Cambridge IGCSE Mathematics (9-1) 0980 — 2020 May/June Paper 2 · Variant 2

1 Write down the order of rotational symmetry of the diagram. ................................................. [1]

Mark scheme: Question Answer Marks Partial Marks 1 2 1

2 At noon the temperature in Maseru was 21 °C. At midnight the temperature had fallen by 26 °C. Work out the temperature at midnight. ............................................. °C [1]

Mark scheme: 2 –5 1

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3 C NOT TO x° SCALE 50° A B D AB = BC and ABD is a straight line. Find the value of x. x = ................................................. [2]

Mark scheme: 3 25 2 B1 for 130 seen or M1 for 50 ÷ 2

4 Write down (a) a square number greater than 10, ................................................. [1] (b) an irrational number. ................................................. [1]

Mark scheme: 4(a) Any square number greater than 10 1 4(b) Any irrational number 1

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5 y = mx + c Find the value of y when m =- 3 , x =- 2 and c =- 8 . y = ................................................. [2]

Mark scheme: 5 –2 2 M1 for (–3)(–2) + (–8)

6 11 cm NOT TO SCALE 5 cm 7 cm Calculate the area of the trapezium. .......................................... cm2 [2]

Mark scheme: 6 45 2 11 + 7 M1 for × 5 oe 2

7 A B On the Venn diagram, shade the region A + B . [1]

Mark scheme: 7 Intersection shaded 1

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8 Write 2 - 4 as a decimal. ................................................. [1]

Mark scheme: 8 0.0625 1

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9 North NOT TO SCALE North A B The bearing of B from A is 105°. Find the bearing of A from B. ................................................. [2]

Mark scheme: 9 285 2 M1 for 180 + 105 or 75 or 105 seen in correct position at B

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10 Simplify. p 4pq 2q # t ................................................. [2]

Mark scheme: 10 2 p 2 2 B1 for correct unsimplified answer t

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11 Without using a calculator, work out 1 - . 4 12 You must show all your working and give your answer as a fraction in its simplest form. ................................................. [3]

Mark scheme: 11 7 9 B1 4 12 21 2 M1 1 − 12 12 5 5 A1 6 6

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12 Roberto buys a toy for $5.00 . He then sells it for $4.60 . Calculate his percentage loss. ............................................. % [2]

Mark scheme: 12 8 2 5 − 4.60 4.60 M1 for [× 100 ] or × 100 5 5

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13 Simplify 8t 8 ' 4t 4 . ................................................. [2]

Mark scheme: 13 2t 4 2 B1 for 2tn or kt4 (n,k ≠ 0)

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14 Solve the equation. 1 - x = 5 3 x = ................................................. [2]

Mark scheme: 14 –14 2 M1 for 1 – x = 3 × 5 or better x 1 or = 5 − or better 3 3

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15 Ella’s height is 175 cm, correct to the nearest 5 cm. Write down the upper bound of Ella’s height. ............................................ cm [1]

Mark scheme: 15 177.5 1

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16 Calculate ( 3 # 10 - 3 ) 3 . Give your answer in standard form. ................................................. [1]

Mark scheme: 16 2.7 × 10–8 1

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17 A train of length 105 m takes 11 seconds to pass completely through a station of length 225 m. Calculate the speed of the train in km/h. ........................................ km/h [3]

Mark scheme: 17 108 3 M1 for (105 + 225) ÷ 11 60 × 60 M1 for their speed × 1000

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18 y 8 T 6 4 U 2 x 0 2 4 6 8 Describe fully the single transformation that maps triangle T onto triangle U. ............................................................................................................................................................. ............................................................................................................................................................. [3]

Mark scheme: 18 Enlargement 3 B1 for each 1 [scale factor] − 2 [centre] (3, 4)

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19 Make y the subject of the formula. h 2 = x 2 + 2y 2 y = ................................................. [3]

Mark scheme: 19 2 2 3 M1 for correct rearrangement for y or y2 term h − x [±] M1 for correct square root 2 M1 for correct division by 2 or 2

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20 D C 20° NOT TO SCALE O 131° B T A A, B, C and D lie on the circle, centre O. TA is a tangent to the circle at A. Angle ABC = 131° and angle ADB = 20°. Find (a) angle ADC, Angle ADC = ................................................. [1] (b) angle AOC, Angle AOC = ................................................. [1] (c) angle BAT, Angle BAT = ................................................. [1] (d) angle OAB. Angle OAB = ................................................. [1]

Mark scheme: 20(a) 49 1 20(b) 98 1 FT 2 × their (a) 20(c) 20 1 20(d) 70 1 FT 90 – their (c)

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21 Simplify. (a) ( 5x 4 ) 3 ................................................. [2] 3 8 (b) ( 256x 256 ) ................................................. [2]

Mark scheme: 21(a) 125x12 2 B1 for 125xk or kx12 21(b) 8x96 2 B1 for 8xk or kx96

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22 p is directly proportional to ( q + 2)2 . When q = 1, p = 1. Find p when q = 10 . p = ................................................. [3]

Mark scheme: 22 16 3 M1 for p = k ( q + 2) 2 M1 for p = (their k )(10 + 2) 2 OR p 1 M2 for = oe (10 + 2) 2 (1 + 2) 2

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23 y 8 7 6 5 4 3 2 1 x 0 1 2 3 4 5 6 7 8 (a) By drawing suitable lines and shading unwanted regions, find the region, R, where x H 2 , y H x and 2x + y G 8 . [5] (b) Find the largest value of x + y in the region R. ................................................. [1]

Mark scheme: 23(a) Correct lines and correct region 5 B2 for 2x + y = 8 correctly ruled clear or B1 for ruled line with negative gradient B1 for y = x correctly ruled B1 for x = 2 correctly ruled 23(b) 6 1

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24 P NOT TO 8 cm SCALE 6.4 cm Q The diagram shows a sector of a circle of radius 8 cm. The length of the arc PQ is 6.4 cm. Find the area of the sector. .......................................... cm2 [4]

Mark scheme: 24 25.6 or 25.59 to 25.60… 4 6.4 2 M3 for × π × 8 2 × π × 8 x 6.4 or M2 for = oe 360 2 × π × 8 x or M1 for × 2 × π × 8 = 6.4 oe 360

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25 Simplify. 2x 2 + x - 15 ax + 3a - 2bx - 6b ................................................. [5]

Mark scheme: 25 2 x − 5 5 B2 for (2x – 5)(x + 3) final answer or B1 for (2x + p)(x + q) where pq = –15 or a − 2b p + 2q = 1 B2 for (x + 3)(a – 2b) or B1 for x(a – 2b) + 3(a – 2b) or a(x + 3) – 2b(x + 3)

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26 3 y 2 = 6 x and y = n x . Find the value of n. n = ................................................. [2] Question 27 is printed on the next page.

Mark scheme: 26 4 2 2 1 M1 for y 3 = x 6 or y 2 = x or y 4 = x

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27 H G NOT TO SCALE E F 6 cm C D 6 cm A B 8 cm The diagram shows a cuboid. AB = 8cm , AD = 6cm and DH = 6cm . Calculate angle HAF. Angle HAF = ................................................. [6]

Mark scheme: 27 64.9 or 64.89 to 64.90 6 + 72 − 100 B5 for [cos =]100 2 × 10 × 72 OR M1 for 82 + 62 M1 for 62 + 62 ( theirAF ) 2 + (theirAH ) 2 − (theirHF ) 2 M2 for 2 × (theirAF ) × ( theirAH ) or M1 for (theirHF)2 = (theirAF)2 + (their AH)2 – 2 × (theirAF) × (their AH) cos(HAF) AF, AH etc from correct method

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