Mathematics 0580/43 May/June 2019: question paper and mark scheme
Cambridge IGCSE Mathematics (0580), Paper 4 (Extended), May/June 2019. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
Here is part of a train timetable for a journey from London to Marseille.
All times given are in local time.
The local time in Marseille is 1 hour ahead of the local time in London.
London 07 19
Ashford 07 55
Lyon 13 00
Avignon 14 08
Marseille 14 46
(a) (i) Work out the total journey time from London to Marseille.
Give your answer in hours and minutes.
…… h …… min [2]
(ii) The distance from London to Ashford is 90 km.
The local time in London is the same as the local time in Ashford.
Work out the average speed, in km/h, of the train between London and Ashford.
…… km/h [3]
(iii) During the journey, the train takes 35 seconds to completely cross a bridge.
The average speed of the train during this crossing is 90 km/h.
The length of the train is 95 metres.
Calculate the length, in metres, of this bridge.
…… m [4]
(b) The fares for the train journey are shown in the table below.
From London to Marseille Standard fare Premier fare
Adult $84 $140
Child $60 $96
(i) For the standard fare, write the ratio adult fare : child fare in its simplest form.
…… : …… [1]
(ii) For an adult, find the percentage increase in the cost of the standard fare to the premier fare.
…… % [3]
(iii) For one journey from London to Marseille, the ratio
number of adults : number of children = 11 : 2.
There were 220 adults in total on this journey.
All of the children and 70% of the adults paid the standard fare.
The remaining adults paid the premier fare.
Calculate the total of the fares paid by the adults and the children.
$ …… [5]
(c) There were 3.08 # 105 passengers that made this journey in 2018.
This was a 12% decrease in the number of passengers that made this journey in 2017.
Find the number of passengers that made this journey in 2017.
Give your answer in standard form.
…… [3]
Mark scheme
1(a)(i) 6h 27 mins 2 B1 for answer .............h 27 mins
1(a)(ii) 150 km/h 3 90
M2 for × 60
36
90
or M1 for
their time
or B1 for 36 [mins] seen
1(a)(iii) 780 4 35
M3 for 90 × × 1000 – 95 oe
3600
or
35
M2 for 90 × × 1000 oe
3600
or B1 for figs 875
35
or M1 for 90 × seen
3600
1000
or for 90 × oe
3600
If 0 scored, SC1 for their distance (> 95) – 95
1(b)(i) 7:5 1
1(b)(ii) 66.7 or 66.66 to 66.67 3 140 − 84
M2 for [× 100] oe
84
140
or for × 100 oe
84
140
or M1 for oe
84
1(b)(iii) 24 576 5 M4 for complete method,
40 × 60 + 0.7 × 220 × 84 + 0.3 × 220 × 140 oe
OR
B1 for 40 [children]
M1 for 0.7 × 220 × 84 oe
M1 for 0.3 × 220 × 140 oe
B1 for 2400 or 12936 or 9240 nfww
1(c) 3.5 × 105 nfww 3 100 − 12
M2 for 3.08 × 105 ÷ oe
100
or M1 for 3.08 [× 105] associated with
(100–12)%
1 1 M2 M1 for
π × 1.752 × 6 − π × their 0.43752 × 1.5
1 1 3 OR
π × 1.752 × 6 × 1 − oe 1
3
(a) Solve.
5x - 17 = 7x + 3
x = …… [2]
(b) Find the integer values of n that satisfy this inequality.
- 7 1 4n G 8
…… [3]
(c) Simplify.
(i) a3 # a6
…… [1]
(ii) (5xy 2) 3
…… [2]
1
-
f p
27x 12 3
(iii)
64y 3
…… [3]
Mark scheme
2(a) –10 2 M1 for –17 – 3 = 7x – 5x oe or better
2(b) −1, 0, 1, 2 final answer 3 B2 for 3 correct values and no incorrect values
or 4 correct values and one incorrect value
7
or M2 for − < n - 2 oe
4
7
or M1 for − < n - k or k < n - 2 oe
4
2(c)(i) a9 1
2(c)(ii) 125x3y6 final answer 2 B1 for 2 correct elements if in form kxnym
2(c)(iii) 4 y[ ] 3
1 [ −1]
final answer
3x 4
B2 for [1] oe seen
3x 4 4y
OR
B1 for 3x4 or 4y[1]
and
1
64 y 3 3
M1 for 12
oe
27 x
64 y[1] 0.333 x −4
If 0 scored, SC1 for or seen
27 x 4 0.25 y −1
y
7
6
5
4
3
A
2
1
–6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 x
–1
–2
B
–3
–4
–5
(a) On the grid, draw the image of
-3
(i) triangle A after a translation by the vector e o , [2]
2
(ii) triangle A after a reflection in the line y = x. [2]
(b) Describe fully the single transformation that maps triangle A onto triangle B.
…… [3]
(c) (i) Find the matrix that represents an enlargement, scale factor - 2, centre (0, 0).
f p [2]
(ii) Calculate the determinant of the matrix in part (c)(i).
…… [1]
Mark scheme
3(a)(i) Image at (–5, 4), (–2, 4), (–4, 6) 2 −3 k
B1 for translation by or
k 2
3(a)(ii) Image at (2, 1), (4, –1), (2, –2) 2 B1 for reflection in y = –x
or y = x drawn
3(b) Rotation 3 B1 for each
90°[ anticlockwise] oe
(1, –1)
3(c)(i) −2 0 2 B1 for 2 by 2 matrix with one correct row or
column
0 −2
3(c)(ii) Strict FT their (c)(i) 1 Answer not equal to zero
FT their (c)(i) only if 2 by 2
3 3 1 1
π × 1.752 × 6 or π × their 0.43752 × 1.5
OR 3 3
3 4 M1 for 1– oe
4
18.94 or 18.939 to18.944… A1
4(b)(ii)(b) 95 final answer 3 B2 for 94.5 or 94.69 to 94.722
OR
M2 for 18.9 ×103 ÷ 200 oe
or M1 for 18.9 × 103 or 200 ÷ 103
or figs 189..÷ 200 or 18.9.. ÷ figs 2
(a)
5.6 cm
NOT TO
10 cm
SCALE
The diagram shows a hemispherical bowl of radius 5.6 cm and a cylindrical tin of height 10 cm.
(i) Show that the volume of the bowl is 368 cm3, correct to the nearest cm3.
4
[The volume, V, of a sphere with radius r is V = rr 3. ]
3
[2]
(ii) The tin is completely full of soup.
When all the soup is poured into the empty bowl, 80% of the volume of the bowl is filled.
Calculate the radius of the tin.
…… cm [4]
(b)
NOT TO
SCALE
6 cm
1.75 cm
The diagram shows a cone with radius 1.75 cm and height 6 cm.
(i) Calculate the total surface area of the cone.
[The curved surface area, A, of a cone with radius r and slant height l is A = rrl.]
…… cm2 [5]
(ii)
NOT TO
SCALE
4.5 cm
1.75 cm
The cone contains salt to a depth of 4.5 cm.
The top layer of the salt forms a circle that is parallel to the base of the cone.
(a) Show that the volume of the salt inside the cone is 18.9 cm3, correct to 1 decimal place.
1
[The volume, V, of a cone with radius r and height h is V = rr 2 h.]
3
[4]
(b) The salt is removed from the cone at a constant rate of 200 mm3 per second.
Calculate the time taken for the cone to be completely emptied.
Give your answer in seconds, correct to the nearest second.
…… s [3]
2
Mark scheme
4(a)(i) 1 4 M1
× × π × 5.63
2 3
367.8... to 367.9 A1
4(a)(ii) 3.06 or 3.060 to 3.061... 4 M1 for 0.8 × 368 [= 294.4]
their 294.4
M2 for [r2 =] oe
10π
or M1 for πr2 × 10 = their 294.4 oe
4(b)(i) 44[.0] or 43.98 to 43.99 nfww 5 25
B2 for [slant height = ] oe
4
or M1 for [l2 = ] 62 + 1.752 oe
M2 for π × 1.75 × their l + π × 1.752
or M1 for π × 1.75 × their l or π × 1.752
4(b)(ii)(a) 1 B1
SF = oe soi
4
The diagram shows the graph of y = f (x) where f (x) = x 2 - - 2, x ! 0 . x y
12
10
8
6
4
2
–3 –2 –1 0 1 2 3 x
–2
–4
–6
–8
– 10
– 12
(a) Use the graph to find
(i) f (1) ,
…… [1]
(ii) ff (- 2) .
…… [2]
(b) On the grid opposite, draw a suitable straight line to solve the equation
2 x 2 - - 7 =- 3x for - 3 G x G 3. x
x = …… or x = …… [4]
(c) By drawing a suitable tangent, find an estimate of the gradient of the curve at x = - 2.
…… [3]
(d) (i) Complete the table for y = g (x) where g (x) = 2 -x for - 3 G x G 3.
x -3 -2 -1 0 1 2 3
y 2 1 0.5 0.125
[3]
(ii) On the grid opposite, draw the graph of y = g (x) . [3]
(iii) Use your graph to find the positive solution to the equation f (x) = g (x) .
x = …… [1]
Mark scheme
5(a)(i) –3 1
5(a)(ii) 6.2 to 6.4 oe 2 M1 for 3 seen or used
5(b) y = 5 – 3x ruled 2 B1 for y = 5 – 3x soi or ruled line with gradient
– 3 or with y – intercept at 5 (but not y = 5)
or
B1FT for incorrect line equation/expression
shown in working and their line correctly drawn
– 0.3 to – 0.2 2 B1 for each, dep on y = 5 – 3x drawn
1.65 to 1.8 or FT their line provided equation/expression
shown in working, dep on B1FT for line
5(c) Tangent ruled at x = −2 1 B1 for correct tangent
–4.5 to –2.5 2 Dep on B1 for tangent or close attempt at
tangent at x = –2
M1 for rise/run also dep on tangent drawn or
close attempt at correct tangent
Must see correct or implied calculation from a
drawn tangent
5(d)(i) 8, 4, 0.25 oe 3 B1 for each
5(d)(ii) Correct graph 3 B2FT for 6 or 7 correct plots
or B1FT for 4 or 5 correct plots
5(d)(iii) 1.8 to 1.9 1
The table shows the time, t seconds, taken by each of 120 boys to solve a puzzle.
Time
20 1 t G 30 30 1 t G 35 35 1 t G 40 40 1 t G 60 60 1 t G 100
(t seconds)
Frequency 38 27 21 16 18
(a) Calculate an estimate of the mean time.
…… s [4]
(b) On the grid, complete the histogram to show the information in the frequency table.
6
5
4
Frequency density
3
2
1
0
20 30 40 50 60 70 80 90 100 t
Time (seconds)
[4]
Mark scheme
6(a) 40.5 or 40.45[8..] or 40.46 nfww 4 M1 for 25, 32.5, 37.5, 50, 80 soi
M1 for Σft
M1 dep for their Σft ÷ 120
6(b) Fully correct histogram 4 B1 for each correct bar
If 0 scored, SC1 for frequency densities of 5.4,
4.2, 0.8 and 0.45 seen
6 9 8 7 9 8 7
5 4 3 4 3 2
or M2 for × × or × ×
9 8 7 9 8 7
5 4 3 4 3 2
or M1 for , , seen or , , seen
9 8 7 9 8 7
3 3
5 +4
If 0 scored, SC1 for oe
729
A straight line joins the points A (-2, -3) and C (1, 9).
(a) Find the equation of the line AC in the form y = mx + c.
y = …… [3]
(b) Calculate the acute angle between AC and the x-axis.
…… [2]
(c) ABCD is a kite, where AC is the longer diagonal of the kite.
B is the point (3.5, 2).
(i) Find the equation of the line BD in the form y = mx + c.
y = …… [3]
(ii) The diagonals AC and BD intersect at (-0.5, 3).
Work out the co-ordinates of D.
(…… , ……) [2]
Mark scheme
7(a) [y = ] 4x + 5 3 B2 for answer [y =] 4x + c oe (c can be numeric
or algebraic)
OR
y − 9 9 − (−3)
M2 for = oe
x − 1 1 − (−2)
OR
9 − −3
M1 for oe or for
1 − −2
M1 for correct substitution of (–2, –3) or (1, 9)
into y = (their m)x + c oe
7(b) 76[.0] or 75.96... 2 M1 for tan[ ] = 4 oe
7(c)(i) 1 23 3 1
[y =] − x + oe B2FT for [y =] − x+c
4 8 their gradient from (a)
oe
(c can be numeric or algebraic)
OR
y−2 1
M2 for =− oe
x − 3.5 their gradient from (a)
OR
1
M1 for − soi
their gradient from (a)
M1 for correct substitution of (3.5, 2)
into y = (their m)x + c oe
7(c)(ii) (–4.5, 4) 2 −8
B1 for each value or for seen
2
(a) Angelo has a bag containing 3 white counters and x black counters.
He takes two counters at random from the bag, without replacement.
(i) Complete the following statement.
The probability that Angelo takes two black counters is
x # . x+3 [2]
7
(ii) The probability that Angelo takes two black counters is .
15
(a) Show that 4x2 - 25x - 21 = 0.
[4]
(b) Solve by factorisation.
4x2 - 25x - 21 = 0
x = …… or x = …… [3]
(c) Write down the number of black counters in the bag.
…… [1]
(b) Esme has a bag with 5 green counters and 4 red counters.
She takes three counters at random from the bag without replacement.
Work out the probability that the three counters are all the same colour.
…… [4]
Mark scheme
8(a)(i) x −1 2 B1 for either numerator or denominator correct
x+2
8(a)(ii)(a) x x −1 7 B1 7
× = FT their (a)(i) =
x + 3 x + 2 15 15
15x(x – 1) = 7(x + 3)(x + 2) M1 Removes all algebraic fractions
FT their equation if in comparable form
15x2 – 15x = 7x2 + 21x + 14x + 42 M1 Correctly expands all brackets
FT their equation if in comparable form
[8x2 – 50x – 42 = 0] A1
With no errors or omissions seen and one further
4x2 – 25x – 21 = 0 stage seen after final M1
8(a)(ii)(b) (4x + 3)(x – 7) [= 0] M2 M1 for
4x(x – 7) + 3(x – 7) or x (4x + 3) – 7(4x + 3)
or for (4x + a)(x + b) where either ab = –21 or
4b + a = –25
If 0 scored, SC1 for 4x + 3 and x – 7 seen but
not in factorised form
3 B1
7 and −
4
8(a)(ii)(c) 7 1 FT their positive solution
8(b) 1 4 5 4 3 4 3 2
oe M3 for × × + × ×
(a)
C
NOT TO
SCALE
A D B
58 m
In the diagram, BC is a vertical wall standing on horizontal ground AB.
D is the point on AB where AD = 58 m.
The angle of elevation of C from A is 26°.
The angle of elevation of C from D is 72°.
(i) Show that AC = 76.7 m, correct to 1 decimal place.
[5]
(ii) Calculate BD.
BD = …… m [3]
(b) Triangle EFG has an area of 70 m2.
EF : FG = 1 : 2 and angle EFG = 40°.
(i) Calculate EF.
EF = …… m [4]
(ii) A different triangle PQR also has an area of 70 m2.
PQ : QR = 1 : 2 and PQ = EF.
Find angle PQR.
Angle PQR = …… [1]
Question 10 is printed on the next page.
Mark scheme
9(a)(i) ∠ ACD = 46 soi B2 B1 for angle ADC = 108 or angle DCB = 18
or
∠CDE = 44 soi
58sin108 M2 sin108 sin their 46
M1 for = oe
sin their 46 x 58
76.68… nfww A1
9(a)(ii) 10.9 or 10.91 to 10.94 3 B2 for [AB =] 68.9 or 68.91 to 68.94
or M2 for a correct explicit statement for AB or
BD
AB
or M1 for = cos26 oe
76.7
9(b)(i) 10.4 or 10.43 to 10.44 4 70
M3 for oe
sin 40
or M2 for x2 × sin 40 = 70 oe
or M1 for 1 x × 2x × sin 40 = 70
2
9(b)(ii) 140 1
(a) 19, 15, 11, 7, ....
(i) Write down the next two terms of the sequence.
…… , …… [2]
(ii) Find the nth term of this sequence.
…… [2]
(iii) Find the value of n when the nth term is -65.
n = …… [2]
(b) Another sequence has nth term 2n2 + 5n - 15.
Find the difference between the 4th term and the 5th term of this sequence.
…… [2]
Mark scheme
10(a)(i) 3, –1 2 B1 for each
10(a)(ii) 23 – 4n oe final answer 2 M1 for k – 4n or 23 – jn (j ≠ 0)
10(a)(iii) 22 2 M1 for their (a)(ii) = –65
10(b) 23 2 B1 for 37 or 60
Checked your answers against these? Let AI mark the whole paper, on your own working.
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.
Check your answers against it
Done this paper? Upload your script and CIE Buddy's AI marker marks it question by question against this exact scheme, puts the ticks on your own working and says what each lost mark needed.
Where can I find the 0580/43 May/June 2019 mark scheme?+
On this page, free: the mark scheme for IGCSE Mathematics 0580/43, May/June 2019 (0580_s19_ms_43), and the question paper (0580_s19_qp_43), to read or download, with each question's marking points written out underneath.
How many marks is 0580/43 out of?+
Mathematics 0580/43 for May/June 2019 is out of 130 marks, with 2 hours 30 minutes to answer it.
Can I get my 0580/43 answers marked?+
Yes. Upload your answers to 0580/43/M/J/19 as a PDF or photos, and CIE Buddy marks every question against this paper's official mark scheme, using Cambridge's marking rules. You get your score, each lost mark explained on your own script, and the topics to revise. It's a guide, not an official Cambridge grade.