Cambridge IGCSE Mathematics - International 0607 — 2025 Oct/Nov Paper 4 · Variant 2

0607/42/O/N/25 · 17 questions · 75 marks · ≈84 min

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Mark scheme9 pages

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Questions as text

Q1 · By writing each number correct to 1 significant figure, find an estimate for 44.9 2

1 (a) By writing each number correct to 1 significant figure, find an estimate for 44.9 2. 092 - . 140 ( 3. 7 - 1. 85) You must show your working. ................................................. [2] (b) Calculate. 44 .9 2. 092 - 140 ( 3. 7 - 1 .85 ) ................................................. [1]

Mark scheme: Question Answer Marks Partial Marks 1(a) 40 2 2 M1 Allow one rounding error. − 100 ( 4 − 2 ) 2 A1 From correct working 1(b) 1.43 or 1.433 to 1.434 1

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Q2 · The table shows the marks scored by 100 students in a test

2 The table shows the marks scored by 100 students in a test. Mark 1–10 11–20 21–40 41–60 61–80 81–90 91–100 Number of students 2 8 17 21 14 25 13 (a) One of these students is chosen at random. Find the probability that this student scored more than 90 marks. ................................................. [1] (b) Write down the group that contains the median. ................................................. [1] (c) Calculate an estimate for the mean. ................................................. [2]

Mark scheme: 2(a) 0.13 oe 1 2(b) 61 - 80 1 2(c) 60.8 2 M1 for 5 midpoints soi If 0 scored SC1 for 60.3

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Question 3

3 Simplify. 3x - 4y + x - 5y ................................................. [2]

Mark scheme: 3 4 x − 9 y final answer 2 M1 for 4 x − ky or kx − 9 y

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Q4 · The interior angles of a quadrilateral are 67°, 112°, x° and (x - 7 )°

4 The interior angles of a quadrilateral are 67°, 112°, x° and (x - 7 )° . Find the value of x. x = ................................................ [2]

Mark scheme: 4 94 2 M1 for 360 = 67 + 112 + x + x − 7 oe

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Q5 · Calculate the area of a circle which has a diameter of 6.4 cm

5 Calculate the area of a circle which has a diameter of 6.4 cm. .......................................... cm2 [2]

Mark scheme: 5 32.2 or 32.16 to 32.17… 2 M1 for π  3.2 2 oe

Q6 · Share 630 in the ratio 5 : 7

6 Share 630 in the ratio 5 : 7. ........................ , ....................... [2]

Mark scheme: 6 262.5, 367.5 final answer 2 630 M1 for oe 5 + 7

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Question 7

7 Factorise. 3ax + 4 by - 3 ay - 4 bx ................................................. [2]

Mark scheme: 7 ( x − y )(3a − 4b) final answer 2 M1 for 3a ( x − y ) − 4b( x − y ) or x (3a − 4b) − y (3a − 4b)

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Q8 · Y 4 0 x – 4 4 – 4 1 (a) f ( x) = - 1 ( 2x - 1)( x + 1) (i) On the diagram, sketch the…

8 y 4 0 x – 4 4 – 4 1 (a) f ( x) = - 1 ( 2x - 1)( x + 1) (i) On the diagram, sketch the graph of y = f ( x) for values of x between -4 and 4. [3] (ii) Write down the x-intercepts. ................................................. [2] (iii) Write down the equations of the asymptotes parallel to the y-axis. ............................................................................... [2] (b) g ( x) = 0. 5 ( x + 1) On the diagram, sketch the graph of y = g ( x) for values of x between -4 and 4. [1] (c) Solve the inequality f ( x) H g ( x) . ............................................................................................................................................................. [3]

Mark scheme: 8(a)(i) Correct sketch 3 B1 for correct outer branches both crossing x-axis B1 for middle branch in correct position B1 graph in 3 sections with no excessive overlaps (except if penalised already in second B1) or gaps or curlback If sketch not correct max of 2 marks 8(a)(ii) –1.28 –1.281 to –1.280 2 B1 for each 0.781 0.7807 to 0.7808 or for –1.3 and 0.78 8(a)(iii) x = –1 2 B1 for each x = 0.5 oe 8(b) Correct line 1 8(c) x − 2.84 –2.837 to –2.836 3 B1 for each, strict inequality on the −1.33  x −1 –1.327… asymptote values, only penalised once. If 0 scored SC1 for 2 correct intersections 0.5 x 0.664 0.6640… –2.8…, –1.3…, 0.66… seen in an inequality

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Q9 · A bag contains 5 black balls and 7 white balls

9 A bag contains 5 black balls and 7 white balls. One ball is chosen at random and not replaced. A second ball is then chosen at random. Find the probability that both balls are black. ................................................. [2]

Mark scheme: 9 5 2  5 4  oe M1 for    33  12 11 

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Q10 · The table shows the marks of 10 students in a French test and in a Spanish test

10 The table shows the marks of 10 students in a French test and in a Spanish test. French mark (x) 27 32 36 44 57 65 78 86 89 93 Spanish mark (y) 28 12 40 47 59 68 75 82 83 89 (a) Find the equation of the regression line for y in terms of x. y = ................................................ [2] (b) Use your equation to estimate the Spanish mark when the French mark is 5. ................................................. [1] (c) Is your answer to part (b) likely to be a reliable estimate of the Spanish mark? Give a reason for your answer. ............................................................................................................................................................ ..................................................................................................................................................... [1]

Mark scheme: 10(a) y = –1.81 + 0.99[0] x 2 B1 for y = –1.81 + kx or y = k + 0.99[0] x or y = –1.8 + 0.99 x 10(b) 3.1 or 3.14… or 3 1 FT their (a) must be > 0 10(c) No, outside range of data 1

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Q11 · B NOT TO SCALE 8 cm A C Angle ABC = 90° and angle BAC = 65°

11 B NOT TO SCALE 8 cm A C Angle ABC = 90° and angle BAC = 65°. Calculate BC. BC = ............................................ cm [2]

Mark scheme: 11 17.2 or 17.15 to 17.16 2 BC M1 for tan65 = oe 8

Question 12

12 Solve. 2x - 5 5 1 - 4x - = 3 6 2 x = ................................................ [4]

Mark scheme: 12 9 1 4 M1 for correctly eliminating fractions or 1 or 1.125 M1 for correctly expanding their brackets 8 8 M1 for correctly collecting their terms into ax = b

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Question 13

13 Factorise. 3x ( 2x - y) 2 + 4 ( 2x - y) 3 ................................................. [3]

Mark scheme: 13 (2 x − y ) 2 (11x − 4 y ) final answer 3 M2 for (2 x − y ) 2 (3 x + 8 x − 4 y ) or M1 for (2 x − y ) 2 (3 x + 4(2 x − y ))

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Q14 · B 47° NOT TO SCALE D 82° 12° 12.3 cm 9.8 cm A C ABC and ADC are triangles

14 B 47° NOT TO SCALE D 82° 12° 12.3 cm 9.8 cm A C ABC and ADC are triangles. AD = 12.3 cm and CD = 9.8 cm. Angle ABC = 47°, angle ADC = 82° and angle BAD = 12°. (a) Find AC. AC = ........................................... cm [3] (b) Show that angle DAC = 41.6°, correct to 1 decimal place. [3] (c) BC = 16.1 cm. Find the area of the shaded quadrilateral ABCD. .......................................... cm2 [4]

Mark scheme: 14(a) 14.6 or 14.62…nfww 3 M2 for AC = 12.32 + 9.82 −2 12.3  9.8  cos82 or M1 for AC 2 = 12.32 + 9.82 −2 12.3  9.8  cos82 Use of angle DAC = 41.6o is 0 marks unless fully correct method to derive it is shown in this part 14(b) 9.8  sin82 M2 sin DAC sin82 sin DAC = oe M1 for = oe their 14.6 9.8 their14.6 41.58 to 41.59 A1 No errors seen 14(c) 55.8 to 56.1 4 B1 for 79.4… M1 for 0.5  their 14.6 16.1 their sin79.4 oe M1 for 0.5 12.3  9.8  sin82 oe

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Question 15

15 Erik invests $x. (a) He receives compound interest at a rate of 7.5% each year. (i) At the end of 5 years, the value of Erik’s investment is $11 485. Show that x = 8000 correct to the nearest dollar. [3] (ii) Find the number of complete years it takes for the total value of his investment of $8000 to be first greater than $16 000. ................................................. [4] (b) The compound interest rate of 7.5% each year is equivalent to a compound interest rate of y % each month. Find the value of y. y = ................................................ [3]

Mark scheme: 15(a)(i) 11485 M2 M1 for x  1.0755 = 11485 oe x = oe 1.0755 7999.9[…] A1 15(a)(ii) 10 nfww 4 B3 for answers which round to 9.58 OR log2 M3 for [ n = ]log1.075 oe or suitable graph showing correct soln or correct trials as far as 9 and 10 n 16000 or M2 for 1.075 = oe 8000 or graph of y = 1.075 n and y = k or at least 3 correct trials (n > 5) or M1 for 8000  1.075n = 16000 oe or graph of y = 1.075 n or at least 2 correct trials (n > 5) 15(b) 0.604 or 0.6044 to 0.6045 3 M2 for 121.075 oe or M1 for [...]12 = 1.075 oe

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Q16 · A spinner has 24 equal sections

16 A spinner has 24 equal sections. Each section is coloured red or blue. There are x red sections and all other sections are blue. The spinner is equally likely to land on any of its sections. The spinner is spun twice and the colour that the spinner lands on each time is recorded. The probability that the spinner lands on blue exactly once is 4. 9 Find the possible values of x. x = .................. or x = .................. [5] Questions 17 is printed on the next page.

Mark scheme: 16 8, 16 final answer 5 x 24 − x 4 M2 for  2   = oe 24 24 9 24 −x or B1 for seen 24 M2 for x 2 − 24 x + 128 = 0 oe 4  24  24 or M1 for x (24 − x ) = oe  2   9 If 0 scored SC1 for a single final answer of either 8 or 16

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Q17 · A solid cone has base radius r and vertical height 3r

17 A solid cone has base radius r and vertical height 3r. The total surface area of the cone is 209.22 cm2. (a) Find r. r = ........................................... cm [4] (b) A mathematically similar cone has a total surface area of 1882.98 cm2. Find the radius of this cone. ............................................ cm [3]

Mark scheme: 17(a) 4 4 209.22 M3 for  r =  oe π(1 + 10) or M2 for 209.22 = πr 2 + πr r 2 + (3r ) 2 oe or M1 for l 2 = r 2 + (3r ) 2 oe 17(b) 12 3 1882.98 M2 for x =  their r oe 209.22 1882.98  x  2 or M1 for = oe   209.22  their r  1882.98 209.22 or oe or oe 209.22 1882.98

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Cambridge’s own grade thresholds for 2025 Oct/Nov, Paper 4 · Variant 2. A higher threshold means an easier paper — the bar moves with how the cohort did.

A53/75
B39/75
C26/75
D21/75
E16/75