Cambridge IGCSE Mathematics - International 0607 — 2019 Oct/Nov Paper 2 · Variant 2
0607/22/O/N/19 · 15 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 · 8 27 49 51 53 55 99 From this list write down the square number
1 8 27 49 51 53 55 99 From this list write down the square number. .................................................... [1]
Mark scheme: Question Answer Marks Partial Marks 1 49 1
Q2 · Change 3.2 metres into millimetres
2 Change 3.2 metres into millimetres. ..............................................mm [1]
Mark scheme: 2 3200 1
Q3 · Write each number in standard form
3 Write each number in standard form. (a) 28 010 .................................................... [1] (b) 0.100 209 .................................................... [1]
Mark scheme: 3(a) 2.801[0] × 10 4 1 3(b) 1.00209 × 10− 1 1
Q4 · Each interior angle of a regular polygon is 170°
4 Each interior angle of a regular polygon is 170°. Find the number of sides of this polygon. .................................................... [3]
Mark scheme: 4 36 3 360 M2 for 180 − 170 or M1 for 180 – 170
Q5 · Xian walks 8 km in 1 hours
5 Xian walks 8 km in 1 hours. 2 She then runs 10 km in 45 minutes. Find her average speed in km/h for the whole journey. ............................................km/h [3]
Mark scheme: 5 8 3 total distance M1 for oe total time B1 for 2.25 oe seen
Q6 · A 28° NOT TO SCALE x° B C AB = AC Find the value of x
6 A 28° NOT TO SCALE x° B C AB = AC Find the value of x. x = ................................................... [2]
Mark scheme: 6 76 2 M1 for (180 − 28) ÷ 2
Q7 · The lengths of the sides of a right-angled triangle are 6 cm, 8 cm and 10 cm
7 The lengths of the sides of a right-angled triangle are 6 cm, 8 cm and 10 cm. Find the tangent of the smallest angle. .................................................... [1]
Mark scheme: 7 0.75 oe 1
Q8 · Magda buys 6 apples and 4 oranges for a total cost of $4.18
8 Magda buys 6 apples and 4 oranges for a total cost of $4.18 . Oranges cost $0.52 each. Find the cost of one apple. $ ................................................... [3]
Mark scheme: 8 0.35 3 M2 for (4.18 − 4 × 0.52) ÷ 6 or M1 for 4 × 0.52
Q9 · The mean of five numbers is 16
9 The mean of five numbers is 16. When two extra numbers are included the mean of the seven numbers is 20. Find the mean of the two extra numbers. .................................................... [2]
Mark scheme: 9 30 2 M1 for 7 × 20 −×5 16
Q10 · The point A has co-ordinates (1, - 5) and the point B has co-ordinates (9, 1)
10 The point A has co-ordinates (1, - 5) and the point B has co-ordinates (9, 1) . Find the equation of the perpendicular bisector of AB in the form y = mx + c . y = ................................................... [5]
Mark scheme: 10 4 14 5 1 −−( 5) y = − x + M1 for oe 3 3 9 − 1 1 M1 for grad = − their gradient 9 + 1 −+5 1 M1 for midpoint , oe 2 2 M1 for subst (their midpoint) and (their gradient) into y = mx + c
Question 11
11 Factorise completely. 8x 2 - 18 .................................................... [2]
Mark scheme: 11 2(2 x − 3)(2 x + 3) 2 M1 for 2(4 x 2 − 9) or (4 x − 6)(2 x + 3) or (2 x − 3)(4 x + 6)
Question 12
12 (a) Simplify. 300 - 27 .................................................... [2] (b) Rationalise the denominator and simplify your answer. 14 3 - 2 .................................................... [3]
Mark scheme: 12(a) 7 3 2 M1 for 10 3 or 3 3 12(b) 2(3 + 2) or 6 + 2 2 3 14(3 + 2) M2 for (3 − 2)(3 + 2) (3 + 2) or M1 for × (3 + 2)
Question 13
13 Solve the equation. 3 log x - log 4 = 4 log 2 x = ................................................... [3] Questions 14 and 15 are printed on the next page.
Mark scheme: 13 4 3 M2 for log x 3 = log(2 4 × 4) or M1 for one correct use of rules of logs
Q14 · Rearrange the formula to make x the subject
14 Rearrange the formula to make x the subject. x y = 1 - 3x - 5 x = ................................................... [4]
Mark scheme: 14 5 y − 5 5 − 5 y 4 M1 for correctly eliminating fractions x = or x = M1 for correctly collecting their x terms 3 y − 2 2 − 3 y M1 for correct final division of their terms
Q15 · An archer fires three arrows at a target
15 An archer fires three arrows at a target. 3 The probability that the archer hits the target with each arrow is . 5 Find the probability that the archer hits the target exactly twice. .................................................... [3]
Mark scheme: 15 54 3 2 3 2 oe M2 for 3 × × oe 125 5 5 2 2 3 × or M1 for 5 5
What was in this paper
The subtopics covered by these 15 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 2019 Oct/Nov, Paper 2 · Variant 2. A higher threshold means an easier paper — the bar moves with how the cohort did.