Cambridge IGCSE Mathematics 0580 — 2009 Oct/Nov Paper 1 · Variant 1

1 Insert one pair of brackets to make the following equation correct. For Examiner's 2 × 8 − 5 − 4 = 15 Use [1]

Mark scheme: Qu. Answers Mark Part Marks 1 2 × 8 − (5 − 4) = 15 1

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2 Write the following numbers in order starting with the smallest. 2 0.283 28 % 7 Answer < < [1]

Mark scheme: 2 28% < 0.283 < 2 1 7

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3 Find the volume of a cube with sides of 2.3 cm. Answer cm3 [1]

Mark scheme: 3 12.2 or 12.17 or 12.167 1

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4 North NOT TO SCALE North B 72° A The diagram shows the position of two airports, A and B. The bearing of B from A is 072°. Work out the bearing of A from B. Answer [2]

Mark scheme: 4 252 2 W1 for 108 or 72 correctly shown on the diagram at B. Or M1 for 180 + 72 or 360 − (180 − 72) soi

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5 The number of spectators, N, at a football match is 16 000, correct to the nearest thousand. For Complete the statement for N in the answer space. Examiner's Use Answer Y N < [2]

Mark scheme: 5 15500 Y N < 16500 1, 1 If zero, SC1 for correct but reversed

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6 Work out the value of 3 3 ×1 1 . 4 7 Show all your working and leave your answer as a fraction. Answer [2]

Mark scheme: 6 15 8 M1 and seen 4 7 120 8 120 oe fraction or 4 oe A1 isw incorrect cancelling after oe 28 28 28 Final answer is a decimal, maximum M1.

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7 A B C Using a straight edge and compasses only, construct the locus of points which are equidistant from AB and from BC. Show clearly all your construction arcs. [2]

Mark scheme: 7 Correct angle bisector (±2°) with two 2 W1 correct bisector without arcs or incorrect arcs pairs of correct arcs. or absent arcs. Line (±2 mm) from B. Line (±2 mm) from B.

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5 For 8 4 8 25 0.3333 Examiner's 2 Use From the list above, write down (a) a prime number, Answer(a) [1] (b) an irrational number. Answer(b) [1]

Mark scheme: 8 (a) 25 or 5 1 (b) 8 isw 1

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9 A train sets off at 11 53 on a journey to Mumbai. The journey takes 2 hours 30 minutes. (a) Write down the time when the train arrives in Mumbai. Answer(a) [1] (b) The distance to Mumbai is 235 kilometres. Calculate the average speed of the train. Answer(b) km/h [2]

Mark scheme: 9 (a) 14 23 isw or 2.23 pm isw. 1 Not 02 23 or 2 23 alone. Not 14h(ours)23 (b) 94 2cao M1 for 235 ÷ 2.5 (or 2h 30min or 150) Method mark is for formula with values.

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10 Solve the simultaneous equations For Examiner's 5x − y = 15, Use 7x − 5y = 3. Answer x = y = [3]

Mark scheme: 10 (x =) 4 and (y =) 5 www 3 M1 for complete correct method for one value A1 for 1 correct answer. ww both correct W3 ww one correct W0 Reversed answer, look in working to be convinced of transcription error.

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11 15 10 Miles 5 0 5 10 15 20 25 Kilometres Distance can be measured in miles or kilometres. 24 kilometres is approximately equal to 15 miles. (a) Draw a straight line on the grid to show the conversion between kilometres and miles. [2] (b) Use your graph to estimate the number of kilometres equal to 12 miles. Answer (b) km [1]

Mark scheme: 11 (a) Ruled line from (0, 0) to (24, 15) 2 W1 for correct freehand or short of (24, 15) but End point between (23.5, 15) and within allowed limits and to at least 12 miles. (24.5, 15). If zero SC1 Ruled line from (0, 0) to Start point within 1 mm of (0, 0) (23.5, 15) or to (24.5, 15) (b) 18.8 to 19.6 1ft Answer in range. If 0 or W1 gained in part (a) follow through line with positive gradient only ± 1 mm IGCSE – October/November 2009 0580 11

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12 For Examiner's Use NOT TO SCALE 4 cm 7 cm The diagram shows a triangular prism of length 7 cm. The cross-section is an equilateral triangle of side 4 cm. Complete an accurate net of the prism. One rectangular face has been drawn for you. [3]

Mark scheme: 12 Correct net layout 1 2 rectangles and 2 equilateral triangles (one on each side) in correct position to make a net. 2 accurate, 7 cm by 4 cm, rectangles on 1 top and bottom. 2 accurate equilateral triangles at the 1 within 2 mm of central grid line sides (height 3.3 cm to 3.7 cm)

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13 For y Examiner's G Use 5 4 3 2 F 1 x –3 –2 –1 0 1 2 3 4 5 –1 –2 –3 –4 The points F and G are shown on the grid. (a) Write down the co-ordinates of the point F. Answer(a)( , ) [1] (b) Write as a column vector. Answer(b) = ( ) [1]  _2  (c) GH =   . Mark and label the point H on the grid. [1]  _7 

Mark scheme: 13 (a) (–2, 1) 1 All coordinates/components reversed.  4  ie (a) (1, −2), (b)   , (c) (−3, 3)  6   6  (b)   1 mark 0, 0, SC1  4  (c) H at (2, –2) 1

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14 (a) Find the value of p when p3 = −27. For Examiner's Use Answer(a) p = [1] −1 1 (b) Find the value of q when q = . 6 Answer(b) q = [1] (c) Simplify 8s 2 ÷ 2s −1. Answer(c) [2]

Mark scheme: 14 (a) –3 final answer 1 (b) 6 final answer 1 4 (c) 4s3 or − 3 final answer 2 W1 for 4sn (n ≠ 0) or ks3 (k ≠ 0) seen s

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md 15 J = 3 (a) Find the value of d when J = 35 and m = 7. Answer(a) d = [2] (b) Make d the subject of the formula. Answer(b) d = [2]

Mark scheme: 15 7d (a) 15 2 M1 for 35 = or better. 3 3 J J d (b) (d =) 2 M1 for 3J = md or = m m 3 3

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16 As the earth rotates, a point on the equator moves round at a speed of 1669.8 kilometres/hour. For Examiner's (a) Write down this number in standard form, correct to 3 significant figures. Use Answer(a) [2] (b) Change 1669.8 kilometres/hour into metres/second. Answer(b) m/s [2]

Mark scheme: 16 (a) 1.67 × 103 2 W1 for 1.67 × 10n (n ≠ 0) or 1.(…..) × 103 as answer If zero SC1 for figs 167 in answer. (b) 464 or 463.8(3…..) 2 M1 for 1669.8 × 1000 ÷ 3600

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17 (a) Factorise 5x2 + 4xy. Answer (a) [1] (b) Simplify completely 7(2x + y) − 3(3x − 2y). Answer (b) [3]

Mark scheme: 17 (a) x(5x + 4y) final answer 1 Ignore check by expansion. (b) 5x + 13y www 3 W1 for 14x + 7y and W1 for –9x + 6y If zero ww SC1 for 5x or (+)13y in answer

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18 For S Examiner's Use T NOT TO SCALE W 105° 38° P Q R The lines PS and QT intersect at W. PQR is a straight line. Angle SPR = 38° and angle TQR = 105°. Write down the size of the following angles. In each case give a reason for your answer. (a) Angle PQW = because [2] (b) Angle PWQ = because [2] (c) Angle TWS = because [2]

Mark scheme: 18 (a) 75 Angle(s) (on a straight) line (=) 1, 1 Or reference to straight line and 180 180 (b) 67 Angle(s) (in a) triangle (sum to) 1ft,1 or exterior angle (of triangle is) sum of interior 180 (opposite) angles (c) 67 (vertically) opposite 1ft,1 IGCSE – October/November 2009 0580 11

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19 For Examiner's Use Silver Other Yellow Red The accurate pie chart shows information about the colours of 240 cars in a car park. (a) The sector angle for silver cars is 90°. Calculate the number of silver cars in the car park. Answer(a) [1] (b) There are 36 yellow cars in the car park. Showing all your working, calculate the sector angle for yellow cars. Answer(b) [2] (c) (i) Measure and write down the sector angle for red cars. Answer(c)(i) [1] (ii) Calculate the percentage of red cars in the car park. Answer(c)(ii) % [2]

Mark scheme: 19 (a) 60 1 (b) 36 ÷ 240 × 360 oe M1 oe e.g. 36 × 90 ÷ 60 54 A1 W2 54 with some relevant working shown (c) (i) 116 to 118 1 (ii) 32.5 or their (c) (i) ÷ 3.6 2ft M1 for their (c) (i) ÷ 360 × 100 Or for their (c) (i) × (60 ÷ 90) ÷ 240 × 100 Allow revised angle in range 116 – 118 seen with working

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