Cambridge IGCSE Mathematics 0580 — 2023 May/June Paper 2 · Variant 3
0580/23/M/J/23 · 23 questions · 70 marks · ≈79 min
The question paper and its mark scheme, free to read here and free to download. This is Cambridge’s own paper, exactly as it was sat.
Question paper12 pages
Mark scheme8 pages
Answers below. Sit the paper first if you are practising.
Questions as text
Question 1
1 (a) Complete the statement. The diagram has rotational symmetry of order ........................ . [1] (b) On the diagram, draw all the lines of symmetry. [2]
Mark scheme: Question Answer Marks Partial Marks 1(a) 2 1 1(b) 2 B1 for one correct line and no extras or two correct lines and one extra
Q2 · A film lasts for 2 hours 50 minutes
2 A film lasts for 2 hours 50 minutes. The film ends at 23 05. Find the time the film starts. ................................................. [1]
Mark scheme: 2 20 15 or [0]8.15pm 1
Q3 · NOT TO 5 cm SCALE 3 cm 8 cm Find the total surface area of the cuboid
3 NOT TO 5 cm SCALE 3 cm 8 cm Find the total surface area of the cuboid. .......................................... cm2 [3]
Mark scheme: 3 158 3 M2 for 2 8 5 8 3 5 3 or M1 for 8 5 or 8 3 or 5 3
Q4 · V = u - 9.8t Find the value of v when u = 4 and t =- 7
4 v = u - 9.8t Find the value of v when u = 4 and t =- 7 . v = ................................................ [2]
Mark scheme: 4 72.6 2 M1 for 4 9.8 7 or better
Q5 · Simplify d 8 ' d 2
5 Simplify d 8 ' d 2 . ................................................. [1]
Mark scheme: 5 d 6 1
Q6 · At the end of the day, a shopkeeper has 12 tins of cat food left
6 At the end of the day, a shopkeeper has 12 tins of cat food left. 3 This is of the number he had at the beginning of the day. 13 Calculate the number of tins he had at the beginning of the day. ................................................. [2]
Mark scheme: 6 52 2 3 3 M1 for 12 x oe or better e.g. 12 ÷ oe 13 13
Q7 · A spinner has five sides
7 A spinner has five sides. Each side is painted red, blue, green, yellow or orange. The table shows some of the probabilities of the spinner landing on each colour. Colour Red Blue Green Yellow Orange Probability 0.3 0.16 0.18 0.25 (a) Complete the table. [2] (b) Dan spins the spinner once. Find the probability that the spinner lands on red or blue. ................................................. [2]
Mark scheme: 7(a) 0.11 oe 2 M1 for 1 – (0.3 + 0.16 + 0.18 + 0.25) oe or B1 for 0.89 oe 7(b) 0.46 oe 2 M1 for 0.3 + 0.16
Q8 · Y 5 A 4 3 2 1 – 1 0 1 2 3 4 5 6 7 x – 1 B – 2 – 3 – 4 Describe fully the single…
8 y 5 A 4 3 2 1 – 1 0 1 2 3 4 5 6 7 x – 1 B – 2 – 3 – 4 Describe fully the single transformation that maps triangle A onto triangle B. ............................................................................................................................................................. ............................................................................................................................................................. [3]
Mark scheme: 8 Rotation 3 B1 for each (0,0) oe 90° clockwise oe
Q9 · The distance–time graph shows information about Kai’s journey from home to the office
9 The distance–time graph shows information about Kai’s journey from home to the office. 70 Office 60 50 40 Distance (km) 30 20 10 Home 0 10 00 11 00 12 00 13 00 14 00 Time (a) Calculate the average speed, in km/h, for Kai’s journey from home to the office. ........................................ km/h [2] (b) When Kai arrives at the office, he finds his meeting is cancelled. He immediately returns home at a constant speed of 50 km/h. Complete the distance–time graph to show his journey home. [1]
Mark scheme: 9(a) 32.5 2 65 their distance M1 for or theirtime 2 9(b) correct ruled line 1 from (12 00, 65) to (13 18, 0)
Q10 · 110 Without using a calculator, work out 5 + 2
11 110 Without using a calculator, work out 5 + 2 . 12 4 You must show all your working and give your answer as a mixed number in its simplest form. ................................................. [3]
Mark scheme: 10 M1 Accept with other correct common k 27 11 [5] and denominators 12 12 12 71 f 27 f 3 e.g. 24, 36, 48 such as and oe or [2] 12 f 12 f 71 c 12 oe 12 12 1 A2 1 8 cao A1 for fraction equivalent to 8 6 6 49 k 1k 7 e.g. or 8 or 7 6 k 6 k 6
Q11 · = { a, b, e, g, l, m, o, r, t, y } P = { a, b, e, g, l, r } Q = { e, g, m, o, r, t, y }…
11 (a) = { a, b, e, g, l, m, o, r, t, y } P = { a, b, e, g, l, r } Q = { e, g, m, o, r, t, y } P Q Complete the Venn diagram. [2] (b) A B Shade the region Al + B . [1]
Mark scheme: 11(a) 2 B1 for 1 region correct a b e m o g t y l r 11(b) 1
Q12 · 4 12 The position vector of A is and BA = e 3 o e 8 o
5 4 12 The position vector of A is and BA = e 3 o e 8 o. Show that OB = 5.1 , correct to 1 decimal place. [3]
Mark scheme: 12 2 M2 1 21 5 M1 for 5 or (5 – 4)2 + (3 − 8)2 e 2 2 or e f from their OB f or their B = (e, f) or only 1 25 Correct working leading to A1 Dep. on M2 or M1 for only 1 25 5.09[9..]
Q13 · Calculate 42 + 3 .04
13 Calculate 42 + 3 .04 . ................................................. [1]
Mark scheme: 13 8.03 or 8.032 to 8.033 1
Q14 · Write .0581o o as a fraction
14 Write .0581o o as a fraction. You must show all your working and give your answer in its simplest form. ................................................. [3]
Mark scheme: 14 581.81...− 5.81...oe M1 32 A2 576 cao A1 for oe 55 990 k 32 If M0 scored SC1 for or for answer 990 55 with insufficient working.
Q15 · The number of trees in a forest is decreasing exponentially at a rate of 1.75% per year
15 The number of trees in a forest is decreasing exponentially at a rate of 1.75% per year. Eleven years ago there were 980 trees. Calculate the number of trees in the forest now. Give your answer correct to the nearest integer. ................................................. [2]
Mark scheme: 15 807 2 11 1.75 M1 for 980 1 oe or better 100
Q16 · The volume of a cylinder is 1970 cm 3
16 The volume of a cylinder is 1970 cm 3. The height of the cylinder is 12.8 cm . Calculate the radius of the cylinder. ............................................ cm [3]
Mark scheme: 16 7.00 or 6.998 to 7.002 3 1970 M2 for [r2 ]= oe or better 12.8 or M1 for 1970 r 2 12.8 or better
Q17 · Rearrange the formula to make m the subject
17 Rearrange the formula to make m the subject. 2 ( m - k) R = m m = ................................................ [4]
Mark scheme: 17 2 k 2 k 4 m or m 2 R R 2 M1 for clearing fractions final answer M1 for expanding brackets (or ÷ 2) M1 for collecting terms in m on one side and terms not in m on the other M1 for dividing by a bracket maximum of 3 if final answer incorrect
Q18 · Y is inversely proportional to the cube root of (x + 5)
18 y is inversely proportional to the cube root of (x + 5). When x = 3, y = 12. Find y when x = 22. y = ................................................ [3]
Mark scheme: 18 8 3 k M1 for y oe 3 x 5 k M1 for substituting their k into y oe 3 22 5 OR M2 for 12 3 3 5 y 3 22 5 oe
Q19 · Solve the equation x 2 + 5x - 7 = 0
19 Solve the equation x 2 + 5x - 7 = 0 . You must show all your working and give your answers correct to 2 decimal places. x = ........................ or x = ........................ [4]
Mark scheme: 19 2 B2 2 5 5 4 1 7 B1 for 5 4 1 7 2 1 p q p q and if in form or r r B1 for p = 5 and r = 2×1 −6.14 and 1.14 cao B2 B1 for 1 correct answer for −6.1 and 1.1 or −6.140... and 1.140... or 6.14 and −1.14 or correct answers seen in working
Q20 · F ( x) = 6 x - 7 g ( x) = x -3 (a) Find f ( x + 2)
20 f ( x) = 6 x - 7 g ( x) = x -3 (a) Find f ( x + 2) . Give your answer in its simplest form. ................................................. [2] (b) Find f -1 ( x) . f - 1 ( x) = ................................................ [2] (c) Find x when g(x) = f(22) . x = ................................................ [2]
Mark scheme: 20(a) 6 x 5 cao final answer 2 M1 for 6 x 2 7 oe 20(b) x 7 x 7 2 y 7 or final answer M1 for x 6 y 7 or y 7 6 x or x 6 6 6 6 6 20(c) 1 2 M1 for x −3 = 6 22 7 or better or 0.2 5
Question 21
21 Simplify. 2 x 2 + 5 x - 12 4x 2 - 9 ................................................. [4]
Mark scheme: 21 x 4 4 B1 for 2 x 3 2 x 3 final answer 2 x 3 B2 for 2 x 3 x 4 or B1 for (2x + a)(x + b) where ab = −12 or a + 2b = 5 or x 2 x 3 4 2 x 3 or 2 x x 4 3 x 4
Q22 · These are the first four terms of a sequence
22 These are the first four terms of a sequence. 2.75 6 11.25 20 1 3 2 The nth term of this sequence is n + an + bn . 4 Calculate the value of a and the value of b. a = ................................................ b = ................................................ [5]
Mark scheme: 22 1 5 1 3 1 2 [a =] oe B4 for n n 3n seen 2 4 2 OR [b =] 3 1 M2 for any two of a b 2.75 4 1 8 4 a 2b 6 4 1 27 9 a 3b 11.25 4 1 64 16 a 4b 20 4 or M1 for one correct equation M1 for correct method to eliminate 1 variable B1 for 1 correct answer
Q23 · A train travels between two stations
23 A train travels between two stations. The distance between the stations is 220 km, correct to the nearest kilometre. The speed of the train is 125 km/h, correct to the nearest 5 km/h. Calculate the upper bound for the time the journey takes. Give your answer in hours and minutes. .............. h .............. min [4]
Mark scheme: 23 1h 48 min nfww 4 4 9 B3 for 1.8 [hrs], 15 [hrs], 5 [hrs] or 108 [mins] nfww 220 to 221 220 0.5 or M2 for or 125 2.5 120 to125 or M1 for 220 + 0.5 or 220 – 0.5 or 125 + 2.5 or 125 – 2.5
What was in this paper
The subtopics covered by these 23 questions, and how many questions each got. Open one in a new tab to see every Cambridge question on it.
3Ratio and proportion2Surface area and volume2Algebraic fractions1Exponential growth and decay1Fractions, decimals and percentages1Functions1Indices I1Introduction to probability1Limits of accuracy1Powers and roots1Rates1Sequences1Sets1Symmetry1The four operations1Time1Transformations1Vectors in two dimensions1What you needed in this session
Cambridge’s own grade thresholds for 2023 May/June, Paper 2 · Variant 3. A higher threshold means an easier paper — the bar moves with how the cohort did.