Mathematics 0580/43 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) (i) Yasmin and Zak share an amount of money in the ratio 21 : 19.
Yasmin receives $6 more than Zak.
Calculate the total amount of money shared by Yasmin and Zak.
$ …… [2]
(ii) In a sale, all prices are reduced by 15%.
(a) Yasmin buys a blouse with an original price of $40.
Calculate the sale price of the blouse.
$ …… [2]
(b) Zak buys a shirt with a sale price of $29.75 .
Calculate the original price of the shirt.
$ …… [2]
(b) Xavier’s salary increases by 2% each year.
In 2010, his salary was $40 100.
(i) Calculate his salary in 2015.
Give your answer correct to the nearest dollar.
$ …… [3]
(ii) In which year is Xavier’s salary first greater than $47 500?
…… [3]
(c) In January 2020, the population of a town was 5% more than its population in January 2018.
In January 2021, the population of this town was 2% less than its population in January 2020.
Calculate the overall percentage increase in the population from January 2018 to January 2021.
…… % [2]
Mark scheme
1(a)(i) 120 2 M1 for 6 ÷ (21 – 19) oe soi
2x
or for =6
40
1(a)(ii)(a) 34 2 15
M1 for 40 – × 40 oe or better
100
or B1 for 6
1(a)(ii)(b) 35 2 15
M1 for 1 − × p = 29.75 or better
100
1(b)(i) 44 274 cao 3 B2 for 44273 to 44274 or 44270
5
2
or M1 for 40100 × 1 + oe
100
1(b)(ii) 2019 nfww 3 M2 for one correct trial of n = 8 or n = 9
either to find a salary or, if working with
1.02n and 47 500÷ 40 100 [= 1.1845], to
find a value of 1.02n
or B2 for final answer 9 or 4 nfww
or M1 for
n
2
their 44 274 × 1 + = 47 500 oe
100
n
2
or 40 100 × 1 + = 47 500 oe
100
or for at least one trial giving a value
greater than their 44 274
1(c) 2.9 [increase] 2 5 2
M1 for 1 + × 1 − oe
100 100
implied by 1.029 or 102.9[%]
(a) y = px 2 + t
(i) Find the value of y when p = 3, x = 2 and t = -13.
y = …… [2]
(ii) Rearrange the formula to write x in terms of p, t and y.
x = …… [3]
(b) (i) Factorise.
15x 2 - 2x - 8
…… [2]
(ii) Solve the equation.
15x 2 - 2x - 8 = 0
x = …… or x = …… [1]
(c) Factorise completely. x 3 - 16xy 2
…… [3]
(d) Simplify.
2x - 1 - 4ax + 2a
2x 2 - x
…… [4]
Mark scheme
2(a)(i) –1 2 M1 for 3 × 22 – 13 oe
2(a)(ii) y −t 3 M1 for correct rearrangement to isolate x2
[±] oe final answer term
p
M1 for correct division by p
M1 for correct square root
Incorrect answer scores a maximum of
M2
If 0 scored, SC1 for a correctly rearranged
formula with p = 3 and t = – 13
substituted
2(b)(i) (5 x − 4)(3 x + 2) oe final answer 2 B1 for (ax + b)(cx + d )
where either ac = 15 and bd = –8
or ad + bc = –2
or 5x(3x + 2) – 4(3x + 2)
or 3x(5x – 4) + 2(5x – 4)
or correct factors seen and spoiled
2(b)(ii) 4 2 1 FT a factorised quadratic
oe and − oe
5 3
2(c) x( x + 4 y )( x − 4 y ) final answer 3 B2 for ( x 2 + 4 xy )( x − 4 y )
or ( x + 4 y )( x 2 − 4 xy )
or answer in the form x(a + b)(a – b)
or correct answer seen and spoiled
or B1 for x( x 2 − 16 y 2 ) oe
or ( x + 4 y )( x − 4 y )
2(d) 1 − 2a 4 B2 for (2x – 1)(1 – 2a) oe
oe final answer or B1 for 2x – 1 – 2a(2x – 1)
x
or 2x(1 – 2a) – (1 – 2a)
B1 for x(2x – 1)
(a) Zoe’s test scores last term were 6 7 7 7 8 9 9 10 10.
Find
(i) the range,
…… [1]
(ii) the mode,
…… [1]
(iii) the median.
…… [1]
(b) The cumulative frequency diagram shows information about the time taken by each of 200 students to solve a problem.
200
180
160
140
120
Cumulative
100 frequency
80
60
40
20
0
0 2 4 6 8 10 12 14 16 18 20
Time (minutes)
Use the diagram to find an estimate of
(i) the median,
…… min [1]
(ii) the interquartile range.
…… min [2]
(c) The test scores of 200 students are shown in the table.
Score 5 6 7 8 9 10
Frequency 3 10 43 75 48 21
Calculate the mean.
…… [3]
(d) The height, in cm, of each of 200 plants is measured.
The histogram shows the results.
4
3
Frequency
2 density
1
0
50 60 70 80 90 100 110 120 130 140 150 160
Height (cm)
Calculate an estimate of the mean height.
You must show all your working.
…… cm [6]
Mark scheme
3(a)(i) 4 1
3(a)(ii) 7 1
3(a)(iii) 8 1
3(b)(i) 14 1
3(b)(ii) 4 2 B1 for [ l.q. =] 11 or [u.q =] 15
3(c) 8.09 3 M1 for 5 × 3 + 10 × 6 + 43 × 7 + 75 × 8 +
48 × 9 + 21 × 10
M1 dep ÷ 200
3(d) 30, 70, 40, 36, 24 seen B2 B1 for 3 or 4 correct
or
M1 for 1 × (80 – 50), 3.5 × (100 – 80),
4 × (110 – 100), 3.6 × (120 – 110) and
0.6 × (160 – 120) oe
(their 30 × 65 + their 70 × 90 M3 M1 for midpoints soi
+ their 40 × 105 + their 36 × 115 M1 for Σfx , x in interval or boundary of
+ their 24 × 140) ÷ 200 interval
M1 dep on second M1 for ÷ 200
99.75 A1
(a) A is the point (1, 5) and B is the point (3, 9).
M is the midpoint of AB.
(i) Find the coordinates of M.
(…… , …… ) [2]
(ii) Find the equation of the line that is perpendicular to AB and passes through M.
Give your answer in the form y = mx + c .
y = …… [4]
-2 -2
(b) The position vector of P is e o and the position vector of Q is e o.
3 5
(i) Find the vector PQ.
f p [2]
(ii) R is the point such that PR = 3PQ .
Find the position vector of R.
f p [2]
(c)
U
NOT TO
SCALE u
Y
O T t
OT = t, OU = u and UY = 2YT.
(i) Find OY in terms of t and u.
Give your answer in its simplest form.
OY = …… [2]
(ii) Z is on OT and YZ is parallel to UO.
Find OZ in terms of t and/or u.
Give your answer in its simplest form.
OZ = …… [1]
Mark scheme
4(a)(i) (2, 7) 2 B1 for each coordinate
4(a)(ii) 1 4 Correct equivalent in different form
− x + 8 oe scores 3 marks.
2
9−5 4
M1 for gradient of AB = or or 2
3 −1 2
M1 dep for gradient
1
p= −
their grad of AB
M1 (dep on previous M1) for substitution
of their midpoint into y = (their p)x + c
oe
where their p ≠ 0
4(b)(i) 0 2 0 k
B1 for or
2 k 2
ρ
4(b)(ii) −2 2 FT their PQ
9 0
B1FT for
6
4(c)(i) 2 1 1 2 ρ 2
t + u or (2t + u) final answer M1 for UY = ( t –u) oe
3 3 3 3
ρ 1
or TY = (u – t) oe
3
or correct route soi
4(c)(ii) 2 1
t cao
3
Solve the simultaneous equations.
(a) x + 2y = 13 x + 5y = 22
y = …… [2]
(b) y = 2-x y = x 2 + 2x + 2
x = …… y = ……
x = …… y = …… [4]
Mark scheme
5(a) [x =] 7 2 B1 for each
[y =] 3
5(b) [x =] 0, [y =] 2 4 B3 for x = 0 and x = –3
or B2 for x2 + 3x = 0
[x =] –3, [y =] 5 or M1 for 2 – x = x2 + 2x + 2
If 0 scored award B1 for x = 0, y = 2
or x = –3, y = 5 from no/incorrect working
ALTERNATIVE
B3 for y = 2 and y = 5
or B2 for y2 – 7y + 10 = 0
or M1 for y = (2 – y)2 + 2(2 – y) + 2
If 0 scored award B1 for x = 0, y = 2
or x = –3, y = 5 from no/incorrect working
In a class of 24 students, 18 students like homework (H ), 15 students like tests (T ) and 1 student does not like homework and does not like tests.
(a) Complete the Venn diagram to show this information.
H T
1
[2]
(b) Write down the number of students who like both homework and tests.
…… [1]
(c) Find n (H l + T ) .
…… [1]
(d) A student is picked at random from the class.
Write down the probability that this student likes tests but does not like homework.
…… [1]
(e) Two students are picked at random from the class.
Find the probability that both students do not like homework and do not like tests.
…… [1]
(f) Two of the students who like homework are picked at random.
Find the probability that both students also like tests.
…… [3]
Mark scheme
6(a) 2 i.e. 8, 10 and 5 correctly placed
H T B1 for 10 correctly placed
or M1 for 18 – x , x and 15 – x correctly
8 10 5 placed on diagram and x = 10 seen
1
6(b) 10 1 FT their Venn diagram
6(c) 5 1 FT their Venn diagram
6(d) 5 1 FT their 5 on the Venn diagram
oe
24
6(e) 0 1
6(f) 5 3 their10 their 9
oe M2 for ×
17 18 17
their10 their 9
or B1FT for or seen
18 17
25
After 0 scored, SC1 for answer oe
81
(a)
x
–2 1
Write down the inequality in x shown by the number line.
…… [2]
(b) (i) Write x 2 + 4x + 1 in the form (x + p) 2 + q .
…… [2]
(ii) Use your answer to part (b)(i) to solve the equation x 2 + 4x + 1 = 0 .
x = …… or x = …… [2]
(iii) Use your answer to part (b)(i) to write down the coordinates of the minimum point on the graph of y = x 2 + 4x + 1.
(…… , …… ) [2]
(iv) On the diagram, sketch the graph of y = x 2 + 4x + 1.
y
O x
[2]
Mark scheme
7(a) –2 < x ⩽ 1 2 B1 for –2 < x or x ⩽ 1
7(b)(i) ( x + 2 )2 − 3 2 M1 for ( x + 2 )2 + k
7(b)(ii) ( x + 2 )2 = 3 M1 FTdep their (b)(i) for k < 0
–3.73 or –3.732... and B1
–0.268 or –0.2679...
7(b)(iii) (–2, –3) 2 FT their ( x + 2 )2 − 3
B1 for each coordinate
7(b)(iv) Correct sketch 2 Parabola with minimum point in correct
2
quadrant and both x-intercepts negative
and positive y-intercept
1
4 -2
0
0 2 4
B1 for parabola with minimum point.
-1
-2
-3
-4
(a) A solid cuboid measures 20 cm by 12 cm by 5 cm.
(i) Calculate the volume of the cuboid.
…… cm3 [1]
(ii) (a) Calculate the total surface area of the cuboid.
…… cm2 [3]
(b) The surface of the cuboid is painted.
The cost of the paint used is $1.52 .
Find the cost to paint 1cm 2 of the cuboid.
Give your answer in cents.
…… cents [1]
9x
(b) A solid metal cylinder with radius x and height is melted.
2
All the metal is used to make a sphere with radius r.
Find r in terms of x.
4
[The volume, V, of a sphere with radius r is V = rr 3 .]
3
r = …… [3]
(c)
NOT TO
SCALE
20 cm
5 cm 150 cm
The diagram shows a cylinder of length 150 cm on horizontal ground.
The cylinder has radius 20 cm.
The cylinder contains water to a depth of 5 cm, as shown in the diagram.
Calculate the volume of water in the cylinder.
Give your answer in litres.
…… litres [7]
Mark scheme
8(a)(i) 1200 1
8(a)(ii)(a) 800 3 M2 for [2 ×] (20 × 12 + 20 × 5 + 12 × 5)
or M1 for 20 × 12 or 20 × 5 or 12 × 5
8(a)(ii)(b) 0.19 1 FT 152 ÷ their 800
8(b) 3x 3 27 x3 [π ]
or 1.5x B2 for r 3 = or better
2 8[π ]
4 9x
or M1 for πr 3 = πx 2 ×
3 2
8(c) 13.6 or 13.59 to 13.61 7 If chord is AB and O is centre of the cross
section
20 − 5
M2 for 2 × cos −1 oe
20
20 − 5
or M1 for cos = oe
20
theirAOB
M1 for × π × 202
360
1 82.8π
or (20)2
2 180
1
M1 for × 202 × sin(their AOB) oe
2
M1 for their area × 150
M1 for their volume ÷ 1000
(a)
C
NOT TO
SCALE
D 60°
14 cm
45°
35°
A B
Calculate the perimeter of the quadrilateral ABCD.
…… cm [7]
(b)
B
NOT TO
SCALE
A
The diagram shows a cube.
The length of the diagonal AB is 8.5 cm.
(i) Calculate the length of an edge of the cube.
…… cm [3]
(ii) Calculate the angle between AB and the base of the cube.
…… [3]
Mark scheme
9(a) 42.3 or 42.28 to 42.30... 7 AB
M1 for = cos35 oe
14
AD
M1 for = sin 35 oe
14
B1 for [C =] 75
14sin 60
M3 for [BC =] oe
sin their 75
14sin 45
and [DC] oe
sin their 75
14sin 60 14sin 45
or M2 for or oe
sin their 75 sin their 75
sin their 75 sin 60
or M1 for = oe
14 BC
sin their 75 sin 45
or = oe
14 CD
9(b)(i) 4.91 or 4.907... 3 B2 for [l2 =] 24.1 or 24.08...
8.52
or M2 for √3 l = 8.5 or [l =] oe
3
or M1 for l 2 + l 2 + l 2 = 8.52 oe
9(b)(ii) 35.3 or 35.26 to 35.3 nfww 3 their (b)(i)
M2dep for sin (angle) = oe
8.5
or M1 for clear recognition of correct
angle
f (x) = 3x - 2 g (x) = 5x - 7 h (x) = x 2 + x j (x) = 3 x
(a) Find
(i) f(2),
…… [1]
(ii) g(2),
…… [1]
(iii) gf(2).
…… [1]
(b) Find f -1 (x) .
f -1 (x) = …… [2]
(c) Find hf(x), giving your answer in the form ax 2 + bx + c .
…… [3]
(d) Find the derivative of h(x).
…… [1]
(e) (i) Find x when j -1 (x) = 4 .
x = …… [1]
(ii) Simplify j -1 j (x) .
…… [1]
Mark scheme
10(a)(i) 4 1
10(a)(ii) 3 1
10(a)(iii) 13 1 FT 5 × their (a)(i) – 7
10(b) x+2 2 y 2
final answer M1 for y + 2 = 3x or for = x−
3 3 3
or for x = 3y – 2
10(c) 9x2 – 9x + 2 final answer 3 M1 for ( 3x − 2 )2 + 3x − 2
B1 for ( 3x − 2 ) = 9 x 2 − 6 x − 6 x + 4
2
10(d) 2x + 1 1
10(e)(i) 81 1
10(e)(ii) x 1 Not y = x
(a) These are the first four terms of a sequence.
11 7 3 -1
(i) Write down the next term.
…… [1]
(ii) Write down the term to term rule for this sequence.
…… [1]
(iii) Find the nth term of this sequence.
…… [2]
2n
(b) The nth term of a different sequence is . n+1
(i) Find the difference between the 5th term and the 6th term of this sequence.
Give your answer as a fraction.
…… [2]
3
(ii) Is a term in this sequence?
4
Show how you decide.
[3]
Mark scheme
11(a)(i) –5 1
11(a)(ii) Subtract 4 oe 1
11(a)(iii) 15 – 4n oe final answer 2 B1 for k – 4n or 15 – jn j≠0
11(b)(i) 1 2 12 10
or equivalent fraction B1 for and
21 7 6
11(b)(ii) 3 M2 3 2n
n= oe M1 for = oe
5 4 n +1
or or
2n ⩾ n + 1 but 3 < 4. M1 for 2n > n + 1 but 3 < 4
No, n is not an integer oe A1
or
3
No, is less than 1, oe
4
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Mathematics 0580/43 for May/June 2021 is out of 130 marks, with 2 hours 30 minutes to answer it.
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