Cambridge IGCSE Mathematics - International 0607 — 2025 May/June Paper 4 · Variant 1

1 150 students are each asked how many texts they sent the previous day. The results are shown in the table. Number of texts 0 1 2 3 4 5 6 Frequency 18 45 37 24 15 8 3 (a) Find (i) the mode ................................................. [1] (ii) the median ................................................. [1] (iii) the range ................................................. [1] (iv) the upper quartile. ................................................. [1] (b) One of the 150 students is selected at random. Find the probability that this student sent fewer than 3 texts. ................................................. [1]

Mark scheme: Question Answer Marks Partial Marks 1(a)(i) 1 1 1(a)(ii) 2 1 1(a)(iii) 6 1 1(a)(iv) 3 1 1(b) 2 1 oe 3

More questions on Averages and measures of spread

2 (a) Find 42 as a percentage of 60. ..............................................% [1] (b) Aisha’s rate of pay is $17.50 per hour. She receives a pay rise of 8%. Calculate Aisha’s new rate of pay per hour. $ ................................................. [2]

Mark scheme: 2(a) 70 1 2(b) 18.9[0] 2  8  M1 for 17.5  1 +  soi  100  or B1 for 1.4[0]

More questions on Percentages

3 (a) Expand. x 2 ( 3x - 2 ) ................................................. [1] (b) Expand and simplify. 2 ( 3x - 2 ) - 5 ( 1 - 4x) ................................................. [2]

Mark scheme: 3(a) 3 x 3 − 2 x 2 final answer 1 3(b) 26x – 9 final answer 2 B1 for 26x + k or kx – 9 or M1 for 6x – 4 – 5 + 20x

More questions on Algebraic manipulation

4 Use set notation to describe each of the shaded regions. U U Q P A B R .................................................. .................................................. [2]

Mark scheme: 4 A  B oe 2 B1 for each ( P  Q )  R oe

More questions on Sets

5 s = # t 2 Find s when u = 4.3, v = 9.5 and t = 8.5 . s = ................................................. [1]

Mark scheme: 5 58.65 1

More questions on Equations

6 8 cm NOT TO SCALE A 5.7 cm 72° B N 3.6 cm C Triangle ABC is the cross-section of a prism. AC = 5.7 cm, NC = 3.6 cm and the length of the prism is 8 cm. Angle ABN = 72° and angle ANC = 90°. (a) AN is the perpendicular height of the triangle. Show that AN = 4.42 cm correct to 3 significant figures. [2] (b) Calculate the volume of the prism. .......................................... cm3 [5]

Mark scheme: 6(a) 3.62 + x 2 = 5.7 2 or better M1 4.419… A1 6(b) 89[.0] or 89.1 or 89.01 to 89.11 5 4.42 M2 for oe tan72 4.42 or M1 for tan72 = oe BN 1 M2 for  ( their BN + 3.6 )  4.42  8 oe 2 1 or M1 for  ( their BN + 3.6 )  4.42 oe 2 If 0 scored, SC1 for answer 63.6 or 63.63 to 63.65

More questions on Surface area and volume

7 Simplify. 2x 3 y 4 # ( 4 x 2 y) 3 ................................................. [3]

Mark scheme: 7 128x 9 y 7 final answer 3 B2 for kx 9 y 7 or 128 kx y 7 or 128 x 9 y k or correct answer seen then spoilt or B1 for one factor correct or for 64x 6 y 3

More questions on Indices II

8 y NOT TO SCALE × A (– 4, 3) x 0 × B (5, –2) Point A is translated to point B. Find AB . f p [2]

Mark scheme: 8  9 2 B1 for each component    −9   −5  If 0 scored SC1 for    5 

More questions on Vectors in two dimensions

9 The value of a car depreciates exponentially by 15% each year. On 1 January 2023 its value was $20 400. (a) Find the value of the car on (i) 1 January 2022 $ ................................................. [2] (ii) 1 January 2025. $ ................................................. [2] (b) Find the year in which the value of the car on 1 January will first be less than $5000. ................................................. [4]

Mark scheme: 9(a)(i) 24 000 2  15  M1 for P ×  1 −  = 20 400 oe or better  100  9(a)(ii) 14 739 2 2  15  M1 for 20 400 × 1 − oe    100  9(b) 2032 4 B3 for 8.65 or 8.651 to 8.652 (20 400) or 9.65 or 9.651 to 9.652 (24 000) or 6.65 or 6.651 to 6.652 (14 739) OR  15   5000  M3 for n log  1 −  = log   oe  100   P  where P = 20 400 or their 24 000 or their 14 739 or good sketch indicating value between 8 and 9 (20 400) or 9 and 10 (their 24 000) or 6 and 7 (their 14 739) or correct trials reaching 8 and 9 (20 400) or 9 and 10 (their 24 000) or 6 and 7 (their 14 739)  15  n 5000 or M2 for  1 −  = oe, same values  100  P of P or suitable graph with n > 1 or at least 3 correct trials  15  n or M1 for P   1 −  = 5000 oe, same  100  values of P soi by at least 2 correct trials with n > 1

More questions on Exponential growth and decay

10 5 cm NOT TO SCALE 12 cm 15 cm The diagram shows a solid cylinder and a solid cone. The radius of the cylinder is 5 cm and its height is 12 cm. The height of the cone is 15 cm. The volume of the cylinder is equal to the volume of the cone. (a) Show that the volume of the cylinder is 300rcm 3. [1] (b) Find the total surface area of the cone. .......................................... cm2 [5]

Mark scheme: 10(a)  25  12 1 10(b) 599 or 599.1 to 599.4 5 B4 for 410.7 to 410.9 seen OR B2 for r = 7.745 to 7.746 1 2 or M1 for  π  r  15 = 300  π 3 M1 for 152 + ( theirr ) 2 M1 for π × (their r)×their l

More questions on Surface area and volume

11 y 20 x – 10 0 10 – 20 x 3 f ( x) = ( x + 2)( x - 3) (a) Sketch the graph of y = f ( x) for values of x between -10 and 10. [3] (b) Find the coordinates of the local minimum. ( ...................... , ...................... ) [2] (c) Write down the equations of the asymptotes to the graph of y = f ( x) that are parallel to the y-axis. ................................................. [2] (d) Solve f ( )x 2 x + 7 . ..................................................................................................................................................... [4]

Mark scheme: 11(a) Correct sketch 3 No gaps or overlaps B1 for either LH branch or RH branch correct shape (Ignore any joining with middle section for this mark) B1 for middle section correct - must pass through origin with no obvious max or min (ignore any joining with outer branches) 11(b) (5.36, 8.87) 2 B1 for each If 0 scored SC1 for 5.35 and 8.86 11(c) x = –2 2 B1 for each x = 3 11(d) –2 < x < –1.78 4 B1 for –1.78 seen 3 < x < 3.94 B1 for 3.94 seen B1 for each inequality If 0 scored SC1 for straight line with positive gradient, with positive y-intercept and cutting curve twice

More questions on Graphs of functions

12 B 68° NOT TO 140 m SCALE 260 m A C The diagram shows a triangular field. AB = 140 m and BC = 260 m. Angle ABC = 68°. (a) Show that AC = 244.8 m correct to 1 decimal place. [3] (b) Giselle walks directly from A to C. Calculate the distance Giselle is from A when she is closest to B. .............................................. m [5]

Mark scheme: 12(a) 2 2 M2 M1 for 140 2 + 260 2 −2 140  260  cos68 140 + 260 −2 140  260  cos68 244.80… A1 12(b) 24.3 to 24.4 5 M2 for complete explicit method for angle A or angle C or BN 260sin68 sin A = oe 244.8 140 2 + 244.82 − 260 2 or cos A = oe 2  140  244.8 140sin68 or sin C = oe 244.8 260 2 + 244.82 − 140 2 or cos C = oe 2  260  244.8 1  140  260  sin68 2 or BN = oe 1  244.8 2 or M1 for implicit method for angle A or angle C or BN sin A sin68 e.g. = oe 260 244.8 or 2602 = 1402 + 244.82 −2 140  244.8  cos A oe sin C sin68 or = oe 140 244.8 or 1402 = 2602 + 244.82 −2 260  244.8  cosC oe 1 1 or  BN  244.8 = 140  260  sin68 oe 2 2 AND M2 dep for 140 cos their A oe or 244.8 – 260 cos their C oe or 140 2 – theirBN 2 oe dependent on at least M1 above or M1 for clear indication of perpendicular from B to AC

More questions on Right-angled triangles

13 Pierre drives from town P to town Q and then to town R. The distance from P to Q is 112 km. The distance from Q to R is 87 km. The average speed from P to Q is x km/h. The average speed from Q to R is ( x + 10 ) km/h. The total time for the journey from P to Q to R is 3 hours. (a) Show that 3x 2 - 169 x - 1120 = 0 . [4] (b) (i) Solve 3x 2 - 169 x - 1120 = 0 . You must show your working. x = .................. or x = .................. [3] (ii) Find the average speed from Q to R. ......................................... km/h [1]

Mark scheme: 13(a) 112 87 M2 112 87 + = 3 oe B1 for or x x + 10 x x + 10 112(x + 10) + 87x = 3x(x + 10) oe M1 Clearing two algebraic fractions Completion to 3x2 – 169x – 1120 = 0 A1 with no errors or omissions 13(b)(i) Correct sketch of 3x2 – 169x – 1120 M1 With negative and positive x-intercept or ( −−)( 169)  ( −169) 2 − 4(3)( −1120) 2  3 62.3 or 62.32… B2 B1 for each –5.99 or –5.990… 13(b)(ii) 72.3 or 72.32… 1 FT 10 + their positive answer to (i)

More questions on Equations

14 Ahmed records the mass of each of 50 pumpkins. The results are shown in the table. Mass (m kg) 0 1 m G 2 2 1 m G 5 5 1 m G 10 10 1 m G 15 15 1 m G 25 Frequency 13 17 10 7 3 (a) Calculate an estimate of the mean. ............................................. kg [2] (b) Ahmed picks two of the 50 pumpkins at random. Find the probability that one of these pumpkins has a mass greater than 10 kg and the other pumpkin has a mass of not more than 2 kg. ................................................. [3]

Mark scheme: 14(a) 5.9 2 M1 for at least 3 correct midpoints soi 14(b) 26 3 10 13 oe M2 for [2  oe 245 ]50 49 10 13 13 10 or B1 for and oe or and oe 50 49 50 49 13 If 0 scored, SC1 for answer oe 125

More questions on Probability of combined events

15 NOT TO SCALE 25 cm These two bottles are mathematically similar. The capacity of the small bottle is 0.5 litres. The capacity of the large bottle is 1 litre. The height of the large bottle is 25 cm. Calculate the height of the small bottle. .............................................cm [3]

Mark scheme: 15 19.8 or 19.84… 3 3 M2 for 25  2 oe or M1 for 3 1  0.5 or 3 0.5  1 oe  h 3 0.5 or for = oe    25  1

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