Cambridge IGCSE Mathematics - International 0607 — 2024 Oct/Nov Paper 2 · Variant 1
0607/21/O/N/24 · 14 questions · 40 marks · ≈45 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 paper8 pages
Mark scheme6 pages
Answers below. Sit the paper first if you are practising.
Questions as text
Question 1
1 Work out. ( 0 .01) 2 ................................................. [1]
Mark scheme: Question Answer Marks Partial Marks 1 0.0001 1 Accept 1 10−4
Q2 · Write in its lowest terms
2 (a) Write in its lowest terms. 60 ................................................. [1] 5 1 (b) Work out - . 7 14 ................................................. [2]
Mark scheme: 2(a) 2 1 CAO 5 2(b) 9 2 10 oe B1 for 14 14
Q3 · This is a list of ten numbers
3 This is a list of ten numbers. 19 24 16 17 22 14 28 34 20 18 (a) Find the range. ................................................. [1] (b) Find the median. ................................................. [2]
Mark scheme: 3(a) 20 1 3(b) 19.5 oe 2 M1 for ordering first or last 6 or identifying 19 and 20
Q4 · Expand x 3 `8 x - x 2j
4 Expand x 3 `8 x - x 2j . ................................................. [2] .05
Mark scheme: 4 8 x 4 − x 5 final answer 2 B1 for 8 x 4 − kx 5 or kx 4 − x 5
Q5 · Simplify `9x 9 y 4j
5 Simplify `9x 9 y 4j . ................................................. [2]
Mark scheme: 5 3x 4.5 y 2 final answer 2 M1 for 3 kx y 2 or kx 4.5 y 2 or 3 x 4.5 y k
Q6 · A regular polygon has 12 sides
6 A regular polygon has 12 sides. Find the size of an exterior angle of this polygon. ................................................. [2]
Mark scheme: 6 30 2 360 (12 − 2) 180 M1 for or or for 12 12 150 seen
Q7 · Y varies as the square of ( x + 1)
7 y varies as the square of ( x + 1) . When y = 18 , x = 2 . Find y when x = 3 . y = ................................................ [3]
Mark scheme: 7 32 3 M2 for y = 2( x + 1) 2 or 18 y = (2 + 1) 2 (3 + 1) 2 2 18 or M1 for y = k ( x + 1) or 2 (2 + 1)
Question 8
8 Factorise. (a) 6ax - 8by - 3 ay + 16bx ................................................. [2] (b) 5x 2 - 7 x - 6 ................................................. [2]
Mark scheme: 8(a) (2 x − y )(3a + 8b) oe 2 M1 for 3a (2 x − y ) + 8b( −−−y ( 2 x )) oe or 2 x (3a + 8b) − y (8b + 3a ) oe 8(b) (5 x + 3)( x − 2) oe 2 M1 for (5 x + a )( x + b) where ab = –6 or 5b + a = –7 or 5 x ( x − 2) + 3( x − 2) or x (5 x + 3) − 2(5 x + 3)
Q9 · Write in standard form
9 Write in standard form. (a) 760 900 ................................................. [1] (b) 0.080 07 ................................................. [1]
Mark scheme: 9(a) 7.609[00] 10 5 1 9(b) 8.007 10−2 1
Q10 · B NOT TO SCALE O C A 124° D A, B, C, and D lie on a circle, centre O
10 B NOT TO SCALE O C A 124° D A, B, C, and D lie on a circle, centre O. Find (a) angle ABC Angle ABC = ................................................ [1] (b) obtuse angle AOC Angle AOC = ................................................ [1] (c) angle OCA. Angle OCA = ................................................ [1]
Mark scheme: 10(a) 56 1 10(b) 112 1 FT 2their(a) 10(c) 34 1 180 −their(b) FT 2
Question 11
11 (a) Simplify. 2 `5 8 - 7 2j ................................................. [2] (b) Rationalise the denominator. 21 3 - 2 ................................................. [2]
Mark scheme: 11(a) 6 2 B1 for 5 16 or 10 2 or 7 4 or better 11(b) 3(3 + 2) or 9 + 3 2 final answer 2 3 + 2 B1 for 3 + 2
Q12 · Vlad has two unbiased dice, each numbered 1, 2, 3, 4, 5, 6
12 Vlad has two unbiased dice, each numbered 1, 2, 3, 4, 5, 6. Vlad rolls the two dice and records the total score. Find the probability that the total score is (a) 13 ................................................. [1] (b) 11. ................................................. [2]
Mark scheme: 12(a) 0 1 12(b) 2 2 1 1 oe M1 for [ 2] 36 6 6
Q13 · The point A has coordinates (4, -1) and the point B has coordinates (8, -3)
13 The point A has coordinates (4, -1) and the point B has coordinates (8, -3). Find the equation of the perpendicular bisector of the line AB. Give your answer in the form y = mx + c . y = ................................................ [5]
Mark scheme: 13 y = 2 x − 14 5 −−−1 3 M1 for [Grad = ] oe 4 − 8 4 − 8 M1 for [Grad perp = ] − their −−−1 3 B1 for (6, –2) as mid-point M1 for subst (their grad perp) and their (6, –2) into y = mx + c
Q14 · Write as a single fraction in its simplest form
14 Write as a single fraction in its simplest form. 8 3 - 4x - 1 2 x + 1 ................................................. [3]
Mark scheme: 14 4 x + 11 3 B1 for (4 x − 1)(2 x + 1) as denominator final answer (4 x − 1)(2 x + 1) (could be expanded) M1 for 8(2 x + 1) − 3(4 x − 1)
What was in this paper
The subtopics covered by these 14 questions, and how many questions each got. Open one in a new tab to see every Cambridge question on it.
What you needed in this session
Cambridge’s own grade thresholds for 2024 Oct/Nov, Paper 2 · Variant 1. A higher threshold means an easier paper — the bar moves with how the cohort did.