Cambridge IGCSE Mathematics - International 0607 — 2020 Oct/Nov Paper 2 · Variant 1
0607/21/O/N/20 · 13 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 scheme5 pages
Answers below. Sit the paper first if you are practising.
Questions as text
Question 1
1 Work out. 1 + 2 - 3 # 4 ................................................. [1]
Mark scheme: Question Answer Marks Partial Marks 1 –9 1
Question 2
2 Work out. - 48 ' - 8 ................................................. [1]
Mark scheme: 2 6 1
Question 3
3 Simplify fully. 5x 4 12 # 15x ................................................. [2]
Mark scheme: 3 1 2 M1 for an equivalent fraction 9
Question 4
4 Solve. - 3 ( 1 - 4)x = 9 x = ................................................. [3]
Mark scheme: 4 1 3 M2 for 12 x = 9 + 3 or 4x – 1 = 3 oe or M1 for −+3 12x or −[1](1 − 4 x ) = 3 oe
Q5 · Divide 120 in the ratio 3 : 5
5 Divide 120 in the ratio 3 : 5. ....................... , ....................... [2]
Mark scheme: 5 45 and 75 2 M1 for 120 ÷ 8 oe soi 15
Q6 · The mean of 5 numbers is 12
6 The mean of 5 numbers is 12. The mean of 3 of these numbers is 8. Find the mean of the other two numbers. ................................................. [3]
Mark scheme: 6 18 3 M2 for (5 × 12 −×3 8) [ ÷ 2] or M1 for 5 × 12 or 3 × 8
Q7 · Y varies inversely as x
7 y varies inversely as x. When x = 3 , y = 16 . Find x when y = 6 . x = ................................................. [3]
Mark scheme: 7 8 3 48 16 × 3 M2 for 6 = or x = x 6 k or M1 for y = or 16 × 3 = 6 × x x
Q8 · A = b = - 3 - 1 (a) Find a - 3 b
8 a = b = - 3 - 1 (a) Find a - 3 b . [2] f p - 4 (b) Find the magnitude of e- 3o. ................................................. [2]
Mark scheme: 8(a) −10 2 B1 for each 0 8(b) 5 2 M1 for ( − 4) 2 + ( − 3) 2 SOI by 25 If 0 scored, SC1 for –5
Q9 · A shop has a sale and all prices are reduced by 20%
9 A shop has a sale and all prices are reduced by 20%. (a) The original price of a shirt is $16. Find the sale price of the shirt. $ ................................................. [2] (b) The sale price of a dress is $40. Find the original price of the dress. $ ................................................. [2]
Mark scheme: 9(a) 12.8[0] 2 100 − 20 M1 for 16 × oe 100 9(b) 50 2 100 M1 for 40 × 100 − 20
Question 10
10 Factorise. (a) 8x + 14 ................................................. [1] (b) 8ax 2 - 6bx 3 ................................................. [2] (c) 6ax + 9 ay - 8bx - 12by ................................................. [2]
Mark scheme: 10(a) 2(4 x + 7) 1 10(b) 2 x 2 (4a − 3bx ) 2 M1 for a correct partial factorisation e.g. 2(4ax 2 − 3bx3 ) or x(8ax − 6bx 2 ) or x 2(8a − 6bx) or 2 x(4ax − 3bx 2 ) 10(c) (2 x + 3 y )(3a − 4b) 2 M1 for 3a (2 x + 3 y ) − 4b(3 y + 2 x ) or 2 x(3a − 4b) + 3 y (3a − 4b)
Q11 · 1 Work out 4 - 3
211 Work out 4 - 3 . ................................................. [2]
Mark scheme: 11 1 M1 for correct first step [ ± ] 2 − 1 8 1 3 e.g. 3 or 4 4 2 or B1 for 2 or 64 seen
Q12 · The table shows the marks of 80 students in an examination
12 The table shows the marks of 80 students in an examination. Mark (x) Frequency 0 1 x G 10 8 10 1 x G 15 16 15 1 x G 20 25 20 1 x G 30 17 30 1 x G 50 14 (a) On the grid, draw a cumulative frequency curve to show this information. 80 70 60 50 Cumulative 40 frequency 30 20 10 0 x 0 10 20 30 40 50 Mark [4] (b) Use your graph to estimate the median mark of the students. ................................................. [1]
Mark scheme: 12(a) Correct graph passing through 4 B2 for correct vertical plots (0, 0) (10, 8) (15, 24) (20, 49) or B1 for 3 or 4 correct vertical plots or (30, 66) (50, 80) all cf’s seen correct B1 for plotting points at upper group limit B1 for smooth curve 12(b) 17 to 19 1 FT their curve 13 1 11 7 − 1 y = x + 5 M1 for gradient = oe 2 4 1 − 4 − 1 M1 for gradient perp = their gradient B1 for midpoint (2.5, 4) seen M1 for subst their gradient (perp) and their mid-point into y = mx + c
Q13 · A is the point (1, 7) and B is the point (4, 1)
13 A is the point (1, 7) and B is the point (4, 1). Find the equation of the perpendicular bisector of AB in the form y = mx + c . y = ................................................. [5]
Mark scheme: Question Answer Marks Partial Marks 12(a) and (b) are dependent on an increasing curve. 12(a) Correct graph passing through 4 B2 for correct vertical plots (0, 0) (10, 8) (15, 24) (20, 49) or B1 for 3 or 4 correct vertical plots or (30, 66) (50, 80) all cf’s seen correct B1 for plotting points at upper group limit B1 for smooth curve 12(b) 17 to 19 1 FT their curve 13 1 11 7 − 1 y = x + 5 M1 for gradient = oe 2 4 1 − 4 − 1 M1 for gradient perp = their gradient B1 for midpoint (2.5, 4) seen M1 for subst their gradient (perp) and their mid-point into y = mx + c
What was in this paper
The subtopics covered by these 13 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 2020 Oct/Nov, Paper 2 · Variant 1. A higher threshold means an easier paper — the bar moves with how the cohort did.