Cambridge IGCSE Mathematics - International 0607 — 2025 May/June Paper 4 · Variant 3

1 Complete the statements. A rhombus has rotational symmetry of order ...................... A parallelogram has ...................... lines of symmetry. [2]

Mark scheme: Question Answer Marks Partial Marks 1 2 2 B1 for each 0

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2 Convert 0.027 kilometres into centimetres. ............................................ cm [1]

Mark scheme: 2 2700 1

3 Calculate. 3 2.7 1 - 3.5 Give your answer correct to 4 significant figures. ................................................. [2]

Mark scheme: 3 0.6114 cao 2 B1 for 0.611 or 0.61142...

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4 C NOT TO SCALE x° 122° E A B In the diagram, CA = CB and ABE is a straight line. Find the value of x. x = ................................................ [2]

Mark scheme: 4 64 2 B1 for angle A or angle B = 58

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5 Solve. 0 = 4x - 11 x = ................................................ [1]

Mark scheme: 5 11 1 oe 4

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6 NOT TO 6 cm SCALE w° 11 cm Calculate the value of w. w = ................................................ [2]

Mark scheme: 6 28.6 or 28.61... 2 6 M1 for tan = oe 11

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7 Omar records the time he spends revising and his test score for each of 8 tests. The maximum score on each test is 20. The table shows the results. Time revising 15 15 18 25 30 35 35 40 (x minutes) Test score ( y) 11 10 12 13 15 16 18 18 (a) Write down the type of correlation between the time Omar spent revising and his test score. ................................................. [1] (b) (i) Find the equation of the regression line for y in terms of x. y = ................................................ [2] (ii) Use your answer to part (b)(i) to estimate Omar’s test score when he spends 20 minutes revising. ................................................. [1]

Mark scheme: 7(a) positive 1 7(b)(i) [y =] 0.306x + 5.98 2 B1 for 0.306x + k or kx + 5.98 or 0.31x + 6.[0] 7(b)(ii) 12 or 12.09 to 12.10... 1 FT their (b)(i)

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8 y 10 x -3 0 3 -10 3 2 f ( )x = - x x (a) On the diagram, sketch the graph of y = f ( x) for values of x between -3 and 3. [3] (b) Find the zero of f ( )x . ................................................. [1] (c) Find the equation of the asymptote to the graph. ................................................. [1] (d) Find the coordinates of the local maximum point. ( ...................... , ...................... ) [2] (e) The equation f ( )x = k has one solution. Find the range of values of k. ................................................. [1]

Mark scheme: 8(a) Correct sketch 10101010 3 Both branches correct, without excessive gaps, curl backs, overlaps or feathering. 5555 Max point in 3rd quadrant and intersection with x-axis between 1 and 2. 3333 -2-2-2-2 -1-1-1-1 0000 0000 1111 2222 3333 B1 each branch correct shape (if joined B1 B1) -5-5-5-5 8(b) 1.44 or 1.442... 1 8(c) x = 0 1 8(d) (–1.14, –3.93) 2 B1 for each or for (–1.1, –3.9) 8(e) k > –3.93 1 FT their y-coordinate in part (d)

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9 The volume of a sphere is 96.5 cm3 . Calculate the radius of the sphere. ............................................ cm [2]

Mark scheme: 9 2.85 or 2.845... 2 4 M1 for 96.5  π or better 3

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10 The speed of each of 155 cars passing a school gate is recorded. The results are shown in the table. Speed (v km/h) 10 1 v G 20 20 1 v G 30 30 1 v G 35 35 1 v G 40 40 1 v G 60 Frequency 7 23 36 49 40 Calculate an estimate of the mean speed. ......................................... km/h [2]

Mark scheme: 10 36.7 or 36.69... 2 M1 for 3 correct mid values soi

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11 PQ = e o - 4 Find PQ . PQ = ................................................. [2]

Mark scheme: 11 6.4[0] or 6.403... or 41 2 M1 for 25 + ( −4 ) 2

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12 w ∝ x + 1 When x = 3, w = 8. Find x when w = 20. x = ................................................ [3]

Mark scheme: 12 24 3 B2 for w = 4 x + 1 20 8 or M2 for = x + 1 3 + 1 or M1 for 8 = k 3 + 1

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13 NOT TO SCALE 6 cm 15 cm A solid is formed from a cone and a hemisphere. The cone and the hemisphere each have a radius of 6 cm. The cone has a height of 15 cm. Find the volume of the solid. .......................................... cm3 [3]

Mark scheme: 13 1020 or 1017 to 1018. ... 3 1 2 M1 for 3π 6  15 or better 1 4 3 M1 for  π 6 or better 2 3

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14 Priya buys pieces of fruit. The pieces of fruit are apples and oranges. The price of 1 apple is x cents. The price of 1 orange is ( x + 10 ) cents. Priya spends 420 cents on apples. (a) Write down an expression, in terms of x, for the number of apples she buys. ................................................. [1] (b) Priya also spends 420 cents on oranges. She buys 13 pieces of fruit. (i) Show that 13x 2 - 710 x - 4200 = 0 . [3] (ii) Factorise 13x 2 - 710 x - 4200 . ................................................. [2] (iii) Find the cost of one orange. ........................................ cents [2]

Mark scheme: 14(a) 420 1 x 14(b)(i) 420 420 M1 + = 13 x x + 10 420( x + 10) + 420 x = 13x ( x + 10) M1 Correctly clearing 2 fractions with linear denominators Correctly expanding brackets and A1 No errors or omissions leading to 13 x 2 − 710 x − 4200 = 0 14(b)(ii) (13 x + 70 )( x − 60 ) final answer 2 B1 for (13 x + a )( x + b ) where ab = –4200 or a + 13b = –710 or for 13 x( x − 60) + 70( x − 60) or for x (13 x + 70) − 60(13 x + 70) 14(b)(iii) 70 2 FT their +ve x + 10 B1 for correct positive x from their factors

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15 NOT TO 75° SCALE y cm 40° 12 cm Calculate the value of y. y = ................................................ [4]

Mark scheme: 15 11.3 or 11.25 to 11.26 4 B1 for 65 seen 12sin their 65 M2 for sin75 sin75 sin their 65 or M1 for = oe 12 y

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16 (a) Use set notation to describe the shaded region. U A B C ................................................. [1] (b) Shade P + Q l. U P Q [1]

Mark scheme: 16(a) 1 ( A  B )  C oe 16(b) 1

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17 f ( )x = 3 - 2x g ( )x = 1 - 5x (a) Find f ( - 2 ) . ................................................. [1] (b) Find x when f ( )x = 6 . ................................................. [2] (c) Find fg ( )x . Give your answer in its simplest form. ................................................. [2] (d) Find g -1 ( )x . g -1 ( )x = ................................................. [2]

Mark scheme: 17(a) 7 1 17(b) 3 1 2 M1 for –2x = 6 – 3 oe or better – or − or –1.5 2 12 17(c) 1 + 10x final answer 2 M1 for 3 – 2(1 – 5x) 17(d) 1 −x 2 y 1 oe final answer M1 for x = 1 – 5y or 5x = 1 – y or = − x 5 5 5 or y – 1 = –5x or better

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18 (a) Bruce buys a new car on 1 January 2022. On 1 January 2023 the value of the car has decreased by 20%. On 1 January 2024 the value of the car has decreased by 15% of its value on 1 January 2023. Find the overall percentage decrease in the value of the car on 1 January 2024. ..............................................% [2] (b) Sangita buys a car with a value of $20 000. The value of the car decreases exponentially at a rate of 5% per year. Calculate the number of complete years it will take for the value of the car to decrease from $20 000 to $8000. ................................................. [4] (c) Sunil buys a car. The value of the car decreases exponentially at a rate of 6% per year. At the end of 9 years the value of the car is $8022. Calculate the original value of the car. $ ................................................. [2]

Mark scheme: 18(a) 32 2  20  15  M1 for  1 −  1 −   100  100  18(b) 18 4 B3 for 17.9 or 17.86... OR  5   8000  M3 for n log  1 −  = log   oe  100   20000  or good sketch indicating value between 17 and 18 or correct trials reaching 17 and 18  5  n 8000 or M2 for 1 − = oe    100  20000 or suitable graph with n > 1 or at least 3 correct trials, n > 1  5  n or M1 for 20000  1 − = 8000 oe soi by    100  at least 2 trials with n > 1 18(c) 14 000 2 9  6  M1 for [...] ×  1 −  = 8022 oe  100 

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19 NOT TO SCALE The two solids are mathematically similar. The volume of the large solid is 416 cm3 . The volume of the small solid is 52 cm3 . The total surface area of the small solid is 60 cm2 . Calculate the total surface area of the large solid. .......................................... cm2 [3]

Mark scheme: 19 240 3 2  416  M2 for 60  3  oe    52  416 52 or M1 for 3 or 3 oe or better or for 52 416  60 3  52  2 = oe      A   416 

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20 Simplify. 3x - 5 3 xy - 3x - 5y + 5 ................................................. [3]

Mark scheme: 20 1 −1 3 B2 for ( y − 1)(3x − 5) or (1 − y )( 5 − 3 x ) or final answer y − 1 1 −y or B1 for 3 x ( y − 1) − 5( y − 1) or y (3 x − 5) − [1](3 x − 5)

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21 A is the point (-2, 8) and B is the point (1, 9). The perpendicular bisector of the line AB meets the x-axis at the point P. Find the coordinates of the point P. ( ...................... , ...................... ) [6]

Mark scheme: 21  1  6 B5 for −3x + 7 = 0 2 oe  , 0  or −8.5 = −3 ( x + 0.5 ) oe 3   OR B1 for (–0.5, 8.5) seen 9 − 8 M1 for gradient of AB = = m1 1 −−2 1 M1 for gradient of perp = −their m1 (m2) M1 for substituting their midpoint into y = their m2 x + c M1 dep for their ( −3 x + 7 ) = 0 dep on third M1 If 0 scored, SC1 for answer (k, 0)

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