Cambridge IGCSE Mathematics 0580 — 2013 May/June Paper 4 · Variant 1

4 For Examiner′s y Use 9 8 7 6 5 4 3 2 Q 1 x –8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 –1 –2 –3 R –4 –5 –6 (a) Describe fully the single transformation that maps shape Q onto shape R. Answer(a) ................................................................................................................................. [3] 5 (b) (i) Draw the image when shape Q is translated by the vector [2] 4 e o. (ii) Draw the image when shape Q is refl ected in the line x = 2. [2] (iii) Draw the image when shape Q is stretched, factor 3, x-axis invariant. [2] (iv) Find the 2 × 2 matrix that represents a stretch of factor 3, x-axis invariant. Answer(b)(iv) [2] e o 0 1 (c) Describe fully the single transformation represented by the matrix 1 0 e o. Answer(c) .................................................................................................................................. [2] _____________________________________________________________________________________

Mark scheme: 4 (a) Enlargement 1 [centre] (–3, 4) 1 Do not allow column vector for coordinates [scale factor] 3 1  5   k  (b) (i) Image at (1 5), (4, 5), (4, 6), (1, 7) 2 SC1 for translation by   or  4  k     (ii) Image at (5, 1), (8, 1), (8, 3), (5, 2) 2 SC1 for reflection in y = 2 (iii) Image at 2 SC1 for three correct vertices or shape with vertices at (–4, 1) (– 4, 3), (– 1, 3), (– 1, 6), (– 4, 9) and (–1, 1), (–1, 4) and (–4, 7)  1 0   1 0   3 0  (iv)   2 SC1 for   , k ≠ ± 1 or   0 3 0 k 0 1       (c) Reflection 2 B1 B1 independent y = x oe

More questions on Transformations

5 For Examiner′s Height (h cm) 150 < h Ğ 160 160 < h Ğ 165 165 < h Ğ 180 180 < h Ğ 190 Use Frequency 5 9 18 10 The table shows information about the heights of a group of 42 students. (a) Using mid-interval values, calculate an estimate of the mean height of the students. Show your working. Answer(a) ......................................... cm [3] (b) Write down the interval which contains the lower quartile. Answer(b) ............................................... [1] (c) Complete the histogram to show the information in the table. One column has already been drawn for you. 2 Frequency density 1 0 150 155 160 165 170 175 180 185 190 Height (cm) [4] _____________________________________________________________________________________

Mark scheme: 5 (a) 171.25 (or 171 or 171.2 or 171.3) 3 M1 for 5 × 155 + 9 × 162.5 + 18 × 172.5 + 10 × 185 [= 7192.5] www and M1 (dep on M1) for their Σfx ÷ 42 (b) 160 < x ≤ 165 oe 1 (c) Blocks with heights of 1.8, 1.2, 1, with 4 B3 for 2 correct blocks correct interval widths and no gaps or B2 for 1 correct block or B1 for 3 correct frequency densities or heights or 3 correct widths IGCSE – May/June 2013 0580 41 Qu. Answer Marks Part marks 15 7

More questions on Averages and measures of spread

6 For Examiner′s A Use B 30° 52° E 15.7 cm NOT TO SCALE 16.5 cm C 23.4 cm D In the diagram, BCD is a straight line and ABDE is a quadrilateral. Angle BAC = 90°, angle ABC = 30° and angle CAE = 52°. AC = 15.7 cm, CE = 16.5 cm and CD = 23.4 cm. (a) Calculate BC. Answer(a) BC = ......................................... cm [3] (b) Use the sine rule to calculate angle AEC. Show that it rounds to 48.57°, correct to 2 decimal places. Answer(b) [3] (c) (i) Show that angle ECD = 40.6°, correct to 1 decimal place. For Examiner′s Use Answer(c)(i) [2] (ii) Calculate DE. Answer(c)(ii) DE = ......................................... cm [4] (d) Calculate the area of the quadrilateral ABDE. Answer(d) ........................................ cm2 [4] _____________________________________________________________________________________

Mark scheme: 15 7. 6 (a) 31.4 3 M2 for sin 30 or M1 for correct implicit statement 15 7. × sin 52 (b) [sinE =] M2 M1 for correct implicit statement 16 5. 48.573… A1 (c) (i) [∠ACE = ] 180 – 52 – 48.57 M1 [= 79.43] [∠ECD = ] 40.57… A1 (ii) 15.3 or 15.27 to 15.281 www 4 M2 for [(DE)2 =] 16.52 + 23.42 – 2 × 16.5 × 23.4cos(40.6 or 40.57) or M1 for full correct implicit statement A1 for 233 to 234 (d) 466 or 466.34 to 466.5 4 M1 for 0.5 × 15.7 × their 31.4 sin(90 – 30) oe M1 for 0.5 × 15.7 × 16.5 sin(128 – their 48.6 or 48.57) oe M1 for 0.5 × 16.5 × 23.4 sin (40.6 or 40.57) oe IGCSE – May/June 2013 0580 41 Qu. Answer Mark Part marks x + 2 9

More questions on Non-right-angled triangles

7 (a) For Examiner′s A Use NOT TO SCALE (2x + 3) cm (x + 2) cm B C In triangle ABC, AB = (x + 2) cm and AC = (2x + 3) cm. 9 sin ACB = 16 Find the length of BC. Answer(a) BC = ......................................... cm [6] (b) A bag contains 7 white beads and 5 red beads. (i) The mass of a red bead is 2.5 grams more than the mass of a white bead. The total mass of all the 12 beads is 114.5 grams. Find the mass of a white bead and the mass of a red bead. Answer(b)(i) White ............................................ g Red ............................................ g [5] (ii) Two beads are taken out of the bag at random, without replacement. For Examiner′s Use Find the probability that (a) they are both white, Answer(b)(ii)(a) ............................................... [2] (b) one is white and one is red. Answer(b)(ii)(b) ............................................... [3] _____________________________________________________________________________________

Mark scheme: x + 2 9 7 (a) 6.61 (6.614…) www 6 B1 for = oe 2 x + 3 16 M1 for 16(x + 2) = 9(2x + 3) or better A1 for [x =] 2.5 M2 for √{(2 × their x + 3)2 – (their x + 2)2} or M1 for (2 × their x + 3)2 – (their x + 2)2 or SC2 for final answer of 4√13 or 7 15 or better 2 SC1 for final answer of 5√7 or better (b) (i) White = 8.5, red = 11 5 B3 for 7w + 5(w + 2.5) = 114.5 or for 7 ( r − 5.2 ) + 5 r = 114 5. oe B1 for 8.5 or 11 or SC2 for 7w + 5 × w + 2.5 = 114.5 leading to 9.33[3…] or SC1 for 7w + 5 × w + 2.5 = 114.5 OR B1 for r = w + 2.5 oe B1 for 7w + 5r = 114.5 oe M1 for elimination of a variable A1 for 8.5 or 11 42 21 14 7 7 6 (ii) or or or 2 M1 for × 132 66 44 22 12 11 (a) (0.318 or 0.3181 to 0.3182) 70 35 7 5 5 7 (ii) or 3 M2 for × + × or 1 – 132 66 12 11 12 11 (b) 5 4 their (a) – × (0.53[0] or 0.5303…) 12 11 or 7 5 35 M1 for × or 12 11 132 or 70 SC1 for oe from replacement 144 IGCSE – May/June 2013 0580 41 Qu. Answer Mark Part marks

More questions on Percentages

8 (a) For Examiner′s D Use 84° NOT TO SCALE E C x° x° 110° 110° A B In the pentagon ABCDE, angle EAB = angle ABC = 110° and angle CDE = 84°. Angle BCD = angle DEA = x°. (i) Calculate the value of x. Answer(a)(i) x = ........................... [2] (ii) BC = CD. Calculate angle CBD. Answer(a)(ii) Angle CBD = ........................... [1] (iii) This pentagon also has one line of symmetry. Calculate angle ADB. Answer(a)(iii) Angle ADB = ........................... [1] (b) A, B and C lie on a circle centre O. Angle AOC = 3y° and angle ABC = (4y + 4)°. NOT TO Find the value of y. SCALE O 3y° C A (4y + 4)° B Answer(b) y = ........................... [4] (c) For Examiner′s S Use NOT TO SCALE R Q 78° P In the cyclic quadrilateral PQRS, angle SPQ = 78°. (i) Write down the geometrical reason why angle QRS = 102°. Answer(c)(i) ...................................................................................................................... [1] (ii) Angle PRQ : Angle PRS = 1 : 2. Calculate angle PQS. Answer(c)(ii) Angle PQS = ........................... [3] (d) NOT TO 7.2 cm2 SCALE 5 cm2 l cm 6.9 cm The diagram shows two similar fi gures. The areas of the fi gures are 5 cm2 and 7.2 cm2. The lengths of the bases are l cm and 6.9 cm. Calculate the value of l. Answer(d) l = ........................... [3] _____________________________________________________________________________________

Mark scheme: 8 (a) (i) 118 2 M1 for (3 × 180 – 2 × 110 – 84) [÷ 2] or better (ii) 31 1FT FT (180 – their (i)) ÷ 2 (iii) 22 1FT FT 84 – 2 × their (ii) or 2 × their (ii) – 40, only if positive answer and less than 84 (b) 32 4 B2 for 360 – 3y = 2(4y + 4) oe and B1 for 11y = 352 oe or M1 for angle at centre = 2 × angle at circumference soi (c) (i) Opposite angles [cyclic quad] add to 1 180 (ii) 68 3 M1 for [angle PRS =] 102 ÷ 3 × 2 and M1 for angle PQS = angle PRS or angle PRQ = angle PSQ 5 (d) 5.75 3 M2 for 6.9 × oe 2.7 or M1 for evidence of ratio of areas = (ratio of sides)2 or sf = 1.2 2 2

More questions on Angles

9 f(x) = x2 + x – 3 g(x) = 2x + 7 h(x) = 2x Examiner′s Use (a) Solve the equation f(x) = 0. Show all your working and give your answers correct to 2 decimal places. Answer(a) x = ........................ or x = ........................ [4] (b) fg(x) = px2 + qx + r Find the values of p, q and r. Answer(b) p = ..................................... q = ..................................... r = ..................................... [3] (c) Find g –1(x). For Examiner′s Use Answer(c) g –1(x) = ............................................... [2] (d) Find x when h(x) = 0.25. Answer(d) x = ............................................... [1] (e) Find hhh(3). Give your answer in standard form, correct to 4 signifi cant fi gures. Answer(e) ............................................... [4] _____________________________________________________________________________________

Mark scheme: 9 (a) − 1 ± 12 − 4 × 1 × ( − 3 ) 2 B1 for 12 − 4 × 1 × ( − )3 or better 2 + − p q p q and if in the form or r r 2 then B1 for p = –1 and r = 2(1) or better –2.30, 1.30 final answer B1 B1 SC1 for –2.30 and 1.30 seen or –2.3 or –2.303 to – 2.302 and 1.3 or 1.302 to 1.303 or final answer –1.30 and 2.30 (b) 4, 30, 53 3 M1 for (2x + 7)2 + (2x + 7) – 3 and B1 for (2x + 7)2 = 4x2 + 14x + 14x + 49 oe IGCSE – May/June 2013 0580 41 Qu. Answer Mark Part marks (c) x − 7 2 M1 for y – 7 = 2x or x = 2y + 7 or –7 2 then ÷ 2 clearly seen in correct order − 7 y with arrow or better or 2 (d) –2 1 (e) 1.158 × 1077 4 B3 for 1.16 × 1077 or 1.1579… × 1077 or 1.157 × 1077 or B2 for 2256 seen or B1 for 28 seen or 256

More questions on Functions

10 For Examiner′s Use Star 1 Star 2 Star 3 The diagrams show a sequence of stars made of lines and dots. (a) Complete the table for Star 5, Star 7 and Star n. Star 1 Star 2 Star 3 Star 4 Star 5 Star 7 Star n Number of lines 10 20 30 40 Number of dots 11 21 31 41 [4] (b) The sums of the number of dots in two consecutive stars are shown in the table. Star 1 and Star 2 Star 2 and Star 3 Star 3 and Star 4 32 52 72 Find the sum of the number of dots in (i) Star 10 and Star 11, Answer(b)(i) ............................................... [1] (ii) Star n and Star (n + 1), Answer(b)(ii) ............................................... [1] (iii) Star (n + 7) and Star (n + 8). Answer(b)(iii) ............................................... [1] (c) The total number of dots in the fi rst n stars is given by the expression 5n2 + 6n . For Examiner′s Use (i) Show that this expression is correct when n = 3. Answer(c)(i) [2] (ii) Find the total number of dots in the fi rst 10 stars. Answer(c)(ii) ............................................... [1] (d) The total number of dots in the fi rst n stars is 5n2 + 6n . The number of dots in the (n + 1)th star is 10(n + 1) + 1. Add these two expressions to show that the total number of dots in the fi rst (n + 1) stars is 5(n + 1)2 + 6(n + 1) . You must show each step of your working. Answer(d) [4] _____________________________________________________________________________________

Mark scheme: 10 (a) 50, 70 1 10n oe 1 51, 71 1 10n + 1 oe 1 (b) (i) 212 1 (ii) 20n + 12 1 (iii) 20n + 152 1 (c) (i) 5 × 32 + 6 × 3 = 63 1 and 11 + 21 + 31 = 63 or 32 + 31 = 63 or 11 + 52 = 63 1 (ii) 560 1 (d) Complete solution with no errors seen 4 B1 for 5n2 + 6n + 10n + 10 + 1 or and a conclusion better e.g. 5n2 + 6n + 10(n + 1) + 1 B1 for use of 5(n + 1)2 = 5n2 + 10n + 5 oe at any stage = 5n2 + 6n + 10n + 10 + 1 B1 for use of 6n + 6 = 6(n + 1) oe at = 5n2 + 10n + 5 + 6n + 6 any stage = 5n2 + 10n + 5 + 6n + 6 = 5(n + 1)2 + 6(n + 1)

More questions on Indices I