Mathematics 0580/42 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.
(a) The price of a newspaper increased from $0.97 to $1.13 .
Calculate the percentage increase.
…… % [3]
(b) One day, the newspaper had 60 pages of news and advertisements.
The ratio number of pages of news : number of pages of advertisements = 5 : 7.
(i) Calculate the number of pages of advertisements.
…… [2]
(ii) Write the number of pages of advertisements as a percentage of the number of pages of news.
…… % [1]
(c) On holiday Maria paid 2.25 euros for the newspaper when the exchange rate was $1 = 0.9416 euros.
At home Maria paid $1.13 for the newspaper.
Calculate the difference in price.
Give your answer in dollars, correct to the nearest cent.
$ …… [3]
(d) The number of newspapers sold decreases exponentially by x% each year.
Over a period of 21 years the number of newspapers sold decreases from 1 763 000 to 58 000.
Calculate the value of x.
x = …… [3]
(e) Every page of the newspaper is a rectangle measuring 43 cm by 28 cm, both correct to the nearest centimetre.
Calculate the upper bound of the area of a page.
…… cm2 [2]
Mark scheme
1(a) 16.5 or 16.49... 3 1.13 − 0.97 1.13
M2 for [×100] oe or × 100 oe
0.97 0.97
1.13
or M1 for oe
0.97
1(b)(i) 35 2 M1 for 60 ÷ (5 + 7 )
1(b)(ii) 140 1
1(c) $1.26 final answer 3 B2 for 1.259... or 1.26 but not as final answer
or M1 for 2.25 ÷ 0.9416
If 0 scored, SC1 for 1.13 × 0.9416
1(d) 15[.0…] 3 58000
M2 for 21 oe
1763000
or M1 for 58000 = 1763000 (k )
21
1(e) 1239.75 2 B1 for 43 + 0.5 or 28 + 0.5 oe seen
(a)
A C
26°
NOT TO
SCALE
F B x° D
E
AC is parallel to FBD, ABC is an isosceles triangle and CBE is a straight line.
Find the value of x.
x = …… [3]
(b)
S
P
58°
17° T NOT TO
SCALE
y°
Q
The diagram shows a circle with diameter PQ.
SPT is a tangent to the circle at P.
Find the value of y.
y = …… [5]
Mark scheme
2(a) 103 3 M1 for angle ABC or angle ACB = 1 (180 − 26)
2
oe
M1 for angle ABF = 26
or angle CBD or angle FBE = 77 or exterior
angle ACB = 103 correctly identified or in
correct position
2(b) 75 5 B4 for 105 at a or b
or 73 at c and 32 at d
or B3 for 58 at m
or 58 at e and 17 at k
or B2 for 32 at d and 90 soi at (c+k)
or 32 at d and 17 at k
or 73 at c
or B1 for 90 soi at (c + k) or between tangent
and radius
or 32 at d or 17 at k
S
P
d 58°
17° T
m a y° c
b k
e
Q
The probability that Andrei cycles to school is r.
(a) Write down, in terms of r, the probability that Andrei does not cycle to school.
…… [1]
(b) The probability that Benoit does not cycle to school is 1.3 - r.
The probability that both Andrei and Benoit do not cycle to school is 0.4 .
(i) Complete the equation in terms of r.
(……) # (……) = 0.4 [1]
(ii) Show that this equation simplifies to 10r 2 - 23r + 9 = 0 .
[3]
(iii) Solve by factorisation 10r 2 - 23r + 9 = 0 .
r = …… or r = …… [3]
(iv) Find the probability that Benoit does not cycle to school.
…… [1]
Mark scheme
3(a) 1–r 1
3(b)(i) (1 – r) (1.3 – r) [= 0.4] 1 FT their(a) dep on (a) being an expression in r
3(b)(ii) 1.3 – 1.3r – r + r 2 or better nfww M1 FT their (b)(i)
0.9 − 2.3r + r 2 [= 0] M1 Strict FT their expansion to a quadratic then
equating to 0.4 and then collecting to 3 terms on
OR ‘one side’
OR
13 – 13r – 10r + 10r2 = 4 oe Strict FT their expansion to a quadratic = 0.4 all
multiplied by 10
10 r 2 − 23r + 9 = 0 A1 no errors or omissions seen
3(b)(iii) (5r − 9)(2r − 1) [= 0] B2 or B2 for e.g. 5r(2r – 1) – 9(2r – 1) and then
5r – 9 = 0 and 2r – 1 = 0
or B1 for 5r(2r – 1) – 9(2r – 1) [ = 0]
or 2r(5r – 9) – 1(5r – 9) [ = 0]
or (5r + a)(2r + b) [ = 0] where a, b are integers
and ab = +9 or 2a + 5b = – 23
If 0 scored, SC1 for 5r – 9 and 2r – 1 seen but
not in factorised form
9 1 B1
[r =] oe [r =] oe
5 2
3(b)(iv) 4 1
0.8 or oe
5
(a) The equation of a straight line is 2y = 3x + 4 .
(i) Find the gradient of this line.
…… [1]
(ii) Find the co-ordinates of the point where the line crosses the y-axis.
( …… , …… ) [1]
(b) The diagram shows a straight line L.
y
6
4
2
–2 0 2 4 6 x
L
–2
(i) Find the equation of line L.
…… [3]
(ii) Find the equation of the line perpendicular to line L that passes through (9, 3).
…… [3]
(c) A is the point (8, 5) and B is the point (- 4, 1).
(i) Calculate the length of AB.
…… [3]
(ii) Find the co-ordinates of the midpoint of AB.
( …… , …… ) [2]
1 x
Mark scheme
4(a)(i) 1.5 oe 1
4(a)(ii) (0, 2) 1
4(b)(i) y = −2 x + 6 oe final answer 3 B2 for y = −2 x + c oe
or y = mx + 6 oe m ≠ 0
or for answer − 2 x + 6
6
or B1 for [gradient =] − oe or c = + 6 soi
3
4(b)(ii) y = 0.5 x − 1.5 oe final answer 3 B1 for [gradient = ] – 1 divided by their gradient
from (b)(i) evaluated soi
M1 for substitution of (9, 3) into
y = (their m)x+ c seen in working
4(c)(i) 12.6 or 12.64 to 12.65 3
M2 for (8 − −4) 2 + (5 − 1) 2 oe
or M1 for (8 − −4 )2 + (5 − 1)2 oe
4(c)(ii) (2, 3) 2 B1 for each
The table shows some values of y = - for 0.15 G x G 3.5 .
2x 4
x 0.15 0.2 0.5 1 1.5 2 2.5 3 3.5
y 3.30 0.88 - 0.04 - 0.43 - 0.58 - 0.73
(a) Complete the table. [3]
1 x
(b) On the grid, draw the graph of y = - for 0.15 G x G 3.5 .
2x 4
The last two points have been plotted for you.
y
3.5
3.0
2.5
2.0
1.5
1.0
0.5
0
0.5 1.0 1.5 2.0 2.5 3.0 3.5 x
– 0.5
– 1.0
[4]
1 x 1
(c) Use your graph to solve the equation - = for 0.15 G x G 3.5 .
2x 4 2
x = …… [1]
(d) (i) On the grid, draw the line y = 2 - x . [2]
(ii) Write down the x co-ordinates of the points where the line y = 2 - x crosses the graph of
1 x y= - for 0.15 G x G 3.5 .
2x 4
x = …… and x = …… [2]
1 x
(e) Show that the graph of y = - can be used to find the value of 2 for 0.15 G x G 3.5 .
2x 4
[2]
Mark scheme
5(a) 2.45, 0.25, − 0.25 3 B1 for each
5(b) Fully correct smooth curve 4 B3FT for 6 or 7 points
or B2 FT for 4 or 5 points
or B1 FT for 2 or 3 points
5(c) 0.7 to 0.8 1 FT their curve
5(d)(i) Correct ruled line 2 M1 for good freehand,
or ruled line with gradient −1.05 to −0.95
or ruled line through (0, 2) but not line y = 2
5(d)(ii) Both intersections of their (b) and 2 Strict FT intersection of their (b) and
their (d)(i) their (d)(i)
B1FT for one correct
OR
B2 for 0.27 to 0.28 and 2.38 to 2.39
5(e) 1 x M1
Substitutes x = 2 into −
2x 4
OR
Identifies y = 0 oe
OR
Correctly manipulates to a single
fraction
2 − x2
e.g. oe seen
4x
Concludes ‘read the graph at y = 0’ A1
oe
OR
1 x
Manipulates 0 = − oe
2x 4
leading to x 2 = 2
OR
2 − x2
States oe = 0 leading to
4x
x2 = 2
6(a) x 2 + 4 x − 21 final answer 2 B1 for three of x 2 , + 7 x, − 3x, − 21
6(b)(i) 5q 2 (3 p 2 − 5q ) final answer 2 B1 for 5(3 p 2 q 2 − 5q 3 ) or q 2 (15 p 2 − 25q ) or
( )
q 15 p 2 q − 25q 2 or 5q (3 p 2 q − 5q 2 )
or for correct answer seen
6(b)(ii) (2 g + 5k )(2 f + 3h ) final answer 2 B1 for 2 g (2 f + 3h ) + 5k (2 f + 3h ) or
2 f (2 g + 5k ) + 3h(2 g + 5k )
or for correct answer seen
6(b)(iii) (9k + m)(9k − m) final answer 2 M1 for (9 + m)(9 – m)
or for correct answer seen
6(c) 5.5 4 M1 for 5 × 3(x − 4) + x + 2 = 5 × 6
M1 for 15 x − 60 + x + 2 = 30 FT their first step
x+2
or 3 x − 12 + =6
5
If M0M0, SC1 for 3x – 12 + x + 2 = 30 oe
M1dep for 16 x = 88 FT their previous steps
(a) Show that each interior angle of a regular pentagon is 108°.
[2]
(b)
D
E C NOT TO
O SCALE
M
X
A B
The diagram shows a regular pentagon ABCDE.
The vertices of the pentagon lie on a circle, centre O, radius 12 cm.
M is the midpoint of BC.
(i) Find BM.
BM = …… cm [3]
(ii) OMX and ABX are straight lines.
(a) Find BX.
BX = …… cm [3]
(b) Calculate the area of triangle AOX.
…… cm2 [3]
Mark scheme
7(a) 360 M2 360
180 − or or M1 for or (5 − 2 ) × 180
5 5
(5 − 2) × 180 or ( 2 × 5 − 4 ) × 90 or or 90(2 × 5 – 4)
or 3 × 180 ÷ 5
5 5
or 6 × 90 ÷ 5
5 × 180 − 360
or 5 × 180 – 360
5
5 − 2 × 180
If 0 scored, SC1 for
5
7(b)(i) 7.05 or 7.053… 3 M2 for 12 × cos54 oe
or M1 for implicit form
or B1 for length of edge of pentagon
= 14.1 to 14.11
If 0 scored, SC1 for right angle at M
7(b)(ii)(a) 22.8 or 22.81 to 22.83… nfww 3 their (b)(i)
M2 for oe
cos 72
or M1 for implicit form oe
or B1 for AX = 36.9 or 36.93 to 36.94
7(b)(ii)(b) 179 or 179.1 to 179.3… 3 M2 for 1 × 12 × their AX × sin 54 oe
2
or 12 × 12 × their OX × sin108 oe
or 12 × their AX × their OX × sin18
or 12 × 122 × sin 72 + area OBX oe
or
1
2
× 122 × sin 72 + area OMB + area MBX oe
or M1 for a correct method to find area of one
relevant triangle AOB, OMB, MBX, OBX
or ONX seen
(a)
D
64° 53°
16.5 cm
NOT TO
SCALE
12.4 cm
A
95°
B
C
The diagram shows two triangles ABD and BCD.
AD = 16.5 cm and BD = 12.4 cm.
Angle ADB = 64°, angle BDC = 53° and angle DBC = 95°.
(i) Find AB.
AB = …… cm [4]
(ii) Find BC.
BC = …… cm [4]
(b)
y°
3.8 cm
NOT TO
SCALE
7.7 cm
The diagram shows a sector of a circle of radius 3.8 cm.
The arc length is 7.7 cm.
(i) Calculate the value of y.
y = …… [2]
(ii) Calculate the area of the sector.
…… cm2 [2]
Mark scheme
8(a)(i) 15.7 or 15.70... 4 M2 for 16.5 2 + 12.4 2 − 2 × 16.5 × 12.4 × cos 64
or M1 for implicit form
A1 for 246 to 247
8(a)(ii) 18.7 or 18.68 to 18.69 4 B1 for 32 or angle DBM = 37 or
angle CBM = 58
12.4 × sin 53
M2 for oe
sin 32
or M1 for implicit form oe
8(b)(i) 116.1 or 116.08 to 116.09... 2 y
M1 for × 2 × π × 3.8 = 7.7 oe
360
8(b)(ii) 14.6 or 14.61 to 14.63… 2 their (b)(i)
M1 for × π × 3.82 oe
360
100 students were each asked how much money, $m, they spent in one week.
The frequency table shows the results.
Amount ($m) 01mG5 5 1 m G 10 10 1 m G 20 20 1 m G 30 30 1 m G 50
Frequency 16 38 30 9 7
(a) Calculate an estimate of the mean.
$ …… [4]
(b) Complete the cumulative frequency table below.
Amount ($m) mG5 m G 10 m G 20 m G 30 m G 50
Cumulative
16 100 frequency
[2]
(c) On the grid, draw the cumulative frequency diagram.
100
80
60
Cumulative frequency
40
20
0
0 10 20 30 40 50 m
Amount ($)
[3]
(d) Use your cumulative frequency diagram to find an estimate for
(i) the median,
$ …… [1]
(ii) the interquartile range,
$ …… [2]
(iii) the number of students who spent more than $25.
…… [2]
Mark scheme
9(a) 12.8[0] 4 M1 for midpoints soi
M1 for use of ∑fm with m in correct interval
including both boundaries
M1 (dep on 2nd M1) for ∑fm ÷ 100
9(b) 54 84 93 2 B1 for 2 correct or 1 error and 2 correct or FT
9(c) correct diagram with all points 3 B1FT their (b) for plots at 5 correct heights
correctly plotted
B1 for 5 points at upper ends of intervals on
correct vertical line
B1FT (dep on at least B1) for increasing curve
or polygon through 5 points
After 0 scored, SC1FT for 4 correct points
plotted
9(d)(i) 9 to 9.8 final answer 1
9(d)(ii) 8.5 to 11.5 2 B1 for [UQ =] 15.5 to 17.5 or [LQ =] 6 to 7 seen
9(d)(iii) 10, 11 or 12 2 B1 for 88 to 90 seen or for answer between 10
and 12
(a) The volume of a solid metal sphere is 24 430 cm3.
(i) Calculate the radius of the sphere.
4
[The volume, V, of a sphere with radius r is V = rr 3. ]
3
…… cm [3]
(ii) The metal sphere is placed in an empty tank.
The tank is a cylinder with radius 50 cm, standing on its circular base.
Water is poured into the tank to a depth of 60 cm.
Calculate the number of litres of water needed.
…… litres [3]
(b) A different tank is a cuboid measuring 1.8 m by 1.5 m by 1.2 m.
Water flows from a pipe into this empty tank at a rate of 200 cm3 per second.
Find the time it takes to fill the tank.
Give your answer in hours and minutes.
…… hours …… minutes [4]
(c)
NOT TO
SCALE
Area = 295 cm2 Area = 159.5 cm2
17 cm
The diagram shows two mathematically similar shapes with areas 295 cm2 and 159.5 cm2.
The width of the larger shape is 17 cm.
Calculate the width of the smaller shape.
…… cm [3]
Mark scheme
10(a)(i) 18[.0] or 17.99 to 18.00… 3 24430 × 3
M2 for 3 oe
4π
4
or M1 for πr 3 = 24430
3
10(a)(ii) 447 or 446.8 to 446.9... 3 M2 for π × 502 × 60 − 24430 oe
or M1 for π × 50 2 × 60 oe
10(b) 4 [hours] 30 [ mins] nfww 4 B3 for 16200 or 4.5 or 270
figs 18 × figs 15 × figs 12
or M2 for oe
figs 2
or M1 for figs 18 × figs 15 × figs 12 oe
10(c) 12.5 or 12.50… 3 159.5
M2 for 17 × oe
295
159.5 295
or M1 for or seen
295 159.5
159.5 x 2
or for = oe
295 17 2
Diagram 1 Diagram 2 Diagram 3 Diagram 4 Diagram 5
The sequence of diagrams above is made up of small lines and dots.
(a) Complete the table.
Diagram 1 Diagram 2 Diagram 3 Diagram 4 Diagram 5 Diagram 6
Number of
4 10 18 28 small lines
Number of
4 8 13 19 dots
[4]
(b) For Diagram n find an expression, in terms of n, for the number of small lines.
…… [2]
(c) Diagram r has 10 300 small lines.
Find the value of r.
r = …… [2]
(d) The number of dots in Diagram n is an 2 + bn + 1.
Find the value of a and the value of b.
a = ……
b = …… [2]
Mark scheme
11(a) 40 54 4 B1 for each
26 34
11(b) n 2 + 3n or n(n + 3) oe 2 B1 for a quadratic expression
or for 2nd common difference 2
(at least 2 shown)
or for 2 correct equations seen
or for subtracting n2
11(c) 100 2 M1 for their (b) = 10300 seen
11(d) 1 2 B1 for each
[a = ] oe or M1 for one correct equation
2
and or for 2nd difference = 1 soi (at least 2 shown)
5
[b =] oe
2
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Mathematics 0580/42 for May/June 2019 is out of 130 marks, with 2 hours 30 minutes to answer it.
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