Cambridge IGCSE Mathematics - International 0607 — 2025 May/June Paper 4 · Variant 2

1 Write the number 13 205.172 68 (a) correct to 3 significant figures ................................................. [1] (b) correct to the nearest 10. ................................................. [1]

Mark scheme: Question Answer Marks Partial Marks 1(a) 13 200 cao 1 1(b) 13 210 cao 1

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2 Solve. (a) 4x - 3 = 15 x = ................................................ [2] (b) 1 - 2x = 11 - 4 x x = ................................................ [2]

Mark scheme: 2(a) 1 9 2 3 15 4.5 or 4 or M1 for 4 x = 15 + 3 or x − = oe 2 2 4 4 2(b) 5 2 M1 for 4 x − 2 x = 11 − 1 oe or better

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3 Work out. ( 1 .7 ) 2 - 8 .5 ................................................. [1]

Mark scheme: 3 –0.0255 or –0.025 48 to –0.025 47 1

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4 Work out. 7 1 - 12 3 ................................................. [1]

Mark scheme: 4 1 1 or 0.25 4

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5 The exchange rate between pounds (£) and Chinese yuan (¥) is £1 = ¥ 8.3 . Change £549 to Chinese yuan. ¥ ................................................. [1]

Mark scheme: 5 4556.7[0] final answer 1

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6 A = ( x + y) h 2 (a) Find the value of A when x = 7, y = 12.5 and h = 6. A = ................................................ [1] (b) Rearrange the formula to write x in terms of A, y and h. x = ................................................ [2]

Mark scheme: 6(a) 58.5 1 6(b) 2 A 2 A − hy 2 A 1 1 − y or oe M1 for = ( x + y ) or A  = h ( x + y ) h h h 2 2 1  1 1  or A = ( xh + yh ) or A =  x + y  h or better 2  2 2 

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7 22 cm NOT TO SCALE x cm 5 cm x cm The diagram shows a wooden cuboid with a cylinder removed. The cuboid has length 22 cm, width x cm and height x cm. The cylinder has length 22 cm and diameter 5 cm. The volume of wood remaining after the cylinder has been removed is 976 cm3. Find the value of x. x = ................................................ [3]

Mark scheme: 7 8 [.00] or 7.999 to 8.000... 3 2 976 + 22  π  2.5 2 M2 for  x =  oe   22 or M1 for 976 = 22  x 2 − 22  π  2.5 2 oe If 0 scored SC1 for answer 11.1 or 11.08 to 11.09

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8 Xiong flies by plane from Manchester to Hong Kong. The distance from Manchester to Hong Kong is 9000 km. The flight takes 12.5 hours. (a) Calculate the average speed of the plane. ......................................... km/h [1] (b) The plane leaves Manchester at 20 45. The local time in Hong Kong is 8 hours ahead of the local time in Manchester. Find the local time in Hong Kong when the plane lands. ................................................. [2]

Mark scheme: 8(a) 720 1 8(b) 17 15 2 B1 for 41 15 or [0]9 15 or [0]4 45 seen

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9 y 9 8 7 B A 6 5 4 3 2 1 – 9 – 8 – 7 – 6 – 5 – 4 – 3 – 2 – 1 0 1 2 3 4 5 6 7 8 9 x – 1 – 2 – 3 – 4 – 5 – 6 – 7 – 8 – 9 0 (a) Translate triangle A with vector e o. Label the image C. [2] - 8 (b) Describe fully the single transformation that maps triangle A onto triangle B. ..................................................................................................................................................... ..................................................................................................................................................... [2] (c) Rotate triangle B through 90° clockwise with centre of rotation (-6, 2). Label the image D. [2]

Mark scheme: 9(a) Correct triangle 2 0  k  M1 for translation or (2, 0), (2, –3), (4, –3)    k  −8  9(b) Reflection 2 B1 for each x = –2 oe 9(c) Correct triangle 2 M1 for rotation 90 clockwise, incorrect centre (–3, 2), (–3, 4), (0, 2)

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10 The table shows the marks of each of 10 students in a physics exam and in a chemistry exam. Physics mark (x) 9 21 33 41 55 68 75 83 89 96 Chemistry mark (y) 31 46 42 50 50 61 69 68 72 90 (a) Find the mean physics mark. ................................................. [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 the chemistry mark when the physics mark is 46. ................................................. [1] (c) Two of the students who scored more than 35 in the physics exam are chosen at random. Find the probability that they both scored more than 65 in the chemistry exam. ................................................. [3]

Mark scheme: 10(a) 57 1 10(b)(i) y = 0.548 x + 26.6 2 B1 for y = kx + 26.6 or y = 0.548x + k or y = 0.55x + 27 10(b)(ii) 52 or 51.8 to 51.9 1 FT their (b)(i) 10(c) 2 3 4 3 oe M2 for  7 7 6 4 or M1 for seen 7 x x − 1 or  where x < y and y < 10 y y − 1

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11 y 10 – 3 0 3 x – 10 4x - 3 f ( )x = 2x + 1 (a) (i) Sketch the graph of y = f ( )x for values of x between -3 and 3. [3] (ii) Write down the equations of the 2 asymptotes of the graph of y = f ( )x . ...................................... , ...................................... [2] (iii) Write down the coordinates of the points where the graph crosses the axes. ( ...................... , ...................... ) ( ...................... , ...................... ) [2] (b) Solve the inequality. 4 x - 3 2 2 - x 2 x + 1 ............................................................................... [4]

Mark scheme: 11(a)(i) Correct sketch 3 Both branches correct; without branches joined, 10101010 excessive feathering, curl backs, overlaps or gaps. RH branch cutting axes correctly B1 for LH branch correct shape 5555 B1 for RH branch correct shape 0000 3333 -2-2-2-2 -1-1-1-1 0000 1111 2222 3333 -5-5-5-5 -10-10-10-10 11(a)(ii) y = 2 2 B1 for each x = –0.5 oe 11(a)(iii) (0, –3) 2 B1 for each (0.75, 0) oe 11(b) −1.85  x −0.5 4 B1 for –1.85 x  1.35 B1 for 1.35 B1 for their −1.85  x −0.5 B1 for x  their1.35 If 0 scored, SC1 for ruled line with positive y- intercept, negative gradient and intersecting both branches OR SC1 for 2 x 2 + x − 5 oe

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12 Paula bought a house on 1 January 2023. On 1 January 2024 the value of the house increased by 10%. On 1 January 2025 the value of the house increased by 6% of its value on 1 January 2024. The value of the house on 1 January 2025 was $215 710. (a) Calculate the amount Paula paid for the house in 2023. $ ................................................. [3] (b) From 2025 the value of the house increases exponentially at the rate of 6% each year. The value of the house on 1 January 2025 was $215 710. Find the year in which the value of the house on 1 January will first be greater than $400 000. ................................................. [4]

Mark scheme: 12(a) 185 000 3 215710 M2 for oe or better (1 + 0.1)(1 + 0.06) or M1 for A (1 + 0.1) = 215710 or A (1 + 0.06) = 215710 oe 12(b) 2036 4 B3 for 11 or 10.59 to 10.6 OR 400000 log 215710  400000  M3 for [ n = ] or log1.06   log1.06  215710  or suitable graph showing correct solution or correct trials as far as 10 and 11 n 400000 or M2 for 1.06 = oe 215710 or graph of y = 1.06 n and y = k or at least 3 correct trials or M1 for 215710  1.06 n = 400000 oe or graph of y = 1.06 n

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13 Use a graphical method to solve 5 # 0. 8 x = 2x + 7 . Give your answer correct to 2 decimal places. x = ................................................ [3]

Mark scheme: 13 Correct sketch e.g. M2 M1 for exponential graph –0.63 B1

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14 y varies inversely as the square root of ( x - 1 ) . y = 1 when x = 5. (a) Find y in terms of x. y = ................................................ [2] (b) w varies directly as y2. w = 45 when y = 3. Use your answer to part (a) to find w in terms of x. Give your answer in the form w = p ( x - 1 ) q , where p and q are constants. w = ................................................ [4]

Mark scheme: 14(a) 2 4 2 k y = or oe B1 for y = ( x − 1) x − 1 ( x − 1) 14(b) w = 20( x − 1)−1 4  2  2 B3 for w = 5   or better   ( x − 1)   OR B1 for w = 5 y 2  2  2 M1 for w = their 5  their     ( x − 1)  

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15 B 30° NOT TO SCALE 12.4 cm A C The area of triangle ABC is 74.4 cm2. AB = 12.4 cm and angle ABC = 30°. (a) Show that BC = 24 cm. [2] (b) Find AC. AC = ........................................... cm [3] (c) Find obtuse angle CAB. Angle CAB = ................................................ [3] (d) NOT TO Y SCALE X Z Triangle XYZ is similar to triangle ABC. The area of triangle XYZ is 62 cm2. Find YZ. YZ = ........................................... cm [3]

Mark scheme: 15(a) 74.4 = 0.5  12.4  BC  sin30 oe M1 Use of 24 scores M0 74.4 A1 74.4 [=24] oe oe e.g. 0.5  12.4  sin30 3.1 15(b) 14.6 or 14.63 to 14.64 3 2 2 M2 for 12.4 + 24 −2 12.4  24  cos30 or M1 for [ AC 2 ] = 12.4 2 + 24 2 −2 12.4  24  cos30 15(c) 124.7 to 125.3 3 their 14.6 must come from trig 12.4 2 + (their14.6) 2 − 24 2 M2 for [cos A = ] 2  12.4  their14.6 or M1 for 24 2 = (their14.6) 2 + 12.4 2 −2 their14.6  12.4cos A OR 24  sin30 M2 for [sin A] = oe their 14.6 sin A sin30 or M1 for = oe 24 their 14.6 OR 74.4 M2 for [sin A] = 1  12.4  their 14.6 2 1 or M1 for  12.4  their 14.6sin A = 74.4 2 15(d) 21.9 or 21.90 to 21.92 or 4 30 oe 3 62 74.4 M2 for 24  or 24  74.4 62 62 74.4  YZ 2 62 oe or M1 for or or for  = 74.4 62  24  74.4

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16 In this question, all lengths are in centimetres. B C NOT TO SCALE A D E The diagram shows a quadrilateral ABCD. ABCE is a rhombus and CDE is a right-angled triangle. AC = ( 2 3 ) x and BE = 2x . (a) Show that AE = 2x . [2] (b) Find, in terms of x, the perimeter of ABCD. ................................................. [4]

Mark scheme: 16(a) M1 Use of AE = 2x scores M0 2 2 2 x  3 x oe oe e.g. tan EAM = ( M is midpoint of AC) )  AE =  x + ( x 3 2 2 A1 x [AE = ] x + 3x = 2x oe e.g. [ AE ] = = 2x 2 sin30 or 4x = 2x oe 16(b) 7 x + 3 x oe final answer 4 B3 for answer kx + x 3 or 7x + k x 3 OR B3 for CD = x 3 and for ED = x or B2 for CD = x 3 or for ED = x CD or M1 for e.g. = sin60 oe 2 x

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