Cambridge IGCSE Mathematics 0580 — 2022 Feb/March Paper 3 · Variant 2

1 (a) One day, Mahika records the number of teachers and students who cycle to school. Tally Frequency Teachers | | | | Students | | | | | | | | | | | | | | | | | | | (i) Complete the frequency column in the table. [1] (ii) Work out the percentage of people who cycle that are students. ............................................. % [2] (b) Mahika records how 120 students from Year 1 and Year 2 travel to school. Each student walks, cycles or travels by bus. • 48 students are in Year 1. • 77 students walk. • 5 students in Year 2 cycle. • 36 students travel by bus. 4 • of the students who travel by bus are in Year 1. 9 (i) Complete the table. Walk Cycle Bus Total Year 1 Year 2 Total 120 [3] (ii) One of the 120 students is chosen at random. Work out the probability that this student does not travel by bus to school. ................................................. [2] (c) There have been 24 complaints about one of the buses. The complaints are: • The bus is late. • The price is too high. • The bus is crowded. (i) Complete the table. Pie chart Complaint Frequency sector angle Late 10 Price 6 Crowded 8 [2] (ii) Complete the pie chart. [2]

Mark scheme: Question Answer Marks Partial Marks 1(a)(i) 5 23 1 1(a)(ii) 82.1 or 82.14… 2 FT their (a)(i) their 23 M1 for [× 100 ] soi their 5 + their 23 1(b)(i) 30 2 16 48 3 B2 for 9 or 10 correct 47 5 20 72 or B1 for 5, 6, 7 or 8 correct 77 7 36 [120] 1(b)(ii) 7 2 120 − 36 oe M1 for oe 10 120 or B1 for 84 or for their77 +their 7 1(c)(i) 150, 90, 120 2 B1 for one correct sector angle 360 or M1 for × k k =1, 6, 8 or 10 24 1(c)(ii) Correct pie chart drawn 2 FT their table if angles add up to 360 B1FT for one sector correctly drawn

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2 (a) Calculate the interior angle of a regular pentagon. ................................................. [2] (b) The diagram shows three congruent regular pentagons and a triangle. y° NOT TO SCALE x° (i) Work out the value of x. Give a geometrical reason for your answer. x = .................... because .................................................................................................... ............................................................................................................................................. [2] (ii) Work out the value of y. Give a geometrical reason for your answer. y = .................... because .................................................................................................... ............................................................................................................................................. [3] (iii) Find the ratio x : y. Give your answer in its simplest form. ....................... : ....................... [1]

Mark scheme: 2(a) 108 2 360 ( 5 − 2 ) × 180 M1 for 180 − or oe 5 5 2(b)(i) 36 2 B1 for each Angles [at a] point [add to] 360 2(b)(ii) 72 2 FT their (b)(i) 180 −their ( b )( i ) M1 for 2 Angles [in a] triangle add to 180 1 or Base angles [of an] isosceles triangle are equal 2(b)(iii) 1:2 1 FT their (b)(i): their (b)(ii) provided there has been some simplification

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4 (a) Y Z O X X, Y and Z lie on a circle, centre O. (i) Write down the mathematical name of the line (a) OX, ................................................. [1] (b) YZ. ................................................. [1] (ii) Measure the length of OX. ............................................ cm [1] (b) Another circle has a radius of 18 cm. Calculate the circumference of this circle. ............................................ cm [2] (c) In this part, all angles are in degrees. NOT TO B SCALE A 18x - 4y O C 9x + 3y X D Y A, B, C and D lie on a circle, centre O, diameter AC. XY is a tangent to the circle at D. (i) Use the information in the diagram to complete these two simultaneous equations. 9x + 3y = ....................... 18x - 4y = ....................... [2] (ii) Solve your simultaneous equations. You must show all your working. x = ................................................ y = ................................................ [3]

Mark scheme: 4(a)(i)(a) Radius 1 4(a)(i)(b) Chord 1 4(a)(ii) 3.5 1 4(b) 113 or 113.09… to 113.112 2 M1 for 2 × 18 × π oe 4(c)(i) 90 2 B1 for each 90 4(c)(ii) For correctly eliminating one M1 M1FT their two linear equations variable [ x = ]7 A1 [ y = ]9 A1 If M0 scored, SC1 for 2 values satisfying one of their original equations If no working shown, SC1 for two correct answers given

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5 (a) A closed box, in the shape of a cuboid, has length 5 cm, width 4 cm and height 2 cm. (i) Draw a net of the box on the 1cm2 grid. [3] (ii) A container is a cube with volume 1m3. Work out the maximum number of these boxes that can be packed into this container. ................................................. [3] (b) A shop sells three different sized boxes of rice. The boxes all have the same cost per kilogram. NOT TO SCALE Box A Box B Box C 80 rupees $3.50 750 g 1.35 kg (i) Work out the cost in rupees of box B. ...................................... rupees [2] (ii) $1 = 64 rupees. Calculate the mass of box C. Give your answer in kilograms. ............................................ kg [3] (c) Change 75cm3into litres. Give your answer in standard form. ........................................ litres [2]

Mark scheme: 5(a)(i) Correct ruled net 3 B2 for 4 or 5 correct rectangles drawn in correct positions relative to each other or B1 for one correct face drawn 5(a)(ii) 25 000 3 100 × 100 × 100 M2 for oe 2 × 4 × 5 or M1 for 100 × 100 × 100 or 0.02×0.04×0.05 or 2×4×5 or 50, 25, 20 If 0 scored, SC1 for final answer figs 25 5(b)(i) 144 2 figs135 1.35 80 750 M1 for or or or 750 figs750 750 80 oe or B1 for 1350 or 0.75[0] 5(b)(ii) 2.1 3 3.5 × 64 × figs75 M2 for oe 80 80 or M1 for 3.5× 64 or 64 5(c) 7.5 × 10−2 cao 2 75 M1 for oe 1000 If 0 scored, SC1 for correctly converting their number to standard form, provided their number is <1

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6 (a) 6 5 6 5 5 7 3 8 2 The diagram shows a fair 9-sided spinner. The numbers on the spinner are 2, 3, 5, 5, 5, 6, 6, 7 and 8. (i) The spinner is spun once. Write down the probability that the spinner lands on (a) the number 8, ................................................. [1] (b) a number less than 7. ................................................. [1] (ii) The spinner is spun 135 times. Work out the expected number of times the spinner lands on the number 6. ................................................. [1] (b) Hitesh throws a dice 80 times. The results are shown in the table. Number thrown Frequency 1 15 2 16 3 14 4 11 5 9 6 15 (i) Write down the mode. ................................................. [1] (ii) Work out the range. ................................................. [1] (iii) Work out the median. ................................................. [1] (iv) Calculate the mean. ................................................. [3]

Mark scheme: 6(a)(i)(a) 1 1 oe 9 6(a)(i)(b) 7 1 oe 9 6(a)(ii) 30 1 6(b)(i) 2 1 6(b)(ii) 5 1 6(b)(iii) 3 1 6(b)(iv) 3.35 3 M1 for 1 × 15 + 2 × 16 + 3 × 14 + 4 × 11 + 5 × 9 + 6 × 15 M1dep for their  fx ÷ 80

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7 (a) 1 mile = 1.609344 kilometres Change 6 miles into metres. Give your answer correct to the nearest metre. ............................................. m [3] (b) (i) The bearing of a boat from a harbour is 322°. Work out the bearing of the harbour from the boat. ................................................. [2] (ii) The boat is 12 km from the harbour. At 2.30 pm the boat starts to sail to the harbour. The speed of the boat is 5 km/h. Work out the time the boat arrives at the harbour. ................................................. [3] (c) The scale drawing shows the positions of Shakti’s house, S, and Mairi’s house, M, on a map. The scale is 1 cm represents 4 km. North S North M Scale: 1 cm to 4 km (i) Measure the bearing of M from S. ................................................. [1] (ii) North S Scale: 1 cm to 5 km This scale drawing shows another map with Shakti’s house, S, marked on it. The scale of this map is 1 cm represents 5 km. Mark the position of Mairi’s house, M, on this map. [4]

Mark scheme: 7(a) 9656 cao 3 M2 for 6 × 1.609344 × 1000 or M1 for 6 × 1.609344 or B1 for final answer figs 9654 to 9660 If 0 scored, SC1 for their decimal answer correctly rounded to nearest integer 7(b)(i) 142 2 M1 for 322 −180 oe or a clear diagram with both 322 or 38 marked and the reverse bearing to be found 7(b)(ii) 4.54 pm or 16 54 3 12 M1 for soi 5 A1 for 2 h 24 [mins] 7(c)(i) 107 1 7(c)(ii) The position of M correctly marked 4 B1 for SM = 9[cm] soi on the diagram  1  M1 for their SM × 4 ×    5  B1 for 7.2[cm] soi or for bearing of M drawn at 107°

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8 (a) (i) Complete the table of values for y = x 2 + 6x - 160 . x -20 -15 -10 -5 0 5 10 15 y 120 -120 -165 -160 -105 [3] (ii) On the grid, draw the graph of y = x 2 + 6x - 160 for - 20 G x G 15 . y 200 150 100 50 – 20 – 15 – 10 – 5 0 5 10 15 x – 50 – 100 – 150 – 200 [4] (iii) (a) Write down the equation of the line of symmetry of the graph. ................................................. [1] (b) Find the coordinates of the lowest point on the graph. ( ....................... , ......................) [1] (iv) Use your graph to solve the equation x 2 + 6x - 160 = 0 . x = ................. or x = ................. [2] (b) Rearrange the formula y = mx + c to make x the subject. x = ................................................ [2]

Mark scheme: 8(a)(i) −25 0 155 3 B1 for each 8(a)(ii) Completely correct curve 4 B3FT for 7 or 8 correctly plotted points B2FT for 5 or 6 correctly plotted points B1FT for 3 or 4 correctly plotted points 8(a)(iii)(a) x = −3 oe 1 8(a)(iii)(b) ( −3, k) oe 1 FT their graph where − 180  k < − 165 8(a)(iv) 10 −16 2 B1 FT for each 8(b) y − c 2 M1 for correct first step [ x = ] oe final answer m y c y − c = mx or = x + m m

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9 Tarak has two fields. He grows wheat, barley and corn in his fields. (a) P S NOT TO 174 m SCALE 126 m 53° Q 120 m R The diagram shows Tarak’s two triangular fields, PQR and PRS. Angle RPS = 90° and angle PRS = 53°. PQ = 174 m, QR = 120 m and PR = 126 m. (i) Show that angle PRQ = 90°. [2] (ii) Calculate the area of the quadrilateral PQRS. Give your answer correct to 4 significant figures. ............................................ m2 [5] (b) (i) The mass, m tonnes, of wheat grown in 2021 is 4.3 tonnes, correct to 1 decimal place. Complete this statement about the value of m. ......................... G m 1 ......................... [2] (ii) In 2020, 2.6 tonnes of barley is grown. In 2021, 3.25 tonnes of barley is grown. Show that the percentage increase in barley grown from 2020 to 2021 is 25%. [1] (iii) In 2019, 2.4 tonnes of corn is grown. In 2020, 20% more corn is grown than in 2019. In 2021, 20% less corn is grown than in 2020. Calculate the amount of corn grown in 2021. ....................................... tonnes [3]

Mark scheme: 9(a)(i) Complete method shown and 2 M1 for correct Pythagoras evaluated e.g. 120 2 + 126 2  = 174 2    9(a)(ii) 18 090 cao 5 B4 for 18 081 to 18 094.1 OR M2 for 126 × tan53 x or M1 for tan53 = 126 and 1 M1 for × 120 × 126 2 1 or × 126 ×theirPS 2 1 or × 126 × (120 + theirPS ) oe 2 If 0 scored, SC1 for evidence of rounding their answer to 4sf 9(b)(i) 4.25 4.35 2 B1 for each If 0 scored, SC1 for answers correct but reversed 9(b)(ii) Complete method seen 1 3.25 − 2.6 e.g. × 100 [ = 25 ] 2.6 9(b)(iii) 2.304 3 100 + 20 100 − 20 M2 for 2.4 × × oe 100 100 100 + 20 or M1 for 2.4 × oe 100 100 − 20 100 + 20 or B1 for and used 100 100

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