Cambridge IGCSE Mathematics - International 0607 — 2021 Oct/Nov Paper 2 · Variant 3
0607/23/O/N/21 · 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. (a) ( - 2 ) + ( - 3 ) - ( - 4 ) ................................................. [1] (b) ( - 2 ) # ( - 3 ) # ( - 4 ) ................................................. [1]
Mark scheme: Question Answer Marks Partial Marks 1(a) –1 1 1(b) –24 1
Q2 · 91 93 95 97 99 From this list write down a prime number
2 91 93 95 97 99 From this list write down a prime number. ................................................. [1]
Mark scheme: 2 97 1
Q3 · $126 is divided into 3 shares in the ratio 1 : 2 : 4
3 $126 is divided into 3 shares in the ratio 1 : 2 : 4 . Find the value of the largest share. $ ................................................. [2]
Mark scheme: 3 72 2 M1 for 126 ÷ 7 oe
Question 4
4 Solve. (a) 5 - 2x = 0 x = ................................................ [1] (b) - 12 + 2x = 5x - 3 x = ................................................ [2]
Mark scheme: 4(a) 2.5 oe 1 4(b) –3 2 M1 for –12 + 3 = 5x – 2x oe or better
Q5 · There are 640 students in a school
5 There are 640 students in a school. The table shows the favourite colour of each of the students. Favourite colour Blue Green Red Yellow Number of students 120 2x 280 x (a) Find the value of x. x = ................................................ [2] (b) Find the relative frequency of students whose favourite colour is red. Give your answer as a fraction in its lowest terms. ................................................. [2]
Mark scheme: 5(a) 80 2 M1 for (640 − 120 − 280) ÷ (2 + 1) 5(b) 7 2 280 cao M1 for oe 16 640
Question 6
6 (a) Simplify. 75 - 27 ................................................. [2] (b) Rationalise the denominator and simplify your answer. 10 5 - 5 ................................................. [3]
Mark scheme: 6(a) 2 3 2 M1 for 5 3 or 3 3 6(b) 5 + 5 1 3 10(5 + 5) or (5 + 5) M2 for 2 2 25 − 5 (5 + 5) or M1 for × (5 + 5)
Q7 · A is the point (3, 7) and B is the point ( 9, - 1)
7 A is the point (3, 7) and B is the point ( 9, - 1) . Calculate the length AB. AB = ................................................ [3]
Mark scheme: 7 10 3 M2 for (9 − 3) 2 + (7 −−( 1)) 2 oe or M1 for (9 − 3) or (7 −−( 1)) oe
Q8 · A regular polygon has 12 sides
8 (a) A regular polygon has 12 sides. Work out the sum of the interior angles of the polygon. ................................................. [2] (b) The interior angle of a regular polygon is x°. Find an expression, in terms of x, for the number of sides of this polygon. ................................................. [2]
Mark scheme: 8(a) 1800 2 M1 for (2 × 12 − 4) × 90 or (12 − 2) × 180 oe 8(b) 360 2 M1 for 180 −x or 180 n − 360 = nx 180 −x
Q9 · Expand the brackets and simplify
9 Expand the brackets and simplify. 5x (2 - 3 x) - 3x (3x - 2 ) ................................................. [2]
Mark scheme: 9 −24 x 2 + 16 x final answer 2 B1 for −24 x 2 + kx or kx 2 + 16 x as answer or M1 for 10 x − 15 x 2 or −9 x 2 + 6 x
Q10 · Solve the simultaneous equations
10 Solve the simultaneous equations. You must show all your working. 4x + 3y =- 10 3x - 4y = 5 x = ................................................ y = ................................................ [4]
Mark scheme: 10 Correctly equating one set of coefficients M1 Correct method to eliminate one variable M1 [ x = ] − 1 A1 [ y = ] − 2 A1 If 0 scored, SC1 for answers that satisfy one equation
Q11 · F ( )x = , x !
11 f ( )x = , x ! 2.5 2x - 5 (a) Find f ( 2) . ................................................. [1] (b) Solve f ( )x = 5 . ................................................. [2]
Mark scheme: 11(a) –1 1 11(b) 13 2 1 oe M1 for 2 x − 5 = or 5(2 x − 5) = 1 or 5 5 better
Q12 · - =12 2x + 3 2x - 3 bx 2 - c Find the values of a, b and c
- =12 2x + 3 2x - 3 bx 2 - c Find the values of a, b and c. a = ................................................ b = ................................................ c = ................................................. [4]
Mark scheme: 12 [ a = ] − 24 4 B1 for (2 x + 3)(2 x − 3) or better as denominator [ b = ] 4 2 2 M1 for (2 x − 3) − (2 x + 3) seen [ c = ] 9 2 2 B1 for 4 x − 12 x + 9 or 4 x + 12 x + 9 or 4 x 2 − 9 or 4 x × − 6
Q13 · A bag contains 12 discs
13 A bag contains 12 discs. There are 2 red discs, 4 blue discs, 5 green discs and 1 yellow disc. A disc is chosen at random and not replaced. A second disc is then chosen at random. Find the probability that both discs are the same colour. ................................................. [3]
Mark scheme: 13 34 17 3 2 1 4 3 5 4 or M2 for × + × + × 132 66 12 11 12 11 12 11 2 1 4 3 5 4 or M1 for × or × or × 12 11 12 11 12 11
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 2021 Oct/Nov, Paper 2 · Variant 3. A higher threshold means an easier paper — the bar moves with how the cohort did.