Cambridge IGCSE Mathematics - International 0607 — 2025 Feb/March Paper 4 · Variant 2

1 This shape is made using 2 rectangles. 4 NOT TO 2x SCALE 3x 7 (a) Find the perimeter of the shape. Give your answer, in terms of x, in its simplest form. ................................................. [2] (b) Find the area of the shape. Give your answer, in terms of x, in its simplest form. ................................................. [2]

Mark scheme: Question Answer Marks Partial Marks 1(a) 6x + 14 or 2(3x + 7) Final answer 2 M1 for 3x + 4 + 3x + (7 – 4) + 7 oe or B1 for 6x + k (k ≠ 22) or kx + 14 (k ≠ 8) or correct answer spoilt 1(b) 15x 2 M1 for area of one correct rectangle e.g. 21x, 12x, 7x

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2 $Y is shared in the ratio 5 : 2. The difference between the two shares is $108. Work out the value of Y. Y = ................................................ [2]

Mark scheme: 2 252 2 108 5 − 2 M1 for soi by 36 or oe 5 − 2 7

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3 Work out 5 # . 6 27 Give your answer as a mixed number. ................................................. [1]

Mark scheme: 3 16 1 4 27

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4 a = e o b = e o 12 -1 1 (a) Work out a - b . 3 f p [2] (b) Find b . ................................................. [2]

Mark scheme: 4(a)  − 3  2 B1 for either component    5  4(b) 2 M1 for 52 + (–1)2 5.1[0] or 5.099… or answer 26

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5 Work out 25 - 7 .03 # 1.5 . ................................................. [1]

Mark scheme: 5 4.72 or 4.723… 1

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6 P NOT TO SCALE Q 113° R S P, Q, R and S are points on the circle. Find angle PQR. Angle PQR = ................................................ [1]

Mark scheme: 6 67 1

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7 Write 7 230 000 000 in standard form. ................................................. [1]

Mark scheme: 7 7.23 ×109 cao 1

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8 y 2 – 3 0 3 x – 2 1 f ( )x = x 2 + 1 (a) On the diagram, sketch the graph of y = f ( x) for values of x between -3 and 3. [2] (b) Solve f ( )x 1 x 2 + x - 1. ................................................................................ [4]

Mark scheme: 8(a) correct sketch 2 B1 for correct shape but: touching x axis or maximum not on y-axis or too high or too low or distinct curl-up at one or both ends or curvature incorrect on one side. 8(b) x < –1.73 or x < –1.725… 4 B2 for x < –1.73 or x < –1.725… x > 0.852 or x > 0.8524 to 0.8525 or B1 for –1.73 to –1.72 seen B2 for x > 0.852 or x > 0.8524 to 0.8525 or B1 for 0.85 or 0.852 to 0.853 seen If 0 scored SC1 for sketch of positive quadratic

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9 A class of students take a writing test and a speaking test. The scores are shown in the table. Writing score (x) 1 2 3 4 6 7 8 9 10 Speaking score (y) 4 6 3 8 6 7 7 8 9 (a) Complete the scatter diagram. The first 5 points have been plotted for you. y 10 9 8 7 6 Speaking score 5 4 3 2 1 0 x 1 2 3 4 5 6 7 8 9 10 Writing score [2] (b) Find the equation of the line of regression for y in terms of x. y = ................................................ [2]

Mark scheme: 9(a) 2 B1 for 2 correct plots Correct plots 9(b) [y =] 0.459x + 3.89 2 B1 for [y =] 0.459x + k or [y =] kx + 3.89 or [y =] 0.46x + 3.9

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10 The table shows the number of pens in each room in a school. Number of pens 1 2 3 4 5 6 Frequency 14 5 3 x 2 1 The mean number of pens is 2.3 . Find the value of x. x = ................................................ [3]

Mark scheme: 10 5 nfww 3 M2 for 49 + 4x = 2.3 × (25 + x) or M1 for numerator 1 × 14 + 2 × 5 + 3 × 3 + 4x + 5 × 2 + 6 × 1 or better or for denominator 14 + 5 + 3 + x + 2 + 1 or better

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11 Vikram invests $450 at a rate of 3.1% per year compound interest. Calculate the number of complete years that it takes for the value of Vikram’s investment to first be greater than $900. ................................................. [4]

Mark scheme: 11 23 4 B3 for 22.704… or 22.7 OR  3.1  900 M3 for n log 1 + = log oe    100  450 or a good sketch indicating value between 22 and 23 or correct trials reaching 22 and 23  3.1  n 900 or M2 for 1 + = oe    100  450 or suitable graph or at least three correct trials  3.1  n or M1 for 450 1 + = 900 oe soi    100  or at least 2 trials with n > 1

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12 A container is in the shape of a cuboid. The cuboid measures 41 cm by 32 cm by 25 cm. (a) Show that the capacity of the container is 32.8 litres. [2] (b) At 10 40 the container is empty. Water flows into the container at a rate of 2 litres per hour until the container is full. (i) At 12 10 there are x litres of water in the container. Find x as a percentage of 32.8 . ..............................................% [3] (ii) Find the time when the container is full. ................................................. [3]

Mark scheme: 12(a) 41  32  25 M2 M1 for 41 × 32 × 25 or for dividing by 1000 seen 1000 = 32.8 with no errors 12(b)(i) 9.15 or 9.146… 3 B1 for 1.5 oe or 3 seen 2  their1.5[ 100] M1 for 32.8 12(b)(ii) [0]3 04 3 32.8 M1 for 2 M1 FT for correct conversion of their length of time to hours and minutes 16 hours 24 min

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13 f ( )x = 4x - 2 g ( x) = ( x + 1 ) 2 (a) Find f ( 3 ) . ................................................. [1] (b) Find fg ( )x . Simplify your answer. ................................................. [2] (c) Find f -1 ( )x . f -1 ( )x = ................................................. [2] (d) Find ff -1 ( 5 ) . ................................................. [1]

Mark scheme: 13(a) 10 1 13(b) 4x2 + 8x + 2 or 2(2x2 + 4x + 1) 2 M1 for 4(x + 1)2 – 2 or better final answer 13(c) x + 2 2 y 2 oe final answer M1 for x = 4y – 2 or y + 2 = 4x or = x − 4 4 4 13(d) 5 cao 1

More questions on Functions

14 C B 75° NOT TO SCALE 10 cm 65° A D 14 cm Triangle BCD is isosceles. (a) Find the area of triangle ABD. ......................................... cm2 [2] (b) Find the shortest distance from D to AB. ............................................ cm [3] (c) Find the perimeter of ABCD. ............................................ cm [6]

Mark scheme: 14(a) 63.4 or 63.44… 2 1 M1 for  10  14  sin 65 2 14(b) 12.7 or 12.68 to 12.69 nfww 3 x M2 for sin65 = oe 14 1 or their area =  10  x 2 or M1 for recognising shortest distance 14(c) 44.2 or 44.18… to 44.23 6 2 2 M2 for  BD =  10 + 14 −2 10  14cos65 or M1 for 102 + 142 – 2 × 10 × 14 × cos65 their BD sin(theirBDC ) M2 for  BC =  oe sin75 BC their BD or M1 for = oe sin(theirBDC ) sin 75 OR M2 for (theirBD ) 2 + (theirBD ) 2 − 2(theirBD )(theirBD )cos30 or M1 for (theirBD)2 + (theirBD)2 – 2(theirBD)(theirBD)cos30 OR M2 for 2 × theirBD × cos 75 1 2 BC or M1 for cos75 = theirBD M1 for 10 + 14 + their BC + their CD dependent on at least trigonometry used for BC and CD

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15 A 4 cm B NOT TO 154° O SCALE C A and C are points on the circle, centre O. BA and BC are tangents to the circle. The radius of the circle is 4 cm. Angle AOC = 154°. (a) Write down the mathematical name of the quadrilateral ABCO. ................................................. [1] (b) Calculate the area of the shaded segment. ......................................... cm2 [3] (c) Calculate the shortest distance from B to the circumference of the circle. ............................................ cm [3]

Mark scheme: 15(a) kite 1 15(b) 18.[0] or 17.99… 3 154 2 1 M2 for π 4 − 4 4 sin154 oe 360 2 154 2 1 or M1 for π 4 or 4 4 sin154 oe 360 2 15(c) 13.8 or 13.78… 3 4 M2 for − 4 oe cos77 4 M1 for cos77 = oe or better x or M1 for their OB – 4

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16 Rupesh plays a game 3 times. He can either win each game or lose each game. The probability that Rupesh wins each game is 0.2 . Find the probability that Rupesh loses the 3 games. ................................................. [2]

Mark scheme: 16 0.512 2 n M1 for (1 − 0.2 ) oe n > 1

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17 Simplify. 2 `64a 6 b -3j3 ................................................. [3]

Mark scheme: 17 4 2 16a 4 3 B2 for two parts correct in final answer 16a b− or final answer 2 or B1 for one out of three parts correct in final b answer OR 1 2 1 2 12 6 3 M2 for 4a b− seen or 4096a b− ( ) ( ) or M1 for 4a 2 b−or1 4096a12 b−6 seen with no more than one error

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18 - = 1 2x - 5 x + 1 (a) Show that 2x 2 - x - 45 = 0 . [4] (b) Solve by factorising. 2x 2 - x - 45 = 0 x = .................. or x = .................. [3]

Mark scheme: 18(a) 10(x + 1) – 6(2x – 5) or better M1 2 x 2 − 5 x + 2 x − 5 M1 Correct method for clearing M1 fractions Leading to 2 x 2 − x − 45 = 0 A1 No errors or omissions 18(b) (2x + 9)(x – 5) [= 0] M2 M1 for (2x + a)(x + b) where ab = –45 or a + 2b = −1 or 2x(x – 5) + 9(x – 5) [=0] or x(2x + 9) – 5(2x + 9) [=0] [x =] 5 B1 [x =] –4.5 oe

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