Cambridge IGCSE Mathematics (US) 0444 — 2017 Oct/Nov Paper 2 · Variant 3

1 Ahmed drives his car from London to Cambridge. He leaves London at 07 45 and arrives in Cambridge at 10 17. Work out the time, in hours and minutes, that he takes to drive from London to Cambridge. .....................h ................. min [1]

Mark scheme: Question Answer Marks Partial marks 1 2 h 32 min 1

2 Work out. 2 3 + 1 ................................................. [1]

Mark scheme: 2 3 1

3 Write $4 as a percentage of $80. .............................................% [1]

Mark scheme: 3 5 1

4 A quadrilateral has one line of symmetry and no rotational symmetry. Write down the name of this quadrilateral. ................................................. [1]

Mark scheme: 4 kite 1

5 Factor completely. 18x + 27y ................................................. [1]

Mark scheme: 5 9(2x + 3y) final answer 1

6 3 10 = 10 p Find the value of p. p = ................................................ [1]

Mark scheme: 6 2 1 oe 3

7 The bearing of Q from P is 055°. Find the bearing of P from Q. ................................................. [2]

Mark scheme: 7 235 2 M1 for 180 + 55 or diagram with a correct angle seen other than the 55° bearing or diagram with the angle to be worked out clearly indicated.

8 Work out 3.6 # 10 8 + 5.4 # 10 9 . Give your answer in scientific notation. ................................................. [2]

Mark scheme: 8 5.76 × 109 2 M1 for figs 576 or 0.36 × 109 or 54 × 108

9 Solve the inequality. 7 - 8x H 19 + 2x ................................................. [2]

Mark scheme: 9 x ⩽ –1.2 oe final answer 2 B1 for –1.2 oe or M1 for correct step to collect x’s and numbers

10 A model of a house is made using a scale of 1 : 30. The model has a surface area of 6000 cm2. Work out the surface area of the actual house. Give your answer in square meters. ........................................... m2 [3]

Mark scheme: 10 540 3 M2 for 6000 × 302 ÷ 1002 oe M1 for 302 or 0.32 (implied by figs 540) or ÷ 1002

11 Work out the size of one interior angle of a regular 12-sided polygon. ................................................. [3]

Mark scheme: 11 150 3 M2 for (12 – 2) × 180 ÷ 12 or 180 – 360 ÷ 12 or M1 for (12 – 2) × 180 or 360 ÷ 12 soi 30

12 Solve the system of linear equations. 3x + y = 7 2x - 3y = 12 x = ................................................ y = ................................................ [3]

Mark scheme: 12 [x =] 3 3 M1 for correctly eliminating one variable [y =] –2 A1 for x = 3 A1 for y = –2 If zero scored, SC1 for two values satisfying one of the original equations

13 Work out 3 17 1 - 1 4 . Give your answer as a mixed number in its lowest terms. ................................................. [3]

Mark scheme: Question Answer Marks Partial marks 13 22 5 1 1 22 k 5k or 2 – Allow or . Correct step for dealing with B1 7 4 7 4 7 k 4 k mixed numbers 88 35 4 7 M1 4 or 2 or Correct method to find common denominator e.g. 3 28 28 28 28 28 7 or 128 25 25 A1 128 128

14 Solve by factoring. 3x 2 - 7x - 20 = 0 x = ................. or x = .................. [3]

Mark scheme: 14 (3x + 5)(x – 4) [= 0] M2 M1 for (3x + b)(x + a) where ab = –20 or 3a + b = –7 5 A1 If zero scored, SC1 for 2 correct answers from no 4 and − oe working or other methods 3

15 Simplify 3 + 5 2 + 20 . ................................................. [3]

Mark scheme: 15 14 + 8 5 3 B1 for 9 + 3 5 + 3 5 + 5 oe B1 for 2 5

16 A NOT TO SCALE 30° C 5 cm B sin A = 0.8 . Work out the length of BC. BC = .......................................... cm [3]

Mark scheme: 16 8 3 5×0.8 M2 for oe sin30 sin30 sin A or M1 for = oe 5 BC

17 Solve for x. xp 3m + xy = 4 x = ................................................ [4]

Mark scheme: 17 12 m −12 m 4 xp or final answer M1 for 12m + 4xy = xp or 3m = − xy p − 4 y 4 y − p 4 M1 for 12m = xp – 4xy or 3m = x( 4p – y) 3m M1 for 12m = x(p – 4y) or = x p − y 4 12 m M1 for p − 4 y To a maximum of 3 marks for an incorrect answer

18 (a) The nth term of a sequence is 6 – 5n. Write down the first three terms of this sequence. ................ , ............. , ............. [1] (b) The nth term of another sequence is 5n 2 + 3 . Is 608 a term in this sequence? Explain how you decide. ........................ because ....................................................................................................................... [3]

Mark scheme: 18(a) 1, –4 and –9 1 18(b) Yes because 11 is an integer oe 3 B2 for [n =] 11 or M2 for √((608 – 3) ÷ 5) or 5 × 112 + 3 [= 608] or M1 for 5n2 + 3 = 608 oe

19 NOT TO SCALE 12 cm The diagram shows a shape made from a square and a semi-circle. The total area of the shape is kr + c square centimeters. Find the value of k and the value of c. k = ................................................ c = ................................................ [4]

Mark scheme: 19 [k = ] 18 4 B3 for k = 18 [c =] 144 B1 for c = 144 OR M3 for 122 + 1 π62 2 or M2 for 1 π62 2 or M1 for radius = 6 or for 122

20 A NOT TO SCALE Q T a P O B b In the diagram, AQ = QB and QP = PB . OA = a and OB = b . (a) Find the following in terms of a and b. Give your answers in their simplest form. (i) AB ................................................. [1] (ii) QP ................................................. [1] (iii) The position vector of P. ................................................. [2] (b) OBTQ is a parallelogram. Find AT in terms of a and b. Give your answer in its simplest form. ................................................. [2]

Mark scheme: 20(a)(i) –a + b oe 1 20(a)(ii) 1 1 1 FT their (a)(i) – a + b 4 4 20(a)(iii) 1 3 2 M1 for correct unsimplified answer or for a correct a + b oe route 4 4 20(b) 1 3 2 M1 for correct unsimplified answer or for a correct – a + b oe route 2 2

21 The diagram shows the numbers of hummingbirds seen by Ali and Hussein in their gardens each day for 10 days. 9 8 7 Ali’s garden Hussein’s garden 6 5 Number of hummingbirds 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 Day (a) Work out the mean number of hummingbirds seen in Ali’s garden each day. ................................................. [3] (b) Work out the median number of hummingbirds seen in Hussein’s garden each day. ................................................. [2] (c) On one of these days there were 4 times as many hummingbirds seen in Hussein’s garden as in Ali’s garden. Which day was this? Day ................................................ [1]

Mark scheme: 21(a) 3.4 3 M1 for 2 + 5 + 4 + 2 + 1 + 3 + 2 + 7 + 6 + 2 [34] M1 for their 34 ÷ 10 21(b) 5 2 M1 for 5, 5 identified 21(c) [Day] 10 1

22 E D 23° F 19° NOT TO C SCALE 67° B A In the diagram, points A, B, C, D, E and F lie on the circumference of the circle. Angle BFC = 19°, angle ADB = 23° and angle ABE = 67°. Work out (a) angle BEC, Angle BEC = ................................................ [1] (b) angle ABC, Angle ABC = ................................................ [3] (c) angle BCE. Angle BCE = ................................................ [2] Question 23 is printed on the next page.

Mark scheme: 22(a) 19 1 22(b) 138 3 M2 for 180 – (19 + 23) or M1 for angle AEB = 23 or angle AEC = 42 22(c) 90 2 M1 for angle EBC = 71 or angle EAB = 90

23 (a) Write down a cube number that is greater than 100 and less than 1000. ................................................. [1] (b) Write down a prime number between 90 and 100. ................................................. [1] (c) Write the following in order of size, starting with the smallest. 7 0.71 7% 49 10 .................... 1 ....................1 ....................1 .................... [3] smallest 27 - 2 (d) Find the value of 3 . c 8 m ................................................. [2]

Mark scheme: 23(a) 125 or 216 or 343 or 512 or 729 1 23(b) 97 1 23(c) 7 3 B2 for two of 0.07, 0.7, 7 soi 7% < < 0.71 < 49 or M1 for converting at least two values oe 10 23(d) 4 2 M1 for numerator 4 or denominator 9 or for final 9 9 answer 4