Mathematics 0580/43 October/November 2019: question paper and mark scheme
Cambridge IGCSE Mathematics (0580), Paper 4 (Extended), October/November 2019. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
(a) In a cycling club, the number of members are in the ratio males : females = 8 : 3.
The club has 342 females.
(i) Find the total number of members.
…… [2]
(ii) Find the percentage of the total number of members that are female.
…… % [1]
(b) The price of a bicycle is $1020.
Club members receive a 15% discount on this price.
Find how much a club member pays for this bicycle.
$ …… [2]
(c) In 2019, the membership fee of the cycling club is $79.50 .
This is 6% more than last year.
Find the increase in the cost of the membership.
$ …… [3]
(d) Asif cycles a distance of 105 km.
On the first part of his journey he cycles 60 km in 2 hours 24 minutes.
On the second part of his journey he cycles 45 km at 20 km/h.
Find his average speed for the whole journey.
…… km/h [4]
(e) Bryan invested $480 in an account 4 years ago.
The account pays compound interest at a rate of 2.1% per year.
Today, he uses some of the money in this account to buy a bicycle costing $430.
Calculate how much money remains in his account.
$ …… [3]
1
(f) The formula s = at 2 is used to calculate the distance, s, travelled by a bicycle.
2
When a = 3 and t = 10 , each correct to the nearest integer, calculate the lower bound of the distance, s.
…… [2]
Mark scheme
1(a)(i) 1254 2 M1 for 342 ÷ 3
1(a)(ii) 27.3 or 27.27… 1
1(b) 867 2 15
M1 for 1020 × oe
100
15
or 1020 × 1 − oe
100
1(c) 4.5[0] 3 79.5 [ 0]
M2 for [×6] oe
100 + 6
79.5 [ 0]
or × 100 oe
100 + 6
or M1 for 79.5[0] associated with 106[%]
1(d) 22.6 or 22.58… nfww 4 45
M1 for or better
20
and
60 + 45
M2 for
45
their 2h 24 min + their
20
45
or M1 for their + their 2h 24min
20
1(e) 91.6[0] to 91.61 3 2.1
4
M2 for 480 × 1 + − 430 oe
100
4
2.1
OR M1 for 480 × 1 + oe
100
A1 for 522, 521.6[0] to 521.61
1(f) 112.8125 2 B1 for 2.5 or 9.5 seen
(a) The diagram shows a triangle and a quadrilateral.
All angles are in degrees.
NOT TO
3a + 2b SCALE
3b + 10 8a
a + 2b 2a b + 50 4b - 2a
(i) For the triangle, show that 3a + 5b = 170 .
[1]
(ii) For the quadrilateral, show that 9a + 7b = 310 .
[1]
(iii) Solve these simultaneous equations.
Show all your working.
a = ……
b = …… [3]
(iv) Find the size of the smallest angle in the triangle.
…… [1]
(b) Solve the equation 6x - 3 =-12 .
x = …… [2]
(c) Rearrange 2 (4x - y) = 5x - 3 to make y the subject.
y = …… [3]
(d) Simplify. 2
(27x 9) 3
…… [2]
(e) Simplify. x 2 + 5x x 2 - 25
…… [3]
Mark scheme
2(a)(i) 2a + a + 2b + 3b + 10 = 180 1
leading to 3a + 5b = 170 without error or
omission
2(a)(ii) 8a + 3a + 2b + b + 50 + 4b – 2a = 360 1
leading to 9a + 7b = 310 without error or
omission
2(a)(iii) Correct method to eliminate one variable M1
[a =]15 A2 A1 for each correct value
[ b=]25 If 0 scored, SC1 for two values that
satisfy one of the equations or for two
correct answers with no/incorrect working
2(a)(iv) 30 1
2(b) 1 3 2 M1 for 6x = –12 + 3 or better
–1.5 or −1 or −
2 2
2(c) 3x + 3 3 M1 for 8x – 2y = 5x – 3
oe final answer
2 1
or 4 x − y = ( 5 x − 3)
2
M1FT for isolating the y term correctly
2(d) 9x6 2 1
( )
M1 for (3x3)2 or 729x18 3 seen
k 6
or for 9x or kx as final answer
2(e) x 3 M1 for x(x + 5)
final answer nfww M1 for (x – 5)(x + 5)
x−5
The table shows some values for y = x 3 + x 2 - 5x .
x -3 -2 -1.5 -1 0 1 1.5 2 2.5 3
y -3 6 6.4 0 -1.9 2 9.4
(a) Complete the table. [3]
(b) On the grid, draw the graph of y = x 3 + x 2 - 5x for - 3 G x G 3.
y
25
20
15
10
5
-3 -2 -1 0 1 2 3 x
-5
[4]
(c) Use your graph to solve the equation x 3 + x 2 - 5x = 0 .
x = …… or x = …… or x = …… [2]
(d) By drawing a suitable tangent, find an estimate of the gradient of the curve at x = 2 .
…… [3]
(e) Write down the largest value of the integer, k, so that the equation x 3 + x 2 - 5x = k has three solutions for - 3 G x G 3.
k = …… [1]
Mark scheme
3(a) 5, –3, 21 3 B1 for each
3(b) Fully correct curve 4 B3 FT for 9 or 10 points
or B2 FT for 7 or 8 points
or B1 FT for 5 or 6 points
3(c) –2.9 to –2.7 2 B1 for 2 correct values
0
1.7 to 1.9
3(d) Tangent ruled at x = 2 B1
10 to 14 B2 Dep on correct tangent or close attempt at
tangent at x = 2
M1 for rise/run also dep on correct
tangent drawn or close attempt at tangent
Must see correct or implied calculation
from a drawn tangent
3(e) 6 1
B
107 m
C
NOT TO
SCALE
158 m 132 m 86 m
North
116°
D
A
The diagram shows a field, ABCD, on horizontal ground.
(a) There is a vertical post at C.
From B, the angle of elevation of the top of the post is 19°.
Find the height of the post.
…… m [2]
(b) Use the cosine rule to find angle BAC.
Angle BAC = …… [4]
(c) Use the sine rule to find angle CAD.
Angle CAD = …… [3]
(d) Calculate the area of the field.
…… m2 [3]
(e) The bearing of D from A is 070°.
Find the bearing of A from C.
…… [2]
Mark scheme
4(a) 36.8 or 36.84… 2 h h 107
M1 for = tan19 or = oe
107 sin19 sin 71
or better
4(b) 42.1 or 42.12… from cosine rule 4 1582 + 1322 − 107 2
M2 for [ cos BAC = ]
2 × 158 × 132
or M1 for implicit version
30939
A1 for [ cos BAC = ] or 0.7417…
41712
4(c) 35.8 or 35.84… from sine rule 3 86 × sin116
M2 for [ = 0.58557...]
132
sin CAD sin116
or M1 for = oe
86 132
4(d) 9670 or 9669 to 9676 3 1
M2 for × 158 × 132 × sin ( their ( b ) ) oe
2
1
and × 86 × 132 × sin ( 64 − their ( c ) ) oe
2
or M1 for either area
4(e) 214.2 or 214.1… or 214 2 M1 for [180 +]70–their (c) oe
The cumulative frequency diagram shows information about the distance, d km, travelled by each of 60 male cyclists in one weekend.
60
50
40
Cumulative frequency 30
20
10
0 d
0 20 40 60 80 100 120
Distance (km)
(a) Use the cumulative frequency diagram to find an estimate of
(i) the median,
…… km [1]
(ii) the lower quartile,
…… km [1]
(iii) the interquartile range.
…… km [1]
(b) For the same weekend, the interquartile range for the distances travelled by a group of female cyclists is 40 km.
Make one comment comparing the distribution of the distances travelled by the males with the distribution of the distances travelled by the females.
…… [1]
(c) A male cyclist is chosen at random.
Find the probability that he travelled more than 50 km.
…… [2]
(d) (i) Use the cumulative frequency diagram to complete this frequency table.
Distance (d km) Number of male cyclists
0 1 d G 40 18
40 1 d G 50 9
50 1 d G 60
60 1 d G 70
70 1 d G 90
90 1 d G 120 2
[2]
(ii) Calculate an estimate of the mean distance travelled.
…… km [4]
Mark scheme
5(a)(i) 52 1
5(a)(ii) 36 1
5(a)(iii) 26 1 FT 62 – their (a)(ii) evaluated correctly
5(b) Valid comment 1 Strict FT their (a)(iii), e.g. distances for
females are more varied
5(c) 11 2 27
oe M1 for 27 written or answer of oe
20 60
5(d)(i) [18 9] 14 12 5 [2] 2 B1 for 1 correct value
5(d)(ii) 48.75 nfww 4 M1 for midpoints soi
M1 for use of ∑fx with their frequencies
M1 (dep on 2nd M1)
for ∑fx ÷ (60 or by their ∑f )
(a)
A D
NOT TO
128° SCALE
28°
C
O 30°
B
In the diagram, A, B, C and D lie on the circle, centre O.
Angle ADC = 128°, angle ACD = 28° and angle BCO = 30°.
(i) Show that obtuse angle AOC = 104°.
Give a reason for each step of your working.
[3]
(ii) Find angle BAO.
Angle BAO = …… [2]
(iii) Find angle ABD.
Angle ABD = …… [1]
(iv) The radius, OC, of the circle is 9.6 cm.
Calculate the total perimeter of the sector OADC.
…… cm [3]
(b)
NOT TO
SCALE
The diagram shows two mathematically similar solid metal prisms.
The volume of the smaller prism is 648 cm3 and the volume of the larger prism is 2187 cm3.
The area of the cross-section of the smaller prism is 36 cm2.
(i) Calculate the area of the cross-section of the larger prism.
…… cm2 [3]
(ii) The larger prism is melted down into a sphere.
Calculate the radius of the sphere.
4
[The volume, V, of a sphere with radius r is V = rr 3 .]
3
…… cm [3]
Mark scheme
6(a)(i) Angle ABC=52 nfww B1 ALTERNATIVE
[Reflex] angle AOC = 256
Opposite angles in cyclic quad oe B1 Angle at centre=2 × angle at
Angles in opposite segments circumference/arc
[Angle AOC=104] B1 Angles around a point
Angle at centre=2 × angle at
circumference/arc nfww
6(a)(ii) 22 nfww 2 B1 for angle OAC = 38 or
angle CAD = 24
6(a)(iii) 28 1
6(a)(iv) 36.6 or 36.62 to 36.63 nfww 3 B2 for 7.4 or 17.42 to 17.43
104
or M2 for 9.6 × 2 + × 2 × π × 9.6
360
104
or M1 for × 2 × π × 9.6
360
6(b)(i) 81 3 A 3 2187
M2 for = oe or better
36 648
648 3 2187
or for A × × = 2187 oe
36 648
or better
A3 2187 2
or M1 for 3 = oe
36 6482
2187 648
or 3 or 3
648 2187
6(b)(ii) 8.05 or 8.051 to 8.052… 3 2187 × 3
M2 for r 3 = oe
4× π
4πr 3
or M1 for = 2187
3
648 × 3 4πr 3
SC2 for or SC1 for = 648
4× π 3
y
8
7
6
5
4
3
2
A
1
x
-8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6
-1
-2
-3
B
-4
-5
-6
(a) Describe fully the single transformation that maps shape A onto shape B.
…… [2]
(b) On the grid, draw the image of
-3
(i) shape A after a translation by the vector e o , [2]
4
(ii) shape A after a rotation through 180° about (0, 0), [2]
(iii) shape A after an enlargement, scale factor 2, centre (-7, 0). [2]
Mark scheme
7(a) Reflection 2 B1 for each
y = –1
7(b)(i) Image at 2 −3 k
(–6, 5) (–6, 7) (–5, 7) (–4, 5) B1 for translation by or
k 4
7(b)(ii) Image at 2 B1 for shape correct size and orientation
(1, –1) (3, –1) (3, –3) (2, –3) but wrong position
7(b)(iii) Image at 2 B1 for shape correct size and orientation,
(1, 2) (1, 6) (3, 6) (5, 2) wrong position
(a) A bag contains 4 red marbles and 2 yellow marbles.
Behnaz picks two marbles at random without replacement.
Find the probability that
(i) the marbles are both red,
…… [2]
(ii) the marbles are not both red.
…… [1]
(b) Another bag contains 5 blue marbles and 2 green marbles.
Bryn picks one marble at random without replacement.
If this marble is not green, he picks another marble at random without replacement.
He continues until he picks a green marble.
Find the probability that he picks a green marble on his first, second or third attempt.
…… [4]
Mark scheme
8(a)(i) 2 2 4 3
oe M1 for ×
5 6 5
8(a)(ii) 3 1 12
oe FT 1 – their oe
5 30
8(b) 5 4 2 5 2 5 4 2
oe nfww M3 for + × + × × oe
7 7 7 6 7 6 5
5 4 3
or for 1 − × × oe
7 6 5
5 2 5 4 2
or M1 for each of × and × × oe
7 6 7 6 5
or completed tree diagram with
appropriate probabilities shown
f (x) = 2x - 3 g (x) = 9 - x 2 h (x) = 3 x
(a) Find
(i) f (4) ,
…… [1]
(ii) hg (3) ,
…… [2]
(iii) g (2x) in its simplest form,
…… [1]
(iv) fg (x) in its simplest form.
…… [2]
(b) Find f -1 (x) .
f -1 (x) = …… [2]
(c) Find x when 5f (x) = 3.
x = …… [2]
(d) Solve the equation gf (x) =-16 .
x = …… or x = …… [4]
(e) Find x when h -1 (x) =- 2 .
x = …… [1]
Mark scheme
9(a)(i) 5 1
9(a)(ii) 1 2 M1 for h(0) or 39− x2 or better
9(a)(iii) 9 – 4x2 final answer 1
9(a)(iv) 15 – 2x2 final answer 2 M1 for 2(9 – x2) –3 or better
9(b) x+3 2 M1 for x = 2y – 3 or y + 3 = 2x or better
final answer
2 y 3
or = x −
2 2
9(c) 4 9 2 3
1.8 or 1 or M1 for 10 x − 15 = 3 or 2 x − 3 =
5 5 5
9(d) –1 and 4 nfww 4 M1 for 9 – (2x – 3)2= –16
A1 for 4x2 – 12x – 16[= 0] oe
M1 (dep on first M1) for correct factors or
use of formula or completing the square
for their 3-term quadratic
OR
M1 for 9 – y2= –16
A1 for y2 = 25
M1 (dep on first M1) for 2x – 3= ±5
9(e) 1 1
9
Solve.
1 2
- =3 x x+1
Show all your working and give your answers correct to 2 decimal places.
x = …… or x = …… [7]
Mark scheme
10 x + 1 – 2x = 3x(x + 1) M2 M1 for a common denominator of
x(x + 1) seen
or attempt to multiply through by
denominators
x + 1 − 2x
or for =3
x ( x + 1)
3x2 + 4x – 1[= 0] oe nfww A1
−4 ± 42 − 4 × 3 × ( −1) B2 B1FT for 42 − 4 × 3 × ( −1) or better
[ x =]
2×3 2
2
or for x +
3
−4 + q −4 − q
B1FT for or
2×3 2×3
2
2 1 2
or for − ± +
3 3 3
–1.55 and 0.22 final answers B2 B1 for each
or
B1 for –1.548 to –1.549 and 0.215…
or for –1.55 and 0.22 seen in working
or for –0.22 and 1.55 as final answer
or for –1.5 or –1.54 and 0.2 or 0.21 as
final answer
A B
C
Q
NOT TO
P SCALE
O
OAB is a triangle and ABC and PQC are straight lines.
P is the midpoint of OA, Q is the midpoint of PC and OQ : QB = 3 : 1.
OA = 4a and OB = 8b .
(a) Find, in terms of a and/or b, in its simplest form
(i) AB,
AB = …… [1]
(ii) OQ,
OQ = …… [1]
(iii) PQ.
PQ = …… [1]
(b) By using vectors, find the ratio AB : BC.
…… : …… [3]
Mark scheme
11(a)(i) 8b – 4a oe 1
11(a)(ii) 6b 1
11(a)(iii) 6b – 2a or 2(3b – a) 1 FT –2a + their (a)(ii)
JJJG JJJG
11(b) 2 : 1 oe final answer 3 Dep on correct BC or correct AC seen
JJJG
B2 for BC = 4b–2a
JJJG
or M1 for a correct route for BC in terms
of a and b
JJJG
or for a correct route for AC in terms of a
and b
If no/incorrect working seen then SC1 for
final answer of 2 : 1 (oe)
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