Cambridge IGCSE Mathematics (US) 0444 — 2024 May/June Paper 2 · Variant 1

1 y 8 7 6 A 5 4 C 3 2 1 B – 8 – 7 – 6 – 5 – 4 – 3 – 2 – 1 0 1 2 3 4 5 x – 1 – 2 The diagram shows two sides of a parallelogram ABCD. Find the coordinates of point D. ( ...................... , ...................... ) [2]

Mark scheme: Question Answer Marks Partial Marks 1 (–3, 7) 2 B1 for correct diagram or correct coordinates for their point D or for  3, k  or  k , 7 

2 Geetha has a box of toys. She picks a toy at random from the box. The probability that she picks a wooden toy is 0.6 . (a) Work out the probability that she does not pick a wooden toy. ................................................. [1] (b) The box contains three types of toys, wooden, plastic, or metal. Type of toy Wooden Plastic Metal Number of toys 14 14 Probability 0.6 Complete the table. [2]

Mark scheme: 2(a) 0.4 oe 1 2(b) 42 2 B1 for 42 0.2 0.2 B1 for 0.2 and 0.2 If B0 scored SC1 for their two probabilities being half their (a)

3 The table shows some information about two sequences. nth term 5th term Sequence A 60 – 4n Sequence B n2 – 300 (a) Complete the table. [2] (b) In sequence B the kth term is -156 . Find the value of k. k = ................................................. [2]

Mark scheme: 3(a) 40 2 B1 for each –275 3(b) 12 2 M1 for [n2 = ] –156 + 300 or better

4 Find the greatest odd number that is a factor of 140 and a factor of 210. ................................................. [2]

Mark scheme: 4 35 2 B1 for answer 5, 7 or 70 or M1 for 2 × 2 × 5 × 7 and 2 × 3 × 5 × 7 or two correct factor trees or tables or 5 × 7 × k seen

5 Work out 6 # 10 17 # 2.1 # 10 17 . Give your answer in scientific notation. ................................................. [2]

Mark scheme: 5 1.26 × 1035 2 B1 for 12.6 × 1034

6 D A 40° C E B NOT TO SCALE x° The diagram shows 5 kites that are congruent to kite ABCD. Each kite is joined to the next kite along one edge. Angle DAB = 40° and DCE is a straight line. Find the value of x. x = ................................................ [3]

Mark scheme: 6 145 3 M1 for 180 ÷ 6 or any angle congruent to BCD = 30 M1 for (360 – 40 – their 30) ÷ 2 oe

7 12 cm J L NOT TO SCALE 12 cm K The diagram shows a shape made from a triangle JKL and a semicircle with diameter JL. JKL is an isosceles right-angled triangle with JK = JL = 12 cm. (a) Work out the area of this shape. Give your answer in the form a + br . ........................................... cm2 [3] (b) Work out the perimeter of this shape. Give your answer in the form a + b 2 + cr . ............................................ cm [4]

Mark scheme: 7(a) 72  18π 3 1 M1 for  12  12 2  1   12  2 M1 for       2   2  7(b) 12  12 2  6π 4 1 M1 for  12 oe 2 2 2 12 M2 for 12  12 or sin45 or M1 for 12 2  12 2 oe or better 12 or sin 45  oe KL

8 These are the first five terms of a sequence. 11 18 25 32 39 Find an expression for the nth term of the sequence. ................................................. [2]

Mark scheme: 8 7n + 4 oe final answer 2 B1 for 7n + j or kn + 4 k ≠ 0 , or 7n + 4 seen then spoilt

9 The value of a car is $10 000. Each year the value of the car decreases exponentially by 20%. Work out the value of this car after 2 years. $ ................................................. [2]

Mark scheme: 9 6400 2 2  20  M1 for 10 000   1   oe  100 

10 Amir invests $3000 in an account. The account pays simple interest at a rate of r % per year. At the end of 6 years the value of his investment is $3360. Find the value of r. r = ................................................ [3]

Mark scheme: 10 2 3  3360  3000  100  M2 for oe 3000  6  3360  3000   100 or oe 3000  6  3000 6 r or M1 for  3360  3000   100 or B1 for 60 or 0.12 or 1.12 2 If 0 scored SC1 for answer 18 oe 3

11 y 7 6 5 4 3 R 2 1 – 1 0 1 2 3 4 5 6 7 x – 1 Find the inequalities that define the unshaded region, R. .............................................................................................................................................................. [4]

Mark scheme: 11 y < x 4 B1 for y < x x < 6 B1 for x < 6 1 ⩽ y ⩽ 5 B2 for 1 ⩽ y ⩽ 5 or B1 for y ⩾ 1 or y ⩽ 5 If B0 scored, SC2 for y ⩽ x, x ⩽ 6 and 1 < y < 5 oe or SC1 for three correct from y = x, x = 6, y = 1 and y = 5

12 Solve the system of linear equations. You must show all your working. 6x + 2y = 29 3x - 4y = 17 x = ................................................ y = ................................................ [3]

Mark scheme: 12 Correct method to eliminate one variable M1 x = 5, y = –0.5 A2 A1 for x = 5 A1 for y = –0.5 If M0 scored SC1 for 2 values satisfying one of the original equations.

13 p° NOT TO SCALE 4m° 5m° (4m + 38)° The diagram shows a cyclic quadrilateral. Find the value of p. p = ................................................ [3]

Mark scheme: 13 62 3 B2 for m = 20 or M1 for 5m + 4m = 180 soi or p + 4m + 38 = 180 soi

14 NOT TO 45° SCALE 4 cm The area of the shaded sector is kr cm 2 . Find the value of k. k = ................................................. [3]

Mark scheme: 14 14 3 360  45 2 M2 for   4 360 c 2 or M1 for   4 where c < 360 360 7 or B1 for 315 or 8

15 (a) Simplify 20 # 5 . ................................................. [1] 2 (b) 3 + 2 3 = c + k 3 ` j Find the value of c and the value of k. c = ................................................. k = ................................................. [2]

Mark scheme: 15(a) 10 1 15(b) [c =] 21 2 2 M1 for 9  6 3  6 3  2 3 or better [k =] 12   with at least three terms correct

16 Simplify. (a) 1770 ................................................. [1] 1 -2 (b) . b 2 l ................................................. [1]

Mark scheme: 16(a) 1 1 16(b) 4 1

17 9 cm NOT TO 150° SCALE 8 cm Work out the area of the triangle. ........................................... cm2 [3]

Mark scheme: 17 18 3 1 M1 for × 9 × 8 × sin 150 oe 2 B1 for sin150 = 0.5

18 y 8 7 6 5 4 3 2 1 x – 3 – 2.5 – 2 – 1.5 – 1 – 0.5 0 0.5 1 1.5 – 1 – 2 – 3 The diagram shows the graph of y = x 3 + 4x 2 - 2 for - 3 G x G 1.5 . By drawing a suitable straight line, solve the equation x 3 + 4x 2 - 2 = 2x for - 3 G x G 1.5 . x = ....................... or x = ....................... [3]

Mark scheme: 18 y = 2x ruled B1 x = –0.5 to –0.55 B2 B1 for –0.5 to –0.55 x = 0.85 to 0.9 B1 for 0.85 to 0.9

19 Factor completely. (a) 12m 2 - 75t 2 ................................................. [3] (b) xy + 15 + 3y + 5x ................................................. [2]

Mark scheme: 19(a) 3(2m + 5t)(2m – 5t) final answer 3 B2 for (6m + 15t)(2m – 5t) or (2m + 5t)(6m – 15t) or B1 for 3(4m2 – 25t2) or (2m + 5t)(2m – 5t) 19(b) (x + 3)(y + 5) final answer 2 B1 for x(y + 5) + 3(y + 5) or y(x + 3) + 5(x + 3)

20 Solve the equation 4 cosx + 5 = 3 for 0c G x G 360c . x = ....................... or x = ....................... [3]

Mark scheme: 20 120, 240 3 B2 for one correct 1 or M1 for cos x =  oe 2 If M1 or 0 scored, SC1 for two angles with a sum of 360

21 C B x° NOT TO SCALE X y° A D The points A, B, C and D lie on a circle. The chords AC and BD intersect at X. (a) Find, in terms of x and/or y (i) angle AXB ................................................. [1] (ii) angle CDX. ................................................. [1] (b) AB = 4 cm , AX = 3 cm , BX = 1.8 cm and CD = 6 cm . Work out the length of CX. CX = ........................................... cm [2]

Mark scheme: 21(a)(i) 180 – (x + y) or 180 – x – y 1 21(a)(ii) y 1 21(b) 2.7 oe 2 4 1.8 M1 for  oe 6 CX

22 Bag A and bag B each contain red counters and blue counters only. Stephan picks a counter at random from bag A and Jen picks a counter at random from bag B. The probability that Stephan picks a red counter is 2. 5 The probability that Stephan and Jen both pick a red counter is 1. 4 Find the probability that Stephan and Jen both pick a blue counter. ................................................. [4]

Mark scheme: 22 9 4  1 2   2  oe M3 for  1      1   oe 40  4 5   5  OR 1 2 M2 for  4 5 2 1 or M1 for  p  oe 5 4 M1 for (1 –theirP(Jen red))×(1 – 52 ) oe