cm
C D
6.75 cm
In triangle ACD, B lies on AC and E lies on AD such that BE is parallel to CD.
AE = 5.2 cm and ED = 2.6 cm.
Calculate BE.
BE = …… cm [2]
(c) Two solids are mathematically similar.
The smaller solid has height 2 cm and volume 32 cm3.
The larger solid has volume 780 cm3.
Calculate the height of the larger solid.
…… cm [3]
(d)
P Q
NOT TO
SCALE
N
R S
PQ is parallel to RS, PNS is a straight line and N is the midpoint of RQ.
Explain, giving reasons, why triangle PQN is congruent to triangle SRN.
…… [4]
Mark scheme
11(a) 20 2 360 16 × 180
M1 for or 180 −
18 18
11(b) 4.5 2 BE 5.2
M1 for = oe
6.75 5.2 + 2.6
11(c) 5.8[0] or 5.798 to 5.799 3 780
M2 for 2 × 3 oe
32
780 32 23 32
or M1 for 3 or 3 or 3 =
32 780 l 780
11(d) QN = NR [given] B1
Two correct pairs of angles with B2 B1 for any correct pair of angles with reason
reasons from or two correct pairs of angles with no/wrong
reasons
angle PQN = angle SRN
alternate
angle QPN = angle RSN
alternate
angle PNQ = angle SNR
[vertically] opposite
ASA [implies congruent] B1 dep on B1 B2
f (x) = 3 - 2x g (x) = x 2 + 5 h (x) = x 3
(a) Find f (- 5) .
…… [1]
(b) Find ff(x).
Give your answer in its simplest form.
…… [2]
(c) Solve g (x) = f (x) + 37 .
x = …… or x = …… [4]
(d) Find f -1 (x) .
f -1 (x) = …… [2]
(e) Find hf (x) + g (x) .
Give your answer in its simplest form.
…… [5]
Mark scheme
12(a) 13 1
12(b) 4x – 3 final answer 2 M1 for 3 − 2 ( 3 − 2 x )
12(c) −7 5 4 M1 for x 2 + 2 x − 35 [= 0] or x 2 + 2 x = 35
M2 for ( x + 7 )( x − 5 )
or x ( x − 5 ) + 7 ( x − 5 ) or x ( x + 7 ) − 5 ( x + 7 )
or M1 for ( x + a )( x + b ) where a, b are
integers with ab = − 35 or a + b = 2
12(d) 3− x 2 M1 for a correct first step:
oe final answer x = 3 − 2 y or y − 3 = −2 x , 2x = 3 – y or
2
y 3
= −x
2 2
12(e) 32 − 54 x + 37 x 2 − 8 x 3 5 B4 for
final answer 27 − 36 x − 18 x + 24 x 2 + 12 x 2 − 8 x 3 + x2 + 5 oe
OR
B1 for ( 3 − 2x ) + x2 + 5
3
and
B2 for expansion of the 3 brackets, allow one
error
or B1 for correct expansion of 2 of the brackets
with at least 3 terms correct
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Mathematics 0580/41 for May/June 2021 is out of 130 marks, with 2 hours 30 minutes to answer it.
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