Question 11 mark
Question
Write the number two million two thousand and two in figures. …… [1]
Mark scheme
1 2 002 002 1
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Cambridge IGCSE Mathematics (0580), Paper 2 (Extended), May/June 2024. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
0580_s24_qp_230580_s24_ms_230580/23/M/J/24
Each question as it's printed, and the mark scheme's points for it, as text. Diagrams, tables and graphs are in the PDFs above.
Question
Write the number two million two thousand and two in figures. …… [1]
Mark scheme
1 2 002 002 1
Question
Put one pair of brackets into this calculation to make it correct. 5 - 4 # 3 - 9 - 2 = 0 [1]
Mark scheme
2 5 – (4 × 3 – 9) – 2 1
Question
Simplify. 7x - 8y - x - y …… [2]
Mark scheme
3 6x – 9y or 3(2x – 3y) final answer 2 B1 for 6x or – 9y in final answer or 6x – 9y seen then spoilt
Question
The base of a cuboid measures 10 cm by 7 cm. The volume of the cuboid is 280 cm 3 . Calculate the height of the cuboid. …… cm [2]
Mark scheme
4 4 2 M1 for 10 × 7 × [...] = 280 oe or better
Question
In a city, the probability that it will rain today is 0.15 . Find the probability that it will not rain today in this city. …… [1]
Mark scheme
5 0.85 oe 1
Question
Factorise completely. 4x 2 y - 5xy 2 …… [2]
Mark scheme
6 xy(4x – 5y) final answer 2 B1 for y(4x2 – 5xy) or x(4xy – 5y2) or xy(4x – 5y) seen then spoilt
Question
The scale of a map is 1 : 40 000. On the map the distance between two villages is 37 cm. Calculate the actual distance between the two villages. Give your answer in kilometres. …… km [2] 3 1
Mark scheme
7 14.8 2 M1 for 1 cm represents 0.4 km soi or B1 for figs 148 as answer
Question
Without using a calculator, work out - . 7 14 You must show all your working and give your answer as a fraction in its simplest form. …… [2]
Mark scheme
8 6 1 M1 Allow any correct denominator 14k and oe 14 14 5 A1 cao 14
Question
C NOT TO SCALE 8 cm 37° A B The diagram shows a right-angled triangle. Calculate AB. AB = …… cm [2]
Mark scheme
9 6.39 or 6.389... 2 AB M1 for cos37 = oe 8
Question
Find the gradient of the line joining the points ( - 2 , 7) and (3, 1). …… [2]
Mark scheme
10 6 2 1− 7 − oe M1 for oe 5 3 − −2
Question
Solve the simultaneous equations. 5t - 2w = 19 3t + 2w = 5 t = …… w = …… [2]
Mark scheme
11 [t = ] 3 2 B1 for each [w = ] –2
Question
Simplify. 32g 16 (a) 16g 8 …… [2] 3 (b) `625k 8j 4 …… [2]
Mark scheme
12(a) 2g 8 final answer 2 B1 for final answer kg 8 or 2 g k or correct answer seen then spoilt 12(b) 125k 6 final answer 2 B1 for final answer ck 6 or 125k c or correct answer seen then spoilt
Question
(a) A B Shade the region A , Bl . [1] (b) P Q R Use set notation to describe the shaded region. …… [1]
Mark scheme
13(a) 1 13(b) R ( P Q ) or R P′ Q′ oe 1 '
Question
Q P NOT TO R SCALE 50° 48° A B T P, Q, R and T are points on the circle. AB is a tangent to the circle at T. Angle ATP = 50° , angle PTR = 48° and PQ = QR . (a) Find angle PRT. Angle PRT = …… [1] (b) Find angle QPR. Angle QPR = …… [2]
Mark scheme
14(a) 50 1 14(b) 24 2 B1 for angle PQR = 132 soi 180 − (180 − 48 ) or M1 for 2
Question
200 180 160 140 120 Cumulative frequency 100 80 60 40 20 0 0 10 20 30 40 50 Time (seconds) The time taken for each of 200 students to complete a calculation is measured. The cumulative frequency diagram shows the results. Use the diagram to find an estimate for (a) the interquartile range …… s [2] (b) the number of students taking more than 40 seconds to complete the calculation. …… [2]
Mark scheme
15(a) 11 2 B1 for 16 or 27 seen 15(b) 6 2 M1 for 194 seen
Question
A = rr 2 + rdh Rearrange the formula to make h the subject. h = …… [2]
Mark scheme
16 A − π r2 2 A π r2 oe final answer M1 for A − π r 2 = π dh or = +h πd πd πd A or − r 2 = dh π
Question
Work out, giving each answer in standard form. (a) `2.1 # 10 101j # `8 # 10 101j …… [2] (b) `2.1 # 10 101j + `2.1 # 10 100j …… [2]
Mark scheme
17(a) 1.68 × 10203 2 B1 for 16.8 × 10202 17(b) 2.31× 10101 2 B1 for figs 231
Question
B NOT TO SCALE A X V The diagram shows two sides, VA and VB, of a regular polygon. AVX is a straight line. Angle BVX = y° and angle AVB = 11.5y° . Find the number of sides of this polygon. …… [3]
Mark scheme
18 25 3 B2 for [y =] 14.4 oe or M1 for y + 11.5y = 180 or for 360 ÷ their y
Question
y 6 T 4 2 –8 –6 –4 –2 0 2 4 6 8 x –2 W –4 –6 –8 (a) Describe fully the single transformation that maps triangle T onto triangle W. …… [3] (b) Draw the enlargement of triangle T with scale factor -2 and centre of enlargement ( -1, 1). [2]
Mark scheme
19(a) Rotation 3 B1 for each 90° clockwise oe (0, –2) 19(b) Triangle at 2 B1 for enlargement s.f. –2 in wrong position (–5, –1), (–5, –7), (–7, –7)
Question
f (x) = 3 x + 2 (a) Find x when f (x) = 245. x = …… [2] (b) Find x when f - 1 (x) = 7 . x = …… [2] .
Mark scheme
20(a) 5 2 M1 for 3x + 2 = 245 20(b) 2189 2 M1 for x = f(7) or 37 + 2
Question
Write the recurring decimal 0.41 as a fraction in its simplest form. You must show all your working. …… [2]
Mark scheme
21 41.11...– 4.11... oe M1 37 A1 37 cao If M0 scored SC1 for answer with 90 90 insufficient working.
Question
Solve the equation tan x + 3 = 0 for 0° G x G 360° . …… [3]
Mark scheme
22 120, 300 3 B2 for one correct or M1 for tan x = − 3 oe If 0 or M1 scored SC1 for answers with difference of 180
Question
Simplify. 2 3 - y+1 y Give your answer as a single fraction in its simplest form. …… [3]
Mark scheme
23 −y −3 −y − 3 y+3 3 B1 for 2 y − 3 ( y + 1) oe or 2 or − y ( y + 1) y +y y ( y + 1) y+3 B1 for common denominator y( y + 1) or − final answer y2 + y or y 2 + y isw
Question
W V 4 cm NOT TO SCALE D C 5 cm A 15 cm B The diagram shows a triangular prism with cross-section triangle BCV. Angle BCV = 90° , BC = 5cm , CV = 4cm and AB = 15cm . Calculate the angle between AV and the base ABCD. …… [4]
Mark scheme
24 14.2 or 14.19 to 14.20 4 4 M3 for tan = oe 152 + 52 or M2 for 152 + 52 or152 + 52 + 42 or M1 for recognition of angle VAC
Question
Simplify. pt - p - t + 1 1 - t2 …… [4]
Mark scheme
25 1− p 4 oe final answer 1+ t B2 for ( p − 1)( t − 1) oe or B1 for p ( t − 1) − ( t − 1) or t ( p − 1) − ( p − 1) B1 for (1 − t )(1 + t ) oe
Question
S NOT TO P SCALE R p O q Q In the diagram, O is the origin. OP = p and OQ = q . R is the point of intersection of PQ and OS, with PR | RQ = 1 | 2 and OR = RS . Find the position vector of S in terms of p and q. Give your answer in its simplest form. …… [4]
Mark scheme
26 4 2 4 B3 for correct unsimplified answer p + q oe ρ 3 3 1 1 or for OR = p + q − p oe 3 3 ρ 1 or M2 for PR = (–p + q ) oe 3 ρ 2 or QR = (–q + p) oe 3 ρ ρ or M1 for PQ = – p + q oe or QP = – q + p oe or a correct route from O to S.
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Mathematics 0580/23 for May/June 2024 is out of 70 marks, with 1 hour 30 minutes to answer it.
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