Mathematics 0580/43 October/November 2022: question paper and mark scheme
Cambridge IGCSE Mathematics (0580), Paper 4 (Extended), October/November 2022. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
(a) Here are the ingredients needed to make a pasta bake to serve 12 people.
250 g butter
600 g pasta
460 g mushrooms
280 g cheese
800 ml milk
(i) Find the mass of the cheese as a percentage of the mass of the mushrooms.
…… % [1]
(ii) Find the mass of butter needed to make a pasta bake to serve 18 people.
…… g [2]
(iii) Monica has 2.2 litres of milk and 1.5 kg of each other ingredient.
Calculate the greatest number of people she can serve with pasta bake.
…… [3]
(b) In 2019, a packet of pasta cost $2.40 .
This was an increase of 25% of the cost of a packet in 2018.
(i) Work out the cost in 2018.
$ …… [2]
(ii) In 2020, the cost of a packet increased by 15% from the cost in 2019.
Work out the total percentage increase in the cost of a packet from 2018 to 2020.
…… % [3]
(c) width
NOT TO
SCALE
Pasta is sold in packets with width 11.5 cm, correct to the nearest 0.5 cm.
A shop places these packets in a single line on a shelf of length 2 m, correct to the nearest 0.1 m.
Find the maximum number of these packets that will fit along this shelf.
You must show all your working.
…… [3]
Mark scheme
1(a)(i) 60.9 or 60.86 to 60.87 1
1(a)(ii) 375 2 250
M1 for [× 18] oe
12
1(a)(iii) 30 nfww 3 M1 for figs2200 ÷ 800 [× 12]oe
M1 for 1500 ÷ 600 [× 12] oe
1(b)(i) 1.92 2 25
M1 for k × 1 + = 2.4[0] oe or better
100
1(b)(ii) 3 3
43.75 or 43
4
25 15
M2 for 1 + × 1 + [−1] [×100] oe
100 100
25 15
or 1 + × 1 + × 100 [–100]
100 100
15
2.40 × 1 +
or for 100 × 100 [– 100] oe
their(b)(i)
15 25 15
or M1 for 2.40 × 1 + or 1 + × 1 + oe
100 100 100
1(c) 18 nfww 3 200 to 210 200 + 5
M2 for or oe
11.5 − 0.25 11 to 11.5
or
M1 for 200 + 5, 200 – 5, 11.5 + 0.25 or 11.5 – 0.25
(a) Simplify fully.
(i) p 3 # p 11
…… [1]
18m 6
(ii)
3m 2
…… [2]
1
9 27 3 -
(iii) e 27x y o
64
…… [3]
(b) A sequence has nth term 3n 2 .
Write down the first 3 terms of this sequence.
…… , …… , …… [2]
(c) Find the nth term for each of these sequences.
(i) 13, 16, 19, 22, 25, …
…… [2]
(ii) 3, 17, 55, 129, 251, …
…… [2]
(d) Solve.
3x - 22
= 23
4
x = …… [3]
(e) Use the quadratic formula to solve 3x 2 + 8x - 20 = 0 .
Show all your working and give your answers correct to 2 decimal places.
x = …… , x = …… [4]
Mark scheme
2(a)(i) p14 final answer 1
2(a)(ii) 6m4 final answer 2 B1 for 6mk or km4 in final answer or correct answer seen and spoilt
2(a)(iii) 4 4 x −3 y −9 3 B2 for correct answer seen and spoilt
or final answer or 2 correct elements in final answer
3x 3 y 9 3
4 3
or B1 for one of or oe or x3 or y9 seen
3 4
2(b) 3, 12, 27 2 B1 for 12 or 27
2(c)(i) 3n + 10 oe final answer 2 B1 for 3n + k oe or jn + 10 oe (j ≠ 0)
or for correct expression shown in working and then spoilt
2(c)(ii) 2n3 + 1 oe final answer 2 B1 for 3rd diff = 12 (both needed) or for cubic answer
or for correct expression shown in working and then spoilt
2(d) 38 3 M2 for 3x = 4 × 23 + 22
or M1 for 3x – 22 = 4 × 23
3x 22
or for = 23 + oe
4 4
2(e) B2 B1 for 82 − 4(3)(−20) oe
−8 ± 82 − 4(3)(−20)
2×3 −8 + q −8 − q
or oe or oe or both
−8 8 2
(−20) 2×3 2×3
or ± −
2×3 4×32
3
or better
– 4.24, 1.57 final answers B2 B1 for each
If B0, SC1 for answers – 4.2 or –4.23 or –4.240 to – 4.239 and
1.6 or 1.572 to 1.573
or – 4.24 and 1.57 seen in working
or for –1.57 and 4.24 as final answer
The height, h cm, of each of 100 plants is recorded.
The table shows information about the heights of these plants.
Height
10 1 h G 15 15 1 h G 25 25 1 h G 40 40 1 h G 60 60 1 h G 70
(h cm)
Frequency 8 18 28 33 13
(a) Complete the histogram to show this information.
The first two blocks have been drawn for you.
2.5
2
1.5
Frequency density
1
0.5
0
10 20 30 40 50 60 70 h
Height (cm)
[3]
(b) Calculate an estimate of the mean height.
…… cm [4]
Mark scheme
3(a) Correct histogram 3 B1 for each correct block
28 33 13
If 0 scored, SC1 for two of , , or 1.87 or 1.866 to
15 20 10
1.867, 1.65, 1.3
3(b) 38.65 4 M1 for 12.5, 20, 32.5, 50, 65 soi
M1 for Σ fx where x is in the correct interval including
boundaries
M1dep for Σ fx ÷100
y
6
5
4
A
3
2
T
1
–9 –8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 x
–1
–2
–3
–4
–5
–6
–7
–8
(a) Draw the reflection of triangle T in the line y =- 2 . [2]
1
(b) Draw the enlargement of triangle T with scale factor and centre of enlargement ( - 5, - 3). [2]
2
(c) Describe fully the single transformation that maps triangle T onto triangle A.
…… [3]
Mark scheme
4(a) Triangle drawn at (1, – 5), 2 B1 for reflection in any horizontal line
(1, – 7), (5, – 5) If 0 scored, SC1 for reflection in x = – 2
4(b) Triangle drawn at (– 2, 0), 2 B1 for correct size and orientation but wrong position
(– 2, – 1), (0, – 1)
4(c) Rotation 3 B1 for each
90 [anticlockwise] oe
[centre] (– 1, 0)
NOT TO
50 cm SCALE
40 cm
1.2 m
36 cm
The diagram shows a water trough in the shape of a prism.
The prism has a cross-section in the shape of an isosceles trapezium.
The trough is completely filled with water.
(a) Show that the volume of water in the trough is 206.4 litres.
[3]
(b) The water from the trough is emptied at a rate of 600 ml per second.
Calculate the time taken, in minutes and seconds, for the trough to be emptied.
…… minutes …… seconds [3]
(c) All the water from the trough is emptied into a vertical cylindrical tank.
The depth of the water in the tank is 84 cm.
(i) Calculate the radius of the tank.
…… cm [3]
(ii) The tank is 60% full.
Calculate the height of the tank.
…… cm [2]
(d)
M
NOT TO
50 cm SCALE
40 cm
1.2 m
A 36 cm
A steel rod AM is placed inside the empty water trough as shown in the diagram.
A is a vertex at the base of the isosceles trapezium and M is the midpoint of the top edge on the opposite face.
Calculate the length of the steel rod, AM.
AM = …… cm [4]
Mark scheme
5(a) (36 + 50) × 40 M2
× 120 oe
2
(36 + 50) × 40 (0.36 + 0.5) × 0.4
or M1 for oe or oe
(0.36 + 0.5) × 0.4 2 2
× 1.2 oe
2
206400 ÷ 1000 = 206.4 A1 Must see an explicit conversion
or 0.2064 × 1000 = 206.4 nfww
5(b) 5 [minutes] 44 seconds 3 B2 for 344 [seconds] oe 5.73…[mins]
or M1 for figs206.4 ÷ figs 6 oe
5(c)(i) 28[.0] or 27.96 to 27.97 3 figs 2064
M2 for [r2=]
( figs84)π
or M1 for πr 2 × figs84 = figs 2064
5(c)(ii) 140 cao 2 M1 for 0.6h = 84 oe
ALT method
M1 for π× ( their (c)(i)) × h = figs206400 ÷ 0.6 oe
2
5(d) 128 or 127.7 to 127.8 4 B3 for 402 + 1202 + 182 oe
OR
B1 for horizontal length 18 soi
M1 for any correct attempt at 2-dimensional Pythagoras’
182 + 1202, 1202 + 402, 182 + 402
(a) P = 5k 2 - 7
(i) Find the value of P when k = 3.
P = …… [2]
(ii) Rearrange the formula to make k the subject.
k = …… [3]
(b) (i) Solve. x - 3 G 5x + 7
…… [2]
(ii) Show your answer to part (b)(i) on the number line.
x
–6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6
[1]
(c) The line y = 16 is drawn on the grid.
y
40
30
20
10
0 10 20 30 40 x
The region R satisfies the following inequalities.
y H 16 x22 2x + 3y H 72 y G 32 - x
(i) By drawing three more lines and shading the region not required, find and label region R.
[6]
(ii) Find the integer coordinates (x, y) in the region R that give the maximum value of 2x + y .
( …… , …… ) [2]
Mark scheme
6(a)(i) 38 2 M1 for 5 × 32 – 7 oe
6(a)(ii) P+7 3 P 7
[±] oe final answer M1 for P + 7 = 5k2 or
5
= k2 −
5
5
M1 for k = ……. FT their first step
2
M1 for square root to final answer
Max M2 for incorrect answer
6(b)(i) x ⩾ – 2.5 final answer 2
M1 for –4x ⩽ 7 + 3 or better
6(b)(ii) 1 FT their inequality in (b)(i)
–6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6
6(c)(i) x = 2 broken line B1
y = 32 – x solid line B1
2x + 3y = 72 solid line B2 B1 for line passing through (0, 24) or (36, 0)
Correct region indicated cao B2 B1 for region satisfying 3 of the inequalities
1 1
R
1
1
6(c)(ii) (16, 16) 2 M1 for substitution into 2x + y for any integer point in their region
Regan is playing a game with these six number cards.
–3 –2 2 3 5 7
(a) She takes two cards at random, without replacement, and multiplies the two numbers to give a score.
Find the probability that
(i) the score is 35
…… [3]
(ii) the score is a positive number.
…… [3]
(b) Regan now takes three cards at random from the six cards, without replacement, and adds the three numbers to give a total.
Find the probability that her total is 5.
…… [4]
Mark scheme
7(a)(i) 1 3 1 1
oe M2 for 2 × × oe
15 6 5
1 1
or M1 for × oe
6 5
or list or indication of 2 correct pairs
1
If 0 scored, SC1 for answer oe
18
7(a)(ii) 7 3 4 3 1 1 1 1
oe M2 for × + 2 × oe or 14 × oe
15 6 5 6 5 6 5
2 4
or 1 – 2 ×
6 5
4 3 1 1 2 4
or M1 for × or 2 × oe or 2 ×
6 5 6 5 6 5
or correct identification of 14 pairs
5
If 0 scored, SC1 for answer
9
7(b) 1 4 1 1 1 1 1 1
oe nfww M3 for 6 × × + 6 × × oe
10 6 5 4 6 5 4
1 1 1 1 1 1
or M2 for 6 × × oe or 2 × × oe
6 5 4 6 5 4
1 1 1
or M1 for k × × where k is an integer and 1 ⩽ k ⩽ 12 but
6 5 4
not k = 2 or k = 6
or identifies –2, 2 and 5 or –3, 3 and 5 as the 3 cards needed
1
If 0 scored, SC1 for answer
18
A
NOT TO
SCALE
9.5 cm
O
10 cm B
7.7 cm
D C E
A, B and C are points on the circle, centre O.
DE is a tangent to the circle at C.
AC = 10 cm , AB = 9.5 cm and BC = 7.7 cm .
(a) Show that angle ABC = 70.2° , correct to 1 decimal place.
[4]
(b) Find
(i) angle AOC
Angle AOC = …… [1]
(ii) angle ACO
Angle ACO = …… [1]
(iii) angle ACD.
Angle ACD = …… [1]
(c) Calculate the radius, OC, of the circle.
OC = …… cm [3]
(d) Calculate the area of triangle ABC as a percentage of the area of the circle.
…… % [4]
Mark scheme
8(a) 9.52 + 7.7 2 − 102 M2 M1 for 102 = 9.52 + 7.72 – 2×9.5×7.7cosB oe or better
[cos B = ] oe
2 × 9.5 × 7.7
70.206 to 70.207 or 70.21 to 70.22 A2 2477
A1 for oe or 0.339 or 0.3386….
7315
8(b)(i) 140.4 1
8(b)(ii) 19.8 1 FT (180 – their (b)(i)) ÷ 2
8(b)(iii) 70.2 1 FT 90 – their (b)(ii)
8(c) 5.31 or 5.314 to 5.315 3 5
M2 for oe
cos their (b)(ii)
5
or M1 for = cos(their (b)(ii)) oe
r
8(d) 38.8 or 38.9 or 38.78 to 38.85 4 0.5 × 9.5 × 7.7 × sin 70.2
M3 for [×100]
π × (their (c)) 2
OR
M1 for 0.5 × 9.5 × 7.7 × sin70.2
M1 for π × (their (c)) 2
(a) Sketch the following graphs.
On each sketch, indicate any intercepts with the axes.
(i) 3x - 4y = 12
y
O x
[2]
(ii) y = x 2 - 3x - 4
y
O x
[4]
(iii) y = 6x
y
O x
[2]
dy 4
(b) (i) Find the derivative, , of y = 5 + 8x - x 3 . dx 3
…… [2]
4
(ii) Find the gradient of y = 5 + 8x - x 3 at x =-1.
3
…… [2]
4
(iii) A tangent is drawn to the graph of y = 5 + 8x - x 3 .
3
The gradient of the tangent is - 28 .
Find the coordinates of the two possible points where this tangent meets the graph.
( …… , …… )
( …… , …… ) [5]
1 -3
Mark scheme
9(a)(i) Correct sketch of 3x – 4y = 12 with 2 B1 for line with positive gradient
y = –3 and x = 4 indicated on axes
y
4
x
O
–3
9(a)(ii) Correct sketch of y = x2 – 3x – 4 4 B3 for correct sketch with one value omitted or incorrect or for a
with (0, – 4) indicated as y – intercept and x = – 1 and x = 4 poor sketch with all 3 intercepts correct.
indicated as roots
or B2 for roots x = – 1 and x = 4 soi with no extra roots
-1 4 or for correct shape with y = – 4 indicated
or B1 for correct shape
-4 or for (x – 4) (x + 1) shown
or for incorrect sketch with (0, – 4 ) indicated as y – intercept
Minimum in fourth quadrant,
not at x = 0
9(a)(iii) Correct sketch of y = 6x 2 B1 for increasing exponential graph seen on both sides of the
with y-intercept indicated at (0, 1) y-axis.
1
9(b)(i) 8 – 4x2 [+ 0] 2 B1 for two terms correct and one extra incorrect term or for one of
two terms correct or for correct answer seen and spoilt
9(b)(ii) 4 2 M1 for substitution of x = –1 into their (b)(i)
9(b)(iii) (3, –7) and (–3, 17) 5 B4 for (3, –7) or (–3, 17)
or B3 for x = ± 3
or M2 for x2 = 9 or k(x – 3)(x + 3) = 0 oe
or for correct method for solving their (b)(i) = – 28
or M1 for their (b)(i) = – 28
(a) a=e o b =e o
2 5
(i) On the grid, draw and label vector 2a.
[1]
(ii) On the grid, draw and label vector (a - b) .
[2]
(b)
C M B
NOT TO q SCALE
N
O p A
OABC is a trapezium with OA parallel to CB.
M is the midpoint of CB and N is the point on AB such that AN : NB = 1 : 2 .
3
O is the origin, OA = p , OC = q and CB = p .
4
(i) Find, in terms of p and/or q, in its simplest form
(a) OB
OB = …… [1]
(b) AB
AB = …… [2]
(c) MN .
MN = …… [3]
(ii) OA and MN are extended to meet at G.
Find the position vector of G in terms of p.
…… [2]
Mark scheme
10(a)(i) 2a drawn correctly with direction arrow 1
10(a)(ii) a − b drawn correctly with direction arrow 2 4
B1 for seen or implied
−3
or M1 for correctly drawing their a – b with an arrow
10(b)(i)(a) 3 1
q+ p final answer
4
10(b)(i)(b) 1 2 M1 for a correct route
q– p final answer
4
10(b)(i)(c) 13 2 3 3 2
p – q final answer M2 for p – (their (b)(i)(b)) oe
24 3 8 3
3 1
or for – p – q + p + (their (b)(i)(b)) oe
8 3
or M1 for a correct route or for
2
[BN =] – (their (b)(i)(b))
3
1
or [AN = ] (their (b)(i)(b))
3
2 13
or final answer kp – q oe or p –kq oe
3 24
10(b)(ii) 19 2 3
p oe final answer M1 for AG = p ÷ 2 soi
16 8
or for answer kp oe
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On this page, free: the mark scheme for IGCSE Mathematics 0580/43, October/November 2022 (0580_w22_ms_43), and the question paper (0580_w22_qp_43), to read or download, with each question's marking points written out underneath.
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Mathematics 0580/43 for October/November 2022 is out of 130 marks, with 2 hours 30 minutes to answer it.
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