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

1 Simplify. 3a + 7b - 4a + b ................................................. [2]

Mark scheme: Question Answer Marks Partial Marks 1 –a + 8b final answer 2 B1 for –a or [+]8b in final answer or for –a + 8b spoilt

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2 A field, ABC, is in the shape of a triangle. AC = 500 m and BC = 650 m. Using a ruler and compasses only, complete the scale drawing of the field ABC. Leave in your construction arcs. Use a scale of 1 cm to represent 100 m. The side AB has been drawn for you. A B Scale: 1 cm to 100 m [3]

Mark scheme: 2 Correct triangle constructed with 3 B2 for correct triangle with no/incorrect AC = 5 cm and BC = 6.5 cm and arcs intersecting arcs or SC2 for accurate triangle with arcs but sides interchanged or B1 for 6.5 [cm] or 5 [cm] soi

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3 Rangan buys 3.6 kg of potatoes and 2.8 kg of leeks. The total cost is $13.72 . Leeks cost $2.65 per kilogram. Find the cost of 1 kg of potatoes. $ ................................................ [3]

Mark scheme: 3 1.75 3 M2 for (13.72 – 2.8 × 2.65) ÷ 3.6 oe or M1 for 2.8 × 2.65

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4 Aisha records the distance she runs and her average speed. The results are shown in the scatter diagram. 16 15 14 13 12 Average speed 11 (km/h) 10 9 8 7 6 5 0 1 2 3 4 5 6 7 8 9 10 Distance (km) (a) The table shows the results of four more runs. Distance (km) 4.2 5.7 7.1 8.8 Average speed (km/h) 13.4 11.8 9.8 8.3 On the scatter diagram, plot these points. [2] (b) What type of correlation is shown in the scatter diagram? ................................................. [1] (c) On the scatter diagram, draw a line of best fit. [1] (d) Use your line of best fit to estimate her average speed when she runs a distance of 6 km. ........................................ km/h [1]

Mark scheme: 4(a) 4 points correctly plotted 2 B1 for 2 or 3 points correctly plotted 4(b) Negative 1 4(c) Correct ruled line of best fit 1 4(d) 10 to 12 1 FT their straight line of best fit

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5 T = 4.085 # 2.35 By writing each number correct to 1 significant figure, work out an estimate for T. You must show all your working. ................................................. [2]

Mark scheme: 5 50 –10 M1 Allow M1 for 3 out of 4 values correctly rounded or for all correct but with any 4 × 2 trailing zeros 5 A1 50 –10 dep on 4 × 2

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6 Without using a calculator, work out 2 # 2 . 3 4 You must show all your working and give your answer as a mixed number in its simplest form. ................................................. [3]

Mark scheme: 6 8 11 M1 and oe improper fractions 3 4 88 A1 oe improper fraction 12 1 A1 dep on 1st A1 7 cao final answer 8 11 3 If M0 scored SC1 for or oe 3 4 improper fraction

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7 Make x the subject of this formula. 2y = 5x - 7 x = ................................................ [2]

Mark scheme: 7 2 y + 7 2 y 7 2 2 y 7 [ x = ] oe or [ x = ] + oe M1 for 2 y + 7 = 5 x oe or = x – oe 5 5 5 5 5 final answer

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8 (a) 1, 2, 3, 5 and 7 are all common factors of two numbers. Write down the digit that the two numbers must end in. ................................................. [1] (b) Write 84 as a product of its prime factors. ................................................. [2]

Mark scheme: 8(a) 0 1 8(b) 22 × 3 × 7 or 2 × 2 × 3 × 7 2 B1 for 2, 2, 3, 7

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9 (a) Ahmed increases 40 by 300%. From this list, put a ring around the correct calculation. 40 # 1.300 40 # 3 40 # 400 40 # 4 40 # 300 [1] 2 (b) Ahmed finds the magnitude of the vector e- 3o. From this list, put a ring around the correct calculation. 2 2 +- 3 2 2 2 - 3 2 2 2 - 3 2 2 2 + ( - 3 ) 2 2 2 + ( - 3 ) 2 [1]

Mark scheme: 9(a) 40 × 4 1 9(b) 2 2 1 2 + ( − 3 )

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10 A town has a population of 45 000. This population increases exponentially at a rate of 1.6% per year. Find the population of the town at the end of 5 years. Give your answer correct to the nearest hundred. ................................................. [3]

Mark scheme: 10 48 700 cao 3 5  1.6  M1 for 45 000 ×  1 +  oe  100  A1 for 48 710 to 48 720 If A0 scored B1 for their more accurate value correctly rounded to the nearest 100

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11 x = 2 y (-1, 4) NOT TO SCALE (-1, 1) 0 x The diagram shows a rectangle with a line of symmetry at x = 2 . Two vertices of the rectangle are at (-1, 1) and (-1, 4). The shaded region is defined by the inequalities a G x G b and c G y G d . Find the values of a, b, c and d. a = ................................................ b = ................................................ c = ................................................ d = ................................................ [2]

Mark scheme: 11 [a =] –1 2 B1 for two or three correct [b =] 5 [c =] 1 or SC1 for [d =] 4 [a =] x ⩾ –1 [b =] x ⩽ 5 [c =] y ⩾ 1 [d =] y ⩽ 4

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12 The interior angle of a regular polygon with n sides is 156°. Work out the value of n. n = ................................................ [2]

Mark scheme: 12 15 2 360 180 ( n − 2 ) M1 for or =156 oe 180 − 156 n

13 Write the recurring decimal .017o as a fraction in its simplest form. You must show all your working. ................................................. [3]

Mark scheme: 13 17.77… – 1.77… oe M1 M1 for correct working shown 8 A2 16 cao B1 for oe seen 45 90

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14 Find the gradient of a line that is perpendicular to 8y + 4x = 5 . ................................................. [2]

Mark scheme: 14 2 2 5 – 4 x M1 for y= oe or better 8

15 20 Car 18 16 14 Motorbike 12 Speed 10 (m/s) 8 6 4 2 0 0 10 20 30 40 50 60 70 80 90 100 Time (s) The diagram shows the speed–time graph for 100 seconds of the journey of a car and of a motorbike. (a) Find the deceleration of the car between 60 and 100 seconds. ......................................... m/s2 [1] (b) Calculate how much further the car travelled than the motorbike during the 100 seconds. ............................................. m [3]

Mark scheme: 15(a) 0.3 1 15(b) 360 3 M2 for correct complete area statement e.g. 1 18 × 60 + × 40 × (18 + 6) – 12 × 100 2 1 1 or × 6 × (60 + 80) – × 6 × 20 2 2 or for answer 420 or M1 for one area calculation

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16 Factorise 6x 2 + 7x - 20 . ................................................. [2]

Mark scheme: 16 (3x – 4)(2x + 5) final answer 2 B1 for (ax + b)(cx + d) where ac = 6 and ad + bc = 7 or bd = – 20

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17 (a) f ( x) = 3x 2 + a where a is an integer. f ( - 2) = 19 Find the value of a. a = ................................................ [2] (b) g ( x) = 2 x + 7 h ( x) = 3 x - 8 (i) Find gh(x) in its simplest form. ................................................. [2] (ii) Find g -1 ( x) . g -1 ( x) = ................................................ [2]

Mark scheme: 17(a) [a =] 7 2 M1 for 3(–2)2 + a = 19 or better 17(b)(i) 6x – 9 or 3(2x – 3) final answer 2 M1 for 2(3x – 8) + 7 or better 17(b)(ii) x − 7 2 M1 for a correct first step final answer x = 2 y + 7 or y − 7 = 2 x 2 y 7 or = x + 2 2

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18 7 cm NOT TO SCALE 12 cm The diagram shows a solid made from a cylinder and a hemisphere, both of radius 7 cm. The cylinder has length 12 cm. Work out the total surface area of the solid. [The surface area, A, of a sphere with radius r is A = 4rr 2 .] ........................................... cm2 [4]

Mark scheme: 18 990 or 989.58 to 989.73 4 M1 for 4 × π × 72 [÷2] M1 for π × 72 M1 for π × 7 × 2 × 12

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19 In this Venn diagram, shade the region M l , N , P . M N P [1]

Mark scheme: 19 1 ξ M N P

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20 ( 4x - 87)° NOT TO SCALE ( x + 60 ) ° 2x° y° The diagram shows a cyclic quadrilateral. Find the value of y. y = ................................................ [4]

Mark scheme: 20 107 4 B2 for x = 40 or M1 for 2x + x + 60 = 180 oe M1 for correctly substituting their x into 4x – 87 + y = 180 oe or 4x – 87 + x + 60 + y + 2x = 360 oe

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21 P Q R NOT TO S SCALE U T W V The diagram shows a cuboid PQRSTUVW. PV = 17.2 cm The angle between the line PV and the base TUVW of the cuboid is 43°. Calculate PT. PT = ........................................... cm [3]

Mark scheme: 21 11.7 or 11.73… 3 PT M2 for sin43 = oe 17.2 or M1 for identifying angle PVT

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22 Simplify. x 2 - 5x 2x 2 - 50 ................................................. [4] Question 23 is printed on the next page.

Mark scheme: 22 x x 4 B1 for x(x – 5) or final answer B2 for 2(x – 5)(x + 5) or (x – 5)(2x + 10) 2 ( x + 5 ) 2 x + 10 or (2x – 10)(x + 5) or B1 for 2(x2 – 25) or (x – 5)(x + 5)

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23 (a) C B D A NOT TO n SCALE F E m The diagram shows a parallelogram CDEF. FE = m and CE = n . B is the midpoint of CD. FA = 2AC Find an expression, in terms of m and n, for AB. Give your answer in its simplest form. AB = ................................................. [3] (b) GH = 56 ( 2p 5 + q ) JK = 18 ( 2p + q ) Write down two facts about vectors GH and JK . ..................................................................................................................................................... ..................................................................................................................................................... [2]

Mark scheme: 23(a) 5 1 3 B2 for correct unsimplified answer in m – n 1 1 6 3 terms of m and n e.g. (m – n) + m 3 2 or M1 for a correct route or for   FC = m – n or CF = n – m or better e.g.  1 AC = (m – n) 3    23(b) GH = 3 JK oe or GH has a greater 2 B1 for each magnitude   GH and JK are parallel

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