Cambridge IGCSE Mathematics (US) 0444 — 2021 Oct/Nov Paper 2 · Variant 3

1 Write 25 g as a percentage of 125 g. ............................................. % [1]

Mark scheme: Question Answer Marks Partial Marks 1 20 1

2 x° NOT TO SCALE 132° The diagram shows two parallel lines intersecting a straight line. Find the value of x. x = ................................................ [2]

Mark scheme: 2 48 2 B1 for 132 or 48 in the correct position on the diagram or M1 for 180 – 132

3 11 13 15 17 19 From this list, write down the number that is both a prime number and a factor of 78. ................................................. [1]

Mark scheme: 3 13 1

4 (a) = ! 2 1 Put a ring around each of the symbols that make this statement correct. 0.5 .................... 5% [1] (b) Insert one pair of parentheses to make this statement correct. 7 - 3 - 1 + 2 = 7 [1]

Mark scheme: 4(a) ≠and > indicated 1 4(b) 7 – (3 – 1) + 2 = 7 cao 1

5 Nina changes 350 euros into dollars when the exchange rate is 1 euro = $1.10 . Work out the amount Nina receives. $ ................................................ [1]

Mark scheme: 5 385 1

6 Marek buys a computer for $400. He sells it at a loss of 15%. Work out the selling price of this computer. $ ................................................ [2]

Mark scheme: 6 340 2  15  M1 for  1 −  × 400 oe  100  or B1 for 60

7 Simplify. 32g 32 ' 4g 4 ................................................. [2]

Mark scheme: 7 8g 28 final answer 2 B1 for kg 28 or 8 g k as final answer or correct answer seen and spoilt

8 Beatrice walks 8 km at a speed of 4 km/h and then 9 km at a speed of 3 km/h. Work out Beatrice’s average speed for the whole journey. ........................................ km/h [3]

Mark scheme: 8 3.4 3 8 9 B1 for oe or oe 4 3 17 M1 dep on B1 for 8 9 their + their 4 3

9 Simplify 50. ................................................. [1]

Mark scheme: 9 5 2 1

10 These are the first four terms of a sequence. 3 -1 -5 -9 (a) Find the next term in this sequence. ................................................. [1] (b) Find the nth term. ................................................. [2]

Mark scheme: 10(a) –13 1 10(b) –4n + 7 oe final answer 2 B1 for – 4n + k or – jn + 7 (j ≠ 0) or for a correct answer spoilt

11 P = M ( g 2 + h 2 ) (a) Find the value of P when M = 100, g = 3, and h = 2 . P = ................................................ [2] (b) Rearrange the formula to write g in terms of P, M, and h. g = ................................................ [3]

Mark scheme: 11(a) 1300 final answer cao 2 M1 for 100(32 + 22) or B1 for 13 seen 11(b) P 2 3 M1 for correct division by M [±] − h M1 for correct re-arrangement to isolate g M or g2 P − Mh 2 M1 for correct square root of two term or [±] final answer M expression Max 2 marks for an incorrect answer

12 Work out + . 12 4 Give your answer as a mixed number in its simplest form. ................................................. [3]

Mark scheme: 12 11 9 M1 Allow any correct common denominator + oe 12k 12 12 2 A2 20 13 cao A1 for 12 or equivalent improper fraction or mixed number

13 Work out .0042. Give your answer in scientific notation. ................................................. [2]

Mark scheme: 13 1.6 × 10–3 2 B1 for 0.0016 oe

14 (a) Evaluate 34. ................................................. [1] 2 (b) 4 + 5 = p + q 5 ` j Find the value of p and the value of q. p = ................................................ q = ................................................ [2]

Mark scheme: 14(a) 81 1 14(b) [p =] 21 2 B1 for each [q =] 8

15 The cost of a train journey is increased by 20% to a new cost of $84. Work out the original cost of the train journey. $ ................................................ [2]

Mark scheme: 15 70 2 20  M1 for x 1 + = 84 oe or better    100 

16 Jo and Mo share $26. Jo receives $10 more than Mo. Find the ratio Jo’s money : Mo’s money. Give your answer in its simplest form. ........................ : ...................... [3]

Mark scheme: 16 9 : 4 3 B2 for 18 and 8 seen or M1 for x + 10 + x = 26 oe

17 Each interior angle of a regular polygon is 177°. Calculate the number of sides of this polygon. ................................................. [2]

Mark scheme: 17 120 2 M1 for 360 ÷ (180 – 177) oe 180( n − 2) or for = 177 oe n

18 Find the equation of the straight line that passes through the points (2, -2) and (3, 10). Give your answer in the form y = mx + b . y = ................................................ [3]

Mark scheme: 18 [y =] 12x – 26 final answer 3 10 −−2 M1 for oe 3 − 2 M1 for correct substitution of (2, –2) or (3, 10) into y = (their m)x + c oe

19 NOT TO SCALE 8 cm 60° O The diagram shows a sector of a circle, center O, radius 8 cm. The perimeter of the shaded segment is a 3 + b r cm . ` j Find the value of a and the value of b. a = ................................................ b = ................................................ [4]

Mark scheme: 19 [a =] 8 4 B2 for [a =] 8 or M1 for [2 ×] 8 × sin 60 oe 16 16 [b =] oe B2 for [b =] oe 3 3 120 or M1 for × 2 × π × 8 360

20 O P The diagram shows a circle, center O. Using compass and straight edge only, construct a tangent line from the point P to the circle. [3]

Mark scheme: 20 3 Only one tangent needed M1 for correct perpendicular bisector with arcs M1 for arc of circle with centre at P midpoint

21 Simplify fully. 3 243y 10 5 ` j ................................................. [2]

Mark scheme: 21 27y6 final answer 2 B1 for ky6 or 27yk as final answer or correct answer seen and spoilt

22 x varies inversely as the square root of u. When u = 9, x = 2. Find u when x = 12. u = ................................................ [3]

Mark scheme: 22 1 3 M2 for 12 u = 2 9 oe oe 4 OR k M1 for x = oe u their k M1 for x = u

23 Find the least common multiple of 6x 2 and 9x 3. ................................................. [2]

Mark scheme: 23 18x 3 final answer 2 B1 for kx 3 or 18 x k final answer or correct answer seen and spoilt

24 (a) NOT TO 3 cm SCALE x cm 2 cm 6 cm Work out the value of x. x = .................................................. [3] (b) A vertical pole of height 12 m stands on horizontal ground. The angle of elevation of the top of the pole from a point P on the ground is 30˚. Work out the distance from P to the foot of the pole. Give your answer in radical form. .............................................. m [3]

Mark scheme: 24(a) 7 3 M2 for 62 + 22 + 32 oe or M1 for correct Pythagoras in one triangle e.g. 62 + 22 24(b) 12 3 3 1 B1 for tan 30 = oe 3 12 M1 for tan 30 = or better x

25 Simplify. 3x 2 - 18x ax - 6a + 2cx - 12c ................................................. [4] Questions 26 and 27 are printed on the next page.

Mark scheme: 25 3 x 4 B1 for 3x(x – 6) final answer B2 for ( x − 6)( a + 2c ) a + 2c or B1 for a ( x − 6) + 2c ( x − 6) or x ( a + 2c ) − 6( a + 2 c )

26 T NOT TO M t SCALE O R r ORT is a triangle and M is the midpoint of TR. O is the origin, OR = r and OT = t . Find, in terms of r and t, in its simplest form, (a) TR , ................................................. [1] (b) the position vector of M. ................................................. [2]

Mark scheme: 26(a) r − t oe 1 26(b) 1 2 M1 for any unsimplified answer or correct (r + t) oe 1 2 route e.g. OR + RM or r + (t – r) 2

27 Solve x - 31 = 2 . x = ................................................ [2]

Mark scheme: 27 1 2 1 1 1 or 0.125 M1 for = 2 , x-1 = 8 or 3x = oe or 8 3 x 2 better If 0 scored, SC1 for final answer of 8