Cambridge IGCSE Mathematics 0580 — 2010 May/June Paper 4 · Variant 2

3 Examiner's Use 2 1 1 10 2 1 The diagram shows a spinner with six numbered sections. Some of the sections are shaded. Each time the spinner is spun it stops on one of the six sections. It is equally likely that it stops on any one of the sections. (a) The spinner is spun once. Find the probability that it stops on (i) a shaded section, Answer(a)(i) [1] (ii) a section numbered 1, Answer(a)(ii) [1] (iii) a shaded section numbered 1, Answer(a)(iii) [1] (iv) a shaded section or a section numbered 1. Answer(a)(iv) [1] For (b) The spinner is now spun twice. Examiner's Use Find the probability that the total of the two numbers is (i) 20, Answer(b)(i) [2] (ii) 11. Answer(b)(ii) [2] (c) (i) The spinner stops on a shaded section. Find the probability that this section is numbered 2. Answer(c)(i) [1] (ii) The spinner stops on a section numbered 2. Find the probability that this section is shaded. Answer(c)(ii) [1] (d) The spinner is now spun until it stops on a section numbered 2. 16 The probability that this happens on the nth spin is . 243 Find the value of n. Answer(d) n = [2]

Mark scheme: 3 Throughout this question isw any cancelling or changing to other forms, after correct answer seen. Penalty of –1 for 2 sf decimals or percentages. Do not accept ratio or worded forms. 4 (a) (i) oe (0.667) 1 Allow 0.6666 – 0.6667 6 3 (ii) oe 1 6 2 (iii) oe (0.333) 1 Allow 0.3333… 6 5 (iv) oe (0.833) 1 Allow 0.8333… 6 1 1 1 (b) (i) oe (0.0278) 2 Allow 0.02777 – 0.02778, M1 for × 36 6 6 6 3 1 (ii) oe (0.167) www 2 2 Allow 0.1666 – 0.1667, M1 for × × 2 oe 36 6 6 1 (c) (i) oe 1 4 1 (ii) oe 1 2 n 4  2  (d) 5 (but not from rounding) 2 M1 for repeating × oe e.g.   6  3  SC1 f l i −7  k 

More questions on Introduction to probability

4 Examiner's y Use 9 8 7 6 5 V 4 3 2 T 1 x –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 9 –1 –2 –3 –4 –5 U –6 (a) On the grid, draw  −7  (i) the translation of triangle T by the vector   , [2]  3  (ii) the rotation of triangle T about (0, 0), through 90° clockwise. [2] (b) Describe fully the single transformation that maps (i) triangle T onto triangle U, Answer(b)(i) [2] (ii) triangle T onto triangle V. Answer(b)(ii) [3] For (c) Find the 2 by 2 matrix which represents the transformation that maps Examiner's Use (i) triangle T onto triangle U,           Answer(c)(i)   [2] (ii) triangle T onto triangle V,           Answer(c)(ii)   [2] (iii) triangle V onto triangle T.           Answer(c)(iii)   [1]

Mark scheme:  7   k  4 (a) (i) Triangle with vertices (–4, 4), (–1, 4), 2 SC1 for translation  k  or  3  (–1, 6) (ii) Triangle with vertices (1, –3), (1, –6), 2 SC1 two correct vertices or 90° anticlockwise (3, –6) about (0, 0) (b) (i) Reflection only 1 Marks independent but must be single y = –x oe 1 transformation to score any marks (ii) Stretch only 1 Marks independent but must be single x-axis oe invariant 1 transformation to score any marks (factor) 3 1 IGCSE – May/June 2010 0580 42  0 − 1  (c) (i)  − 1 0  2 B1 each column  1 0  (ii)  0 3  ft 2ft ft factor in (b)(ii) only if stretch and can recover to correct matrix SC1ft for right-hand column 1 1 0   n  0  0   1 0 1 0     n (iii) 1 ft to 1 0     3 0 0 n  ft 1ft  n  to  1  or  1   0  0  n ≠ 0, ± 1 1 for , allow 0.33 or better 3 180 2 + 115 2 − 90 2 2

More questions on Transformations

5 Examiner's North Use A NOT TO SCALE 180 km 115 km 90 km T H 30° 70° R The diagram shows some straight line distances between Auckland (A), Hamilton (H), Tauranga (T) and Rotorua (R). AT = 180 km, AH = 115 km and HT = 90 km. (a) Calculate angle HAT. Show that this rounds to 25.0°, correct to 3 significant figures. Answer(a) [4] (b) The bearing of H from A is 150°. Find the bearing of (i) T from A, Answer(b)(i) [1] (ii) A from T. Answer(b)(ii) [1] For (c) Calculate how far T is east of A. Examiner's Use Answer(c) km [3] (d) Angle THR = 30° and angle HRT = 70°. Calculate the distance TR. Answer(d) km [3] (e) On a map the distance representing HT is 4.5cm. The scale of the map is 1 : n. Calculate the value of n. Answer(e) n = [2]

Mark scheme: 180 + 115 90 5 (a) (cos) M2 M1 for correct implicit expression 902 = …… 2 × 180 × 115 24.98 – 24.99 A2 A1 for (cos) = 0.9064… (b) (i) 125(.0….) ft 1ft ft 150 – their (a) (ii) 305(.0….) ft 1ft ft 180 + their (b)(i) (c) 180sin (54.98 to 55) M2 B1 for 54.98 to 55 or 35 to 35.02 soi in correct or 180cos (35 to 35.02) oe position. or 180sin (360 – their (b)(ii)) Provided either angle is acute or 180cos(their (b)(i) – 90) oe 147(.4….) cao www 3 A1 90 sin 30 TR 90 (d) M2 M1 for = or other correct sin 70 sin 30 sin 70 implicit equation 47.9 (47.88 – 47.89) cao www 3 A1 (e) 2 000 000 oe 2 Allow 1 : 2 000 000 as answer. SC1 figs 2 in answer which could be a ratio. 4 3

More questions on Non-right-angled triangles

6 A spherical ball has a radius of 2.4 cm. Examiner's Use (a) Show that the volume of the ball is 57.9 cm3, correct to 3 significant figures. 4 3 [The volume V of a sphere of radius r is V = πr . ] 3 Answer(a) [2] (b) NOT TO SCALE Six spherical balls of radius 2.4 cm fit exactly into a closed box. The box is a cuboid. Find (i) the length, width and height of the box, Answer(b)(i) cm, cm, cm [3] (ii) the volume of the box, Answer(b)(ii) cm3 [1] (iii) the volume of the box not occupied by the balls, Answer(b)(iii) cm3 [1] (iv) the surface area of the box. Answer(b)(iv) cm2 [2] For (c) Examiner's Use NOT TO SCALE The six balls can also fit exactly into a closed cylindrical container, as shown in the diagram. Find (i) the volume of the cylindrical container, Answer(c)(i) cm3 [3] (ii) the volume of the cylindrical container not occupied by the balls, Answer(c)(ii) cm3 [1] (iii) the surface area of the cylindrical container. Answer(c)(iii) cm2 [3]

Mark scheme: 4 3 6 (a) π × 4.2 M1 Must see method 3 57.87 – 57.92 to at least 4 figures A1 (b) (i) 14.4, 9.6, 4.8 1, 1, 1 Any order (ii) 664 (663.5 – 663.6) ft 1ft (iii) 315 or 316 or 317 (315.2 – 316.8) ft 1ft ft their (b)(ii) – 6 × ‘57.9’ (only if positive) (iv) 507 (506.8 – 506.9) ft 2ft M1 for (14.4 × 9.6 + 14.4 × 4.8 + 9.6 × 4.8) × 2 or their 3 lengths. IGCSE – May/June 2010 0580 42 (c) (i) Height seen or implied as 6 × 4.8 M1 or better π × 2.42 × their height M1 Indep 521 (520.8 – 521.3) www 3 A1 (ii) 174 or 173 (173.2 – 174.1) ft 1ft ft their (c)(i) – 6 × ‘57.9’ only if positive (iii) 470 – 471 cao www 3 3 M1 for 2 × π × 2.42 (36.17 to 36.2), and M1 indep for π × 4.8 × their height from (c)(i)

More questions on Surface area and volume

7 200 students were asked how many hours they exercise each week. Examiner's Use The table shows the results. Time (t hours) 0ItY5 5ItY10 10ItY15 15ItY20 20ItY25 25ItY30 30ItY35 35ItY40 Number of 12 15 23 30 40 35 25 20 students (a) Calculate an estimate of the mean. Answer(a) h [4] (b) Use the information in the table above to complete the cumulative frequency table. Time (t hours) t Y=5 t Y=10 t Y=15 t Y=20 t Y=25 t Y=30 t Y=35 t Y=40 Cumulative frequency 12 27 50 80 120 200 [1] For Examiner's Use 200 180 160 140 120frequency 100 80Cumulative 60 40 20 t 0 5 10 15 20 25 30 35 40 Time (t hours) (c) On the grid, draw a cumulative frequency diagram to show the information in the table in part (b). [4] (d) On your cumulative frequency diagram show how to find the lower quartile. [1] (e) Use your cumulative frequency diagram to find (i) the median, Answer(e)(i) [1] (ii) the inter-quartile range, Answer(e)(ii) [1] (iii) the 64th percentile, Answer(e)(iii) [1] (iv) the number of students who exercise for more than 17 hours. Answer(e)(iv) [2]

Mark scheme: 7 (a) 12 × 2.5 + 15 × 7.5 + 23 × 12.5 + 30 × M1 mid-values any three soi 17.5 + 40 × 22.5 + 35 × 27.5 + 25 × M1 Use of Σfx dep on x anywhere in each interval 32.5 + 20 × 37.5 (including lower bound) – allow 2 slips or omissions ÷ 200 M1 Depend on second M 21.9 www 4 A1 (b) 155, 180 1 (c) 8 points plotted ft, ignoring (0, 0) P3ft P2ft for 6 or 7 plotted , P1ft for 4 or 5 plotted Reasonable increasing curve or C1ft Condone starting at (5, 12) and ft only if shape polygon through their 8 points correct. (d) Either horizontal or vertical line at 1 least 1 cm long at y = 50 on the curve (e) (i) 22 – 23 1 (ii) 13.5 – 14.5 1 (iii) 25.5 – 26.5 1 (iv) 136 – 140 must be integer 2 SC1 for 60 – 64 seen and must be integer 2 2

More questions on Statistical charts and diagrams

8 (a) y is 5 less than the square of the sum of p and q. Examiner's Use Write down a formula for y in terms of p and q. Answer(a) y = [2] (b) The cost of a magazine is $x and the cost of a newspaper is $(x – 3). The total cost of 6 magazines and 9 newspapers is $51. Write down and solve an equation in x to find the cost of a magazine. Answer(b) $ [4] For (c) Bus tickets cost $3 for an adult and $2 for a child. Examiner's Use There are a adults and c children on a bus. The total number of people on the bus is 52. The total cost of the 52 tickets is $139. Find the number of adults and the number of children on the bus. Answer(c) Number of adults = Number of children = [5]

Mark scheme: 8 (a) (p + q)2 – 5 oe final answer 2 SC1 for (p + q)2 oe seen (b) 6x + 9(x – 3) = 51 or better B3 B2 for 6x + 9(x – 3) or B1 for 6x or 9(x – 3) 5.2(0) final answer B1 5.2(0) ww is B1 only (c) a + c = 52 oe B1 Condone consistent use of other variables 3a + 2c = 139 oe B1 or M3 for 3a + 2(52 – a) = 139 or 3(52 – c) + 2c = 139 o.e. Correctly eliminating a or c. M1 Allow one numerical slip. 35 A1 If A0, SC1 for 17, 35 17 A1 IGCSE – May/June 2010 0580 42

More questions on Equations

9 (a) Examiner's A 9 cm B Use x° y° NOT TO SCALE X 3 cm C D E 6 cm The lines AB and CDE are parallel. AD and CB intersect at X. AB = 9 cm, CD = 6 cm and DX = 3 cm. (i) Complete the following statement. Triangle ABX is to triangle DCX. [1] (ii) Calculate the length of AX. Answer(a)(ii) AX = cm [2] (iii) The area of triangle DCX is 6 cm2. Calculate the area of triangle ABX. Answer(a)(iii) cm2 [2] (iv) Angle BAX = x° and angle ABX = y°. Find angle AXB and angle XDE in terms of x and/or y. Answer(a)(iv) Angle AXB = Angle XDE = [2] For (b) R Examiner's Use 35° NOT TO SCALE Q O 42° S P P, Q, R and S lie on a circle, centre O. Angle OPS = 42° and angle PRQ = 35°. Calculate (i) angle POS, Answer(b)(i) Angle POS = [1] (ii) angle PRS, Answer(b)(ii) Angle PRS = [1] (iii) angle SPQ, Answer(b)(iii) Angle SPQ = [1] (iv) angle PSQ. Answer(b)(iv) Angle PSQ = [1] (c) The interior angle of a regular polygon is 8 times as large as the exterior angle. Calculate the number of sides of the polygon. Answer(c) [3]

Mark scheme: 9 (a) (i) Similar 1 Allow enlargement AX 9 (ii) 4.5 2 M1 for = oe 3 6 2 2  3   2  (iii) 13.5 cao 2 M1 for   or   oe e.g. using base and  2   3  height but other methods must be complete (iv) 180 – x – y oe B1 180 – x oe B1 (b) (i) 96 1 (ii) 48 ft 1ft ft 0.5 their (b)(i) (iii) 97 ft 1ft ft 145 – their (b)(ii) (iv) 35 1 180( n − 2) (c) 20n = 360 oe or = 160 oe M2 M1 for 9e = 180 oe allow diagram to show this n if reasonably clear or 180(n – 2) = 8 × 360 oe 8× 360  360  360 or M1 for 8 × 360 or or 8  = 180 − n  n  n 18 www 3 A1

More questions on Similarity

10 Examiner's Use The diagrams show some polygons and their diagonals. (a) Complete the table. Number of sides Name of polygon Total number of diagonals 3 triangle 0 4 quadrilateral 2 5 5 6 hexagon 9 7 heptagon 14 8 [3] (b) Write down the total number of diagonals in (i) a decagon (a 10-sided polygon), Answer(b)(i) [1] (ii) a 12-sided polygon. Answer(b)(ii) [1] For 1 Examiner's (c) A polygon with n sides has a total of n ( n − q ) diagonals, where p and q are integers. Use p (i) Find the values of p and q. Answer(c)(i) p = q = [3] (ii) Find the total number of diagonals in a polygon with 100 sides. Answer(c)(ii) [1] (iii) Find the number of sides of a polygon which has a total of 170 diagonals. Answer(c)(iii) [2] (d) A polygon with n + 1 sides has 30 more diagonals than a polygon with n sides. Find n. Answer(d) n = [1]

Mark scheme: 10 (a) Pentagon 1 Octagon 20 1, 1 (b)(i) 35 1 (ii) 54 1 (c)(i) p = 2, q = 3 3 M1 for substituting a value of n e.g. 1 4( 4 −q ) = 2 n ≥ 3 p or M1 for number of diagonals from one vertex is n – 3 (allow in words) and B1 for one correct value. If 0, SC1 for n ( n − 3) seen. 2 (ii) 4850 ft 1ft ft their (c)(i) allow only if ft calculates to a positive integer. (iii) 20 cao 2 SC1 for answer of 17 or M1 for their formula = 170 (d) 31 cao 1

More questions on Ratio and proportion