Cambridge IGCSE Mathematics - International 0607 — 2017 Oct/Nov Paper 2 · Variant 1
0607/21/O/N/17 · 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 scheme4 pages
Answers below. Sit the paper first if you are practising.
Questions as text
Q1 · By rounding each number correct to 1 significant figure, estimate the value of 189.6 # 41…
1 By rounding each number correct to 1 significant figure, estimate the value of 189.6 # 41 .28 . 0. 00509 + 0. 00298 .................................................... [3]
Mark scheme: Question Answer Marks Partial Marks 1 200 × 40 3 M1 for for 3 correct approximations A1 for numerator = 8000 0.005 + 0.003 or denominator = 0.008 1 000 000
Q2 · Written as the product of their prime factors, 7056 = 2 4 # 3 2 # 7 2 and 8232 = 2 3 # 3…
2 Written as the product of their prime factors, 7056 = 2 4 # 3 2 # 7 2 and 8232 = 2 3 # 3 # 7 3 . Giving your answers as the product of prime factors, find (a) the highest common factor (HCF) of 7056 and 8232, .................................................... [1] (b) the lowest common multiple (LCM) of 7056 and 8232, .................................................... [1] (c) 7056. .................................................... [1]
Mark scheme: 2(a) 23 × 3[1] × 72 isw 1 2(b) 24 × 32 × 73 isw 1 2(c) 22 × 3[1] × 7[1] isw 1
Q3 · Show the inequality - 1 1 x G 4 on this number line
3 Show the inequality - 1 1 x G 4 on this number line. x –5 –4 –3 –2 –1 0 1 2 3 4 5 [2]
Mark scheme: 3 2 B1 for correct interval indicated –1 4
Q4 · Work out - , giving your answer as a fraction in its lowest terms
4 Work out - , giving your answer as a fraction in its lowest terms. 8 6 .................................................... [2]
Mark scheme: 4 5 2 9 4 M1for – oe 24 24 24
Q5 · Solve the simultaneous equations
5 Solve the simultaneous equations. x – 3y = 4 5x – 6y = –7 x = .................................................... y = .................................................... [3]
Mark scheme: 5 x = –5 3 M1 for correctly eliminating one variable y = –3 B1 for each answer If zero scored SC1 for correct substitution and evaluation to find the other variable
Q6 · A is the point (3, 6) and B is the point ( - 5, 10)
6 A is the point (3, 6) and B is the point ( - 5, 10) . (a) Work out the co-ordinates of the midpoint of AB. ( ....................... , ....................... ) [2] (b) Find the length of AB, giving your answer in the form a 5. .................................................... [3]
Mark scheme: 6(a) (–1, 8) 2 3 − 5 6 + 10 B1 for each co-ordinate or for , 2 2 6(b) 3 M1 for 82 + 42 4 5 A1 for 80
Q7 · Work out, giving your answer in standard form
7 Work out, giving your answer in standard form. (6. 3 # 10 4) + (5. 6 # 10 5 ) .................................................... [2]
Mark scheme: 7 6.23 × 105 2 B1 for figs 623 or 0.63 × 105 soi or 56 × 104 soi
Q8 · Shade the region indicated in each of these Venn diagrams
8 Shade the region indicated in each of these Venn diagrams. (a) U A B Al + B l [1] (b) U A B A , (B + C ) [1] C (c) U A B A + B + C l [1] C
Mark scheme: 8(a) 1 8(b) 1 8(c) 1
Q9 · NOT TO SCALE O 140° D A 25° B C A, B, C and D are points on a circle centre O
9 NOT TO SCALE O 140° D A 25° B C A, B, C and D are points on a circle centre O. Find (a) angle ACD, Angle ACD = ................................................... [2] (b) angle BAD. Angle BAD = ................................................... [2]
Mark scheme: 9(a) 110 2 B1 for reflex angle AOD = 220 or AXD = 70 9(b) 45 2 FT 155 – their 110 B1 for angle BCD = their 110 + 25
Q10 · Y is inversely proportional to the square root of x
10 y is inversely proportional to the square root of x. When x = 9, y = 12. Find y when x = 100. .................................................... [3]
Mark scheme: 10 3.6 3 9 36 M2 for 12 × oe or y = 100 x y 12 k or M1 for = or y = 9 100 x
Q11 · Factorise x 2 - 3x - 10
11 (a) Factorise x 2 - 3x - 10 . .................................................... [2] (b) Using your answer to part (a), solve x 2 - 3x - 10 2 0 . .................................................... [2] Questions 12 and 13 are printed on the next page.
Mark scheme: 11(a) (x – 5)(x + 2) 2 B1 for (x + a)(x + b) where ab = –10 or a + b = –3 or for x(x – 5) + 2(x – 5) or x(x + 2) – 5(x + 2) 11(b) x < –2, x > 5 2 B1FT for correct 'inequalities' from (a)
Q12 · Rationalise the denominator and simplify
12 Rationalise the denominator and simplify. 14 2 3 + 2 .................................................... [3]
Mark scheme: 12 6 2 – 4 or 2(3 2 – 2) 3 3 - 2 M1 for × final answer 3 - 2 B1 for (3 + 2 )(3 – 2 ) = 7
Q13 · Expand the brackets and simplify
13 Expand the brackets and simplify. (3a - 5b)(2a - 3b) .................................................... [3]
Mark scheme: 13 6a2 – 19ab + 15b2 3 B2 for 6a2 – 9ab – 10ab + 15b2 final answer or B1 for 3 correct terms above
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 2017 Oct/Nov, Paper 2 · Variant 1. A higher threshold means an easier paper — the bar moves with how the cohort did.