Cambridge IGCSE Mathematics (US) 0444 — 2018 Oct/Nov Paper 2 · Variant 3

1 Work out of 99 kg. 11 ........................................... kg [1]

Mark scheme: Question Answer Marks Partial Marks 1 63 1

2 Factor. y - 2y 2 ................................................. [1]

Mark scheme: 2 y 1( − 2 y ) 1

3 Work out $7 as a percentage of $140 . ............................................ % [1]

Mark scheme: 3 5 1

4 Work out 7 - 2 # 4 - 3 . ................................................. [1]

Mark scheme: 4 –4 1

35 Work out 125 2. ................................................. [1]

Mark scheme: 5 25 1

6 (a) Write the number five million, two hundred, seven in figures. ................................................. [1] (b) Write 0.008 13 in scientific notation. ................................................. [1]

Mark scheme: 6(a) 5 000 207 1 6(b) .8 13 × 10 − 3 1

7 Simplify. 2p - q - 3q - 5p ................................................. [2]

Mark scheme: 7 − 3 p − 4 q final answer 2 B1 for −3 p + kq or kp − 4 q final answer

8 Write these numbers correct to 2 significant figures. (a) 0.076 499 ................................................. [1] (b) 10 100 ................................................. [1]

Mark scheme: 8(a) 0.076 1 8(b) 10 000 1

9 Work out ' . 4 3 Give your answer as a fraction. ................................................. [2]

Mark scheme: 9 3 2 Accept equivalent fractions oe 1 3 3 8 8 M1 for × or ÷ oe 4 2 12 12

10 Solve. 3w - 7 = 32 w = ................................................ [2]

Mark scheme: 10 13 2 7 32 M1 for 3 w = 32 + 7 or w − = 3 3

11 A = r rl + r r2 Solve for l. l = ................................................ [2]

Mark scheme: 11 A − π r 2 A 2 M1 for A − πr 2 = πrl or πr 2 − A = − πrl or or − r π r πr A = l + r πr

12 Simplify 243 - 27 . ................................................. [2]

Mark scheme: 12 6 3 2 B1 for 9 3 or 3 3 or 2 27

13 Solve the equation cosx = for 0° G x G 360° . 2 ................................................. [2]

Mark scheme: 13 60, 300 2 B1 for each If zero scored SC1 for two answers that add to 360

14 Find the amplitude and period of 5 sin (4x ) . Amplitude = ................................................ Period = ................................................ [2]

Mark scheme: 14 [Amplitude=] 5 2 B1 for each [Period=] 90

15 A car travels at 108 km/h for 20 seconds. Calculate the distance the car travels. Give your answer in meters. ............................................ m [3] 2

Mark scheme: 15 600 3 108 × 1000 × 20 M2 for oe 60 × 60 108 × 1000 or M1 for oe 60 × 60 or for speed × time

16 (a) Simplify . w3 ................................................. [1] (b) Simplify 3w 3 3 . ^ h ................................................. [2]

Mark scheme: 16(a) 1 − 1 1 or w w 16(b) 27w 9 final answer 2 B1 for kw 9 or 27 w k

17 y varies directly as the square root of x. When x = 9, y = 6 . Find y when x = 25. y = ................................................ [3]

Mark scheme: 17 10 3 M1 for y = k x M1 for y = their k × 25 OR y 25 M2 for = 6 9

18 Write as a single fraction in its simplest form. 1 1 - x x + 1 ................................................. [3]

Mark scheme: 18 1 3 B1 for common denominator x ( x + )1 oe isw oe final answer x ( x + )1 B1 for x + 1 – x seen isw

19 NOT TO SCALE 72° 6 cm The diagram shows a sector of a circle with radius 6 cm and sector angle 72°. The perimeter of this sector is (p + qr ) cm. Find the value of p and the value of q. p = ................................................ q = ................................................ [3]

Mark scheme: 19 [p =] 12 3 B1 for [p =] 12 12 12 [q =] oe B2 for [q =] 5 5 72 or M1 for × π × 2 × 6 oe 360

20 The solutions to the equation x 2 - 2x - 4 = 0 are a + b and a - b , where a and b are integers. Find the value of a and the value of b. a = ................................................ b = ................................................ [4]

Mark scheme: 20 [a =] 1 4 M2 for (x – 1)2 = 5 [b =] 5 or M1 for (x – 1)2 A1 for a = 1 or b = 5 OR p + q p − q if or used r r 2 ± 20 B3 for 2 or B1 for q = (–2)2 – 4(1)(–4) or better and B1 for p = –(–2) and r = 2(1) or better

21 y 8 6 4 2 0 x 2 4 6 8 Write down the three inequalities that define the unshaded region. ................................................ ................................................ ................................................ [4]

Mark scheme: 21 x ⩾ 0 4 B1 for x ⩾0 x + y ⩽ 8 oe B1 for x + y ⩽ 8 oe x + 2y ⩾ 8 oe B2 for x + 2y ⩾ 8 oe 1 or B1 for gradient = – or for x + 2y = k 2

22 Simplify. 2x 2 - x - 1 2x 2 + x ................................................. [4]

Mark scheme: 22 x − 1 1 4 B2 for ( 2 x + 1)( x − )1 or 1− final answer x x or B1 for 2x(x – 1) + [1](x – 1) or x(2x + 1) – [1](2x + 1) or ( 2 x + a )( x + b ) where ab = –1 or a + 2b = –1 B1 for x ( 2 x + )1

23 Q P 10 cm NOT TO SCALE D C 10 cm A B 10 2 cm The diagram shows a triangular prism. AB = 10 2 cm, BC = 10 cm, PC = 10 cm, angle BCP = 90°, and angle QDC = 90°. Calculate the angle between AP and the rectangular base ABCD. ................................................. [4]

Mark scheme: 23 30 4 10 M3 for tan = oe 2 (10 2 ) + (10 )2 2 2 oe or M2 for (10 2 ) + (10) or B1 for recognising angle PAC is required

24 Solve the equations. (a) w = 3 w = ................................................ [1] 2 (b) = 3 u - 1 u = ................................................ [2] 4 + 1 = 2 (c) 2x 1 x = ................................................ [2]

Mark scheme: 24(a) 9 1 24(b) 5 2 2 oe M1 for 2 = 3(u – 1) or for = u − 1 3 3 24(c) 1 2 1 4 1 M1 for x = 16 2

25 Factor completely. (a) px + py - x - y ................................................. [2] (b) 2t 2 - 98m 2 ................................................. [3] Question 26 is printed on the next page.

Mark scheme: 25(a) ( x + y )( p − )1 final answer 2 M1 for p(x + y)–[1](x + y) or x ( p − )1 + y ( p − )1 25(b) 2 (t + 7 m )(t − 7 m ) final answer 3 M2 for ( 2t + 14 m )(t − 7 m ) or (t + 7 m )( 2t − 14 m ) or correct answer seen or M1 for 2(t 2 − 49 m 2 ) or (t + 7 m )(t − 7 m ) or 2(t + 7 )(t − 7 )

26 C P B c X NOT TO SCALE O a A In the diagram, OABC is a parallelogram. OP and CA intersect at X and CP : PB = 2 : 1. OA = a and OC = c . (a) Find OP, in terms of a and c, in its simplest form. OP = ................................................ [2] (b) CX : XA = 2 : 3 (i) Find OX , in terms of a and c, in its simplest form. OX = ................................................ [2] (ii) Find OX : XP. OX : XP = ................... : ................... [2]

Mark scheme: 26(a) 2 2 M1 for correct unsimplified form or correct c + a route 3 26(b)(i) 2 3 2 M1 for correct unsimplified form or correct a + c route 5 5 26(b)(ii) 3 : 2 oe 2 3 B1 for OX = OP oe 5