Cambridge IGCSE Mathematics (US) 0444 — 2021 May/June Paper 4 · Variant 1

1 (a) The total cost of a taxi journey is calculated as • $0.50 per kilometer plus • $0.40 per minute. (i) Calculate the total cost of a journey of 32 km that takes 30 minutes. $ ................................................. [2] (ii) The total cost of a journey of 100 km is $98. Show that the time taken is 2 hours. [3] (b) Three taxi drivers travel a total of 8190 km in the ratio 5 : 2 : 7. Calculate the distance each driver travels. Driver 1 .......................................... km Driver 2 .......................................... km Driver 3 .......................................... km [3] (c) After midnight, the cost of any taxi journey increases by 45%. One journey costs $84.10 after midnight. Calculate the cost of the same journey before midnight. $ ................................................. [2]

Mark scheme: Question Answer Marks Partial Marks 1(a)(i) 28 2 M1 for 32 × 0.50 + 30 × 0.40 1(a)(ii) 98 – 100 × 0.5 M3 M1 for 100 × 0.50 + x × 0.40 = 98 48 ÷ 0.4 = 120 [minutes] = 2 [hrs] M1 for 50 + 0.4 x = 98 or 0.4 x = 48 48 M1 for x = x = 120 [min] = 2 [hr] 0.4 OR M1 for 100 × 0.5 [ = 50 ] M1 for 98 − 50 [ = 48] M1 for 48 ÷ 0.4 = 120 [ min ] = 2 [ hr ] 1(b) 2925 3 B2 for one correct answer 1170 or M1 for 8190 ÷ (5 + 2 + 7) 4095 1(c) 58 2  45  M1 for  1 +  k = 84.1 oe  100 

2 (a) NOT TO SCALE 6.3 cm R cm 2.4 cm The diagram shows a solid cone and a solid hemisphere. The cone has radius 2.4 cm and slant height 6.3 cm. The hemisphere has radius R cm. The total surface area of the cone is equal to the total surface area of the hemisphere. Calculate the value of R. R = ................................................ [4] (b) NOT TO SCALE 16 cm 12 cm 7.6 cm 7.6 cm The diagram shows a solid cone with radius 7.6 cm and height 16 cm. A cut is made parallel to the base of the cone and the top section is removed. The remaining solid has height 12 cm, as shown in the diagram. Calculate the volume of the remaining solid. .......................................... cm3 [4]

Mark scheme: 2(a) 9 1 0.18 or 50 2(b) 1000 M1 1944 × 3600 × 3600 9 ÷ 0.15 = 60 M1 2(c) 1 ruled line to axis with point of contact at 240 240 2(d) 6.9375 4 1 M2 for area = × (130 + 240 ) × 9 oe 2 or M1 for one correct partial area M1dep for their total area ÷ 240

3 (a) The exchange rate is 1 euro = $1.142 . (i) Johann changes $500 into euros. Calculate the number of euros Johann receives. Give your answer correct to the nearest euro. ........................................ euros [2] (ii) Johann buys a computer for $329. The same computer costs 275 euros. Calculate the difference in cost in dollars. $ ................................................. [2] 3 (b) Lucy spends of the money she has saved this month on a book that costs $5.25 . 8 Calculate how much money Lucy has saved this month. $ ................................................. [2] (c) Kamal invests $6130 at a rate of r% per year compound interest. The value of his investment at the end of 5 years is $6669. Calculate the value of r. r = ................................................ [3]

Mark scheme: 3(a) 2.64 or 2.638… 4 2 π × 2.4 2 + π × 2.4 × 6.3 M3 for [ R = ] oe π + 2π or M2 for 2 2 1 2 π× 2.4 + π× 2.4 × 6.3 = πR + × 4πR 2 or M1 for [ π× 2.4 2 ] + π× 2.4 × 6.3 oe 2 1 2 or [ πR ] + × 4πR oe 2 3(b) 953 or 952.6 to 952.8 4   16 − 12  3  1 2 M3 for × π × 7.6 × 16 ×  1 −    3   16     1 2 1 2 or × π× 7.6 × 16 − × π× 1.9 × (16 − 12 ) 3 3 OR  16 − 12  3 B1 for top radius = 1.9 or   oe  16  M2 for 1 2 1 2 × π× 7.6 × 16 − × π× (their 1.9) × (16 − 12 ) 3 3   16 − 12  3  1 2 or × π × 7.6 × 16 ×  1 − their    3   16     1 2 or M1 for × π× 7.6 × 16 3 1 2 or for × π× (their 1.9) × (16 − 12 ) 3

4 (a) (i) Write 0.00307 in scientific notation. ................................................. [1] (ii) Work out 7.8 # 10 200 + 7 .8 # 10 201 , giving your answer in scientific notation. ................................................. [2] (b) Find the least common multiple (LCM) of 48 and 90. ................................................. [2] (c) Expand and simplify. ( a 3 + 3 2) 2 ................................................. [3] (d) Simplify 175 + 700 . You must show all your work. ................................................. [2]

Mark scheme: 4(a)(i) 438 cao 2 500 M1 for 1.142 4(a)(ii) 14.95 2 M1 for [329 –] 275 × 1.142 oe 4(b) 14 2 8 M1 for 5.25 × oe 3 4(c) 1.7[0] or 1.699… 3 6669 M2 for 5 6130 or M1 for 6669 = 6130 ( k ) 5

5 (a) x – 1 5 (i) Write down the inequality shown by this number line. ................................................. [2] (ii) Find the integers that satisfy this inequality. ................................................. [2] (b) Solve the inequality. 3x - 2 2 2 x 4 ................................................. [2] (c) y 6 5 4 3 2 1 0 x 0 1 2 3 4 5 6 On the grid, find and label the region, R, bounded by the following inequalities. x G 5 y H 1 y G x x + y H 5 [5]

Mark scheme: 5(a) 13.5 or 13.47… 4 B1 for angle 102 seen M2 for 10.6 2 + 6.4 2 − 2 × 10.6 × 6.4 × cos (180 − 78 ) OR M1 for 10.6 2 + 6.4 2 − 2 × 10.6 × 6.4 × cos (180 − 78 ) A1 for 181.5… 5(b) 8.68 or 8.682 to 8.683 nfww 4 B1 for angle = 44 10.6 M2 for sin(180 – 58 – 78) × oe sin 58 sin(180 − 58 − 78) sin58 or M1 for = oe x 10.6 5(c) 78.2 or 78.17 to 78.19… 3 1 M2 for × 10.6 × ( 6.4 + their 8.68 ) × sin ( 78 ) 2 oe OR 1 M1 for × 10.6 × 6.4 × sin(180 – 78) oe 2 1 M1 for × 10.6 × their 8.68 × sin78 oe 2

6 A NOT TO SCALE 10.6 cm 58° 78° B C X 6.4 cm The diagram shows triangle ABC. X is a point on BC. AX = 10.6 cm, XC = 6.4 cm, angle ABC = 58°, and angle AXB = 78°. (a) Calculate AC. AC = ........................................... cm [4] (b) Calculate BX. BX = ........................................... cm [4] (c) Calculate the area of triangle ABC. .......................................... cm2 [3]

Mark scheme: 6(a) 1 6(b) 2 B1 for 2 or 3 correct elements or M1 for 34 – x , x and 39 – x correctly placed 28 11 on diagram and x = 28 6 5 6(c)(i) 8 1 6(c)(ii) 11 1 6(c)(iii) 2 1 6(c)(iv) C ∩ S ∩ B′ oe 1 6(c)(v) 19 1 oe 30 6(c)(vi) 2 3 4 3 oe M2 for × 57 19 18 4 or M1 for seen 19 6(c)(vii) Equal numbers 15 1 15 or equal probability oe 30

7 (a) Factor. 3x - 1 - y + 3xy ................................................. [2] (b) Simplify. x 2 - 25 x 2 - x - 20 ................................................. [3] (c) Write as a single fraction in its simplest form. x + 5 x + 8 + x x - 1 ................................................. [3]

Mark scheme: 7(a) x + 5 3 B1 for ( x − 5 )( x + 5 ) final answer x + 4 B1 for ( x − 5 )( x + 4 ) 7(b) 2 x 2 + 12 x − 5 2 x 2 + 12 x − 5 3 B1 for common denominator x ( x − )1 oe or x ( x − 1) x 2 − x B1 for ( x − 1)( x + 5 ) + x ( x + 8 ) or better final answer 7(c)(i) 6 x 2 − 8 x final answer 2 B1 for each term in final answer or M1 for correct answer seen and spoilt 7(c)(ii) 64 2 FT their (c)(i) correctly evaluated provided at least 2 terms but not the original equation M1 for substituting x = 4 into their (c)(i) 7(c)(iii) (0, 6) 4 dy M1 for their derivative = 0 or = 0 soi  4 98  dx  ,  oe  3 27  4 B1 for x = 0 and x = 3 M1dep for substituting one of their x values into y = 2 x 3 − 4 x 2 + 6 soi

8 (a) D NOT TO SCALE A 13 cm 20 cm E F 24 cm B C The diagram shows a prism, ABCDEF. AB = 13 cm, AC = 20 cm, CF = 24 cm, and angle ABC = 90°. (i) Calculate the total surface area of the prism. .......................................... cm2 [6] (ii) Calculate the volume of the prism. .......................................... cm3 [1] (b) A sector of a circle with radius 6 cm has a sector angle of 50°. Calculate the perimeter of the sector. ............................................ cm [3]

Mark scheme: 8(a)(i) 3 22 43 48 50 2 B1 for 4 correct or M1 for one error in adding. 8(a)(ii) correct diagram 3 B1FT their (a)(i) for 5 correct heights B1 for 5 points at upper ends of intervals on correct vertical line B1FT dep on at least B1 for increasing curve through their 5 points After 0 scored, SC1 for 4 of their points correctly plotted 8(a)(iii) 35 to 38 1 8(b) Correct box-and-whisker diagram 4 B1 for median 1.64 drawn B1 for LQ 1.57 drawn B1 for UQ 1.71 drawn If 0 scored SC1 for 1.64, 1,57 or 1.71 seen 1.45 1.57 1.64 1.71 1.83

9 (a) The table shows information about the mass, in kilograms, of each of 50 children. Mass (k kg) 0 1 k G 10 10 1 k G 25 25 1 k G 35 35 1 k G 40 40 1 k G 50 Frequency 3 19 21 5 2 (i) Complete the cumulative frequency table. Mass (k kg) k G 10 k G 25 k G 35 k G 40 k G 50 Cumulative frequency [2] (ii) On the grid, draw a cumulative frequency curve to show this information. 50 40 30 Cumulative frequency 20 10 0 0 10 20 30 40 50 k Mass (kg) [3] (iii) Use your diagram to find an estimate of the number of children with a mass of 32 kg or less. ................................................. [1] (iv) Two of the 50 children are chosen at random. Find the probability that one child has a mass of 25 kg or less and the other child has a mass of more than 25 kg. Give your answer correct to 3 decimal places. ................................................. [3] (b) 300 Frequency density 200 100 0 0.8 0.9 1.0 1.1 1.2 1.3 1.4 1.5 Height (m) The histogram shows information about the heights, in meters, of 170 children. Calculate an estimate of the mean height. ............................................. m [5]

Mark scheme: 9(a) 1350 or 1354…. 6 M2 for 20 2 − 132 or M1 for BC2 + 132 = 202 A1 for 231 or 15.2 or 15.19 to 15.20 M1 for 20 × 24 and 13 × 24 and their 15.2 × 24 M1 for [½ ×] their 15.2 × 13 9(b) 2370 or 2369 to 2371… cao 1 9(c) 24.6 or 24.58 to 24.59 4 13 M3 for sin [...] = oe 20 2 + 24 2 or M2 for 20 2 + 24 2 or 24 2 + 20 2 − 132 or M1 for AF2 = 202 + 242 or 242 + 202 - 132 or M1 for correct angle identified

10 (a) Find the size of an exterior angle of a regular polygon with 18 sides. ................................................. [2] (b) A 5.2 cm NOT TO SCALE B E 2.6 cm C D 6.75 cm In triangle ACD, B lies on AC and E lies on AD such that BE is parallel to CD. AE = 5.2 cm and ED = 2.6 cm. Calculate BE. BE = ........................................... cm [2] (c) Two solids are mathematically similar. The smaller solid has height 2 cm and volume 32 cm3. The larger solid has volume 780 cm3. Calculate the height of the larger solid. ............................................ cm [3] (d) P Q NOT TO SCALE N R S PQ is parallel to RS, PNS is a straight line and N is the midpoint of RQ. Explain, giving reasons, why triangle PQN is congruent to triangle SRN. ..................................................................................................................................................... ..................................................................................................................................................... ..................................................................................................................................................... ..................................................................................................................................................... [4]

Mark scheme: 10(a) 0.75 3 7 3 0.75 3 B2 for 4 or 3 correct or B1 for 2 correct 10(b) correct curve 4 B3FT for 8 or 9 correct plots B2FT for 6 or 7 correct plots B1FT for 4or 5 correct plots 10(c) Accept any integer ≥ 8 1 10(d) 1 B3 1 line y = 4 − x ruled B2 for [ y = ] 4 − x identified 2 2 1 or B1 for ruled line with gradient − 2 or B1 for ruled line through (0, 4) but not y = 4 0.2 to 0.3 4.2 to 4.3 B1

11 f ( )x = 3 - 2 x g ( )x = x 2 + 5 h ( )x = x 3 j ( x) = 3x (a) Find f ( 5) . ................................................. [1] (b) Find f ( j ( 5)) . ................................................. [2] (c) Find f ( f ( x)) . Give your answer in its simplest form. ................................................. [2] (d) Solve g ( x) = f ( x) + 7 . Give your answers in the form p ! q 6 . ................................................. [4] (e) (i) Find f -1 ( )x . f -1 ( )x = ................................................ [2] (ii) Find h -1 ( x) . h -1 ( x) = ................................................ [1] (f) Find x when j -1 ( x) =- 2 . x = ................................................ [2]

Mark scheme: 11(a) 20 2 360 16 × 180 M1 for or 180 − 18 18 11(b) 4.5 2 BE 5.2 M1 for = oe 6.75 5.2 + 2.6 11(c) 5.8[0] or 5.798 to 5.799 3 780 M2 for 2 × 3 oe 32 780 32 2 3 32 or M1 for 3 or 3 or = 32 780 l 3 780 11(d) QN = NR [given] B1 Two correct pairs of angles with B2 B1 for any correct pair of angles with reason reasons from or two correct pairs of angles with no/wrong reasons angle PQN = angle SRN alternate angle QPN = angle RSN alternate angle PNQ = angle SNR [vertically] opposite ASA [implies congruent] B1 dep on B1 B2

12 The table shows some values of y = 3 + 4x - x 2 for - 1 G x G 5 . x -1 -0.5 0 1 2 3 4 4.5 5 y -2 6 6 -2 (a) Complete the table. [3] (b) On the grid, draw the graph of y = 3 + 4x - x 2 for - 1 G x G 5 . y 8 6 4 2 – 1 0 1 2 3 4 5 x – 2 [4] (c) Write down an integer value of k for which the equation 3 + 4x - x 2 = k has no solutions. ................................................. [1] (d) By drawing a suitable tangent, estimate the slope of the curve when x = 3 . ................................................. [2] 9 2(e) By drawing a suitable straight line on the grid, solve the equation - 1 + x - x = 0 . 2 x = ........................ or x = ........................ [4]

Mark scheme: 12(a) 13 1 12(b) 4x – 3 final answer 2 M1 for 3 − 2 ( 3 − 2x ) 12(c) − 7 5 4 M1 for x 2 + 2 x − 35 [ = 0] or x 2 + 2 x = 35 M2 for ( x + 7 )( x − 5 ) or x ( x − 5 ) + 7 ( x − 5 ) or x ( x + 7 ) − 5 ( x + 7 ) or M1 for ( x + a )( x + b ) where a, b are integers with ab = − 35 or a + b = 2 12(d) 3 − x 2 M1 for a correct first step: oe final answer x = 3 − 2 y or y − 3 = −2 x , 2x = 3 – y or 2 y 3 = − x 2 2 12(e) 32 − 54 x + 37 x 2 − 8 x 3 5 B4 for final answer 27 − 36 x − 18 x + 24 x 2 + 12 x 2 − 8 x 3 + x2 + 5 oe OR 3 B1 for ( 3 − 2x ) + x2 + 5 and B2 for expansion of the 3 brackets, allow one error or B1 for correct expansion of 2 of the brackets with at least 3 terms correct