Cambridge IGCSE Mathematics - International 0607 — 2025 Oct/Nov Paper 2 · Variant 1
0607/21/O/N/25 · 19 questions · 75 marks · ≈84 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 paper16 pages
Mark scheme9 pages
Answers below. Sit the paper first if you are practising.
Questions as text
Question 1
1 Complete the table. Fraction Decimal Percentage 3 = 0.3 = 30 10 = 0.09 = 7 = = 20 [2]
Mark scheme: Question Answer Marks Partial Marks 1 9 2 B1 for 2 correct 9 100 0.35 35
Q2 · Find the reciprocal of 1 5
2 Find the reciprocal of 1 5. ................................................. [2]
Mark scheme: 2 5 2 8 1 oe B1 for or 8 5 1.6 or M1 for inverting their improper fraction.
Q3 · By writing each number correct to 1 significant figure, estimate the value of 8.76 # 43.82
3 By writing each number correct to 1 significant figure, estimate the value of 8.76 # 43.82 . 0.592 ................................................. [2]
Mark scheme: 3 9 40 M1 One correct value 0.6 200 A1
Q4 · F E 35° 70° NOT TO SCALE 110° 115° x° y° A B C D ABCD is a straight line
4 F E 35° 70° NOT TO SCALE 110° 115° x° y° A B C D ABCD is a straight line. (a) (i) Find the value of x. ................................................. [1] (ii) Find the value of y. ................................................. [2] (b) Give a reason why BE is not parallel to CF. ............................................................................................................................................................ ..................................................................................................................................................... [1]
Mark scheme: 4(a)(i) 45 1 4(a)(ii) 60 2 B1 for angle ECF = 75 or M1 for 180 – their x – their ECF 4(b) Not corresponding angles oe 1 e.g. angle FCB ≠ 115 angle ABE = 115 or Not alternate angles oe e.g. angle ECF ≠ 70 angle BEC = 70
Q5 · A = e o b = e o - 2 4 Find the vector 2a - 3b
5 a = e o b = e o - 2 4 Find the vector 2a - 3b. f p [2]
Mark scheme: 5 19 2 19 k B1 for or −16 k −16
Q6 · The types of vehicle travelling on a road were recorded
6 The types of vehicle travelling on a road were recorded. The table shows the results for a sample of 200 vehicles. Type of vehicle Car Van Truck Motorcycle Number of vehicles 84 64 40 12 Relative frequency (a) Complete the table. [2] (b) One day 5000 vehicles use the road. Work out an estimate for the number of trucks that use the road that day. ................................................. [1]
Mark scheme: 6(a) 84 64 40 12 2 B1 for 2 correct. oe 200 200 200 200 6(b) 1000 1
Q7 · A quadrilateral has exactly one line of symmetry
7 A quadrilateral has exactly one line of symmetry. It has no pairs of parallel sides. Write down the mathematical name for this quadrilateral. ................................................. [1]
Mark scheme: 7 Kite 1
Q8 · Rearrange each formula to make b the subject
8 Rearrange each formula to make b the subject. (a) c = 5 ab b = ................................................ [2] 3a + 2b (b) P = 5a - b b = ................................................. [3]
Mark scheme: 8(a) c 2 2 c oe final answer B1 for c2 = 5ab or b = 5 a 5a 8(b) 5aP − 3a 3 M1 for clearing fraction oe final answer M1 for expanding and rearranging 2 + P M1 for factorising and dividing Incorrect answer scores M2 maximum
Q9 · A NOT TO SCALE 10 cm B C In the diagram, angle ABC = 90˚
9 A NOT TO SCALE 10 cm B C In the diagram, angle ABC = 90˚. AB = BC and AC = 10 cm. Calculate the area of triangle ABC. ......................................... cm2 [3]
Mark scheme: 9 25 3 M1 for x2 + x2 = 102 or 10 cos 45 or 10 sin 45 1 M1 for their AB their BC oe. 2 If 0 scored SC1 for 45 soi
Q10 · Samina has two pieces of string
10 Samina has two pieces of string. The lengths of the two pieces of string are in the ratio 5 : 3. Samina cuts 4 cm off each piece of string. The lengths of the two remaining pieces of string are in the ratio 2 : 1. Find the original lengths of Samina’s pieces of string. .................................................. cm .................................................. cm [4]
Mark scheme: 10 20 4 B3 for one correct 12 OR x 5 M1 for = oe y 3 x − 4 2 M1 for = oe y − 4 1 B1 for x – 4 or y – 4 seen
Q11 · The numbers 1 to 20 are shown in the Venn diagram
11 The numbers 1 to 20 are shown in the Venn diagram. U A 20 B 5 7 4 12 2 13 8 17 6 3 9 1 10 11 14 15 16 19 C 18 (a) List the elements of A k B . ................................................. [1] (b) Find (i) n ( A j C ) ................................................. [1] (ii) n [( A j B ) l k C ] . ................................................. [1] (c) Two of the 20 numbers are picked at random without replacement. Find the probability that (i) both numbers are in ( A j B ) l k C ................................................. [2] (ii) one number is in A but not B and the other number is in B but not A. ................................................. [3]
Mark scheme: 11(a) 3, 4, 9, 12 1 11(b)(i) 14 1 11(b)(ii) 3 1 11(c)(i) 6 2 their 3 oe B1 for oe 380 20 11(c)(ii) 60 3 5 6 oe M2 for [2] oe 380 20 19 5 6 6 5 or M1 for or or or 20 19 20 19
Q12 · B NOT TO SCALE 7 cm 5 cm A 10 cm C In triangle ABC, AB = 5 cm, BC = 7 cm and AC = 10 cm
12 B NOT TO SCALE 7 cm 5 cm A 10 cm C In triangle ABC, AB = 5 cm, BC = 7 cm and AC = 10 cm. Show by calculation that angle ABC is obtuse. [3] 5 p
Mark scheme: 12 5 2 + 7 2 − 10 2 M2 M1 for 102 = 52 + 72 – 2 × 5 × 7 × [cosABC =] cosABC 2 5 7 or 52 + 72 – 102 or 52 + 72 or 52 + 72 = 74 26 A1 − or –26 and Negative therefore 70 obtuse or 74 < 100 therefore obtuse
Q13 · = 2 6 2 3 Find the value of p
13 (a) = 2 6 2 3 Find the value of p. p = .................................................. [1] 2 (b) Find the value of 64 - 3 . ................................................. [2]
Mark scheme: 13(a) 4 1 13(b) 1 2 1 − 13 B1 for 16 or or 4–2 or 4096 seen. 2 16 64 3
Q14 · Y A B E D C G F O x 1 The lines with equations x = 2 , y = x and x + y = 5 are shown on…
14 y A B E D C G F O x 1 The lines with equations x = 2 , y = x and x + y = 5 are shown on the diagram. 2 These lines divide the space into 7 different regions A, B, C, D, E, F and G. Write down the inequalities which define (a) region A ..................................................................................................................................................... [1] (b) region C ..................................................................................................................................................... [1] (c) region E. ..................................................................................................................................................... [2]
Mark scheme: 14(a) x ⩽ 2 and x + y ⩾ 5 oe 1 14(b) 1 1 y < x and x + y ⩾ 5 oe 2 14(c) 1 2 B1 for 2 correct x ⩽ 2 and y > x and x + y ⩽ 5 oe 2
Q15 · F ( x) = 3x - 1 g ( x) = 3 - 2x h ( x) = x 2 - 2x + 3 (a) Find f ( -3)
15 f ( x) = 3x - 1 g ( x) = 3 - 2x h ( x) = x 2 - 2x + 3 (a) Find f ( -3) . ................................................. [1] (b) Find f -1 ( 10 ) . ................................................. [2] (c) Find and simplify gh(x). ................................................. [2] (d) Find g -1 ( x) . g -1 ( x) = ................................................ [2] (e) Solve h ( x) = f ( x) . x = .................. or x = ................... [4]
Mark scheme: 15(a) –10 1 15(b) 11 2 x + 1 oe M1 for 3x – 1 = 10 or 3 3 15(c) –2x2 + 4x – 3 final answer 2 M1 for 3 – 2 (x2 – 2x + 3) 15(d) 3 − x 2 M1 for x = 3 – 2y or y – 3 = –2x or oe final answer y 3 2 = − x 2 2 15(e) 1, 4 4 9 B3 for (x – 1)(x – 4) or x = 5 2 or B2 for x2 – 5x + 4 [=0] or B1 for x 2 − 2 x + 3 = 3 x − 1 If 0 scored SC1 for one correct solution
Q16 · Show that the interior angle of a regular hexagon is 120°
16 (a) Show that the interior angle of a regular hexagon is 120°. [1] (b) B NOT TO 15 cm SCALE A 10 cm The diagram shows a container in the shape of a prism. The cross-section of the prism is a regular hexagon with side length 10 cm. The height of the prism is 15 cm. A stick has one end at A and rests against B. The stick makes an angle of x with the base. Find tan x. tanx = ................................................ [4]
Mark scheme: 16(a) 360 ÷ 6 or 180(6 – 2) 1 16(b) 15 4 M1 for angle x clearly indicated on oe diagram 20 B2 for diagonal of base = 20 10 10 or M1 for or cos60 sin30
Q17 · Write ( 5 - 2 3 ) 2 in the form a + b 3
17 Write ( 5 - 2 3 ) 2 in the form a + b 3 . ................................................. [2]
Mark scheme: 17 2 B1 for 3 terms correct 37 – 20 3 52 – 10 3 – 10 3 + 2 3 ×2 3 oe
Q18 · NOT TO F SCALE E A B C D C, E and F are points on a circle
18 NOT TO F SCALE E A B C D C, E and F are points on a circle. ABCD is a tangent to the circle at C. EB is parallel to FC. (a) Show that triangle CEB is similar to triangle FCE. ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ..................................................................................................................................................... [3] (b) CB = 6 cm, FE = 8 cm and EB = 5 cm. Calculate CE. ............................................ cm [2] Area of triangle CEB (c) Find . Area of triangle FCE ................................................. [1]
Mark scheme: 18(a) BEC =ECF Alternate angle 3 B2 for 2 correct pairs of angles or one BCE =EFC Alternate Segment correct pair of angles with reason therefore similar [AA or AAA] or B1 for one correct pair of angles 18(b) 20 2 6 5 oe M1 for = oe 3 8 EC 18(c) 36 1 oe 64
Question 19
19 Simplify. 4 a 2 + 4 ab - 15b 2 2a 2 + 2ac - 3ab - 3bc ................................................. [5]
Mark scheme: 19 2 a + 5b 5 B2 for (2a – 3b)(2a + 5b) oe final answer or M1 for a pair of brackets which give 2 a + c terms correct. B2 for (2a – 3b)(a + c) or M1 for 2a(a + c) – 3b(a + c) or a(2a – 3b) + c(2a – 3b)
What was in this paper
The subtopics covered by these 19 questions, and how many questions each got. Open one in a new tab to see every Cambridge question on it.
2Algebraic fractions1Algebraic manipulation1Angles1Estimation1Functions1Geometrical terms1Indices I1Inequalities1Non-right-angled triangles1Probability of combined events1Pythagoras’ theorem1Ratio and proportion1Relative and expected frequencies1Right-angled triangles1Similarity1Surds1Vectors in two dimensions1What you needed in this session
Cambridge’s own grade thresholds for 2025 Oct/Nov, Paper 2 · Variant 1. A higher threshold means an easier paper — the bar moves with how the cohort did.