Cambridge IGCSE Mathematics (with coursework) 0581 — 2009 Oct/Nov Paper 4 · Variant 1

This document consists of 21 printed pages and 3 blank pages. IB09 11_0581_04/3RP © UCLES 2009 [Turn over *0774771730* For Examiner's Use UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education MATHEMATICS 0581/04 Paper 4 (Extended) October/November 2009 2 hours 30 minutes Candidates answer on the Question Paper. Additional Materials: Electronic calculator Geometrical instruments Mathematical tables (optional) Tracing paper (optional) READ THESE INSTRUCTIONS FIRST Write your Centre number, candidate number and name on all the work you hand in. Write in dark blue or black pen. You may use a soft pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer all questions. If working is needed for any question it must be shown below that question. Electronic calculators should be used. If the degree of accuracy is not specified in the question, and if the answer is not exact, give the answer to three significant figures. Give answers in degrees to one decimal place. For π use either your calculator value or 3.142. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question. The total of the marks for this paper is 130. www.XtremePapers.com

2 © UCLES 2009 0581/04/O/N/09 For Examiner's Use 1 Chris goes to a shop to buy meat, vegetables and fruit. (a) (i) The costs of the meat, vegetables and fruit are in the ratio meat : vegetables : fruit = 2 : 2 : 3. The cost of the meat is $2.40. Calculate the total cost of the meat, vegetables and fruit. Answer(a)(i) $ [2] (ii) Chris pays with a $20 note. What percentage of the $20 has he spent? Answer(a)(ii) % [2] (b) The masses of the meat, vegetables and fruit are in the ratio meat : vegetables : fruit = 1 : 8 : 3. The total mass is 9 kg. Calculate the mass of the vegetables. Answer(b) kg [2]

3 © UCLES 2009 0581/04/O/N/09 [Turn over For Examiner's Use For Examiner's Use (c) Calculate the cost per kilogram of the fruit. Answer(c) $ [3] (d) The cost of the meat, $2.40, is an increase of 25% on the cost the previous week. Calculate the cost of the meat the previous week. Answer(d) $ [2]

4 © UCLES 2009 0581/04/O/N/09 For Examiner's Use 2 y x 8 7 6 5 4 3 2 1 –1 –2 –3 –4 –5 –6 0 –1 1 2 3 4 5 6 7 –2 –3 –4 –5 –6 U X V W T (a) Describe fully the single transformation which maps (i) triangle T onto triangle U, Answer(a)(i) [2] (ii) triangle T onto triangle V, Answer(a)(ii) [3]

5 © UCLES 2009 0581/04/O/N/09 [Turn over For Examiner's Use (iii) triangle T onto triangle W, Answer(a)(iii) [3] (iv) triangle U onto triangle X. Answer(a)(iv) [3] (b) Find the matrix representing the transformation which maps (i) triangle U onto triangle V, Answer(b)(i) [2] (ii) triangle U onto triangle X. Answer(b)(ii) [2]

6 © UCLES 2009 0581/04/O/N/09 For Examiner's Use 3 1 1 6 7 11 12 Six cards are numbered 1, 1, 6, 7, 11 and 12. In this question, give all probabilities as fractions. (a) One of the six cards is chosen at random. (i) Which number has a probability of being chosen of 1 3 ? Answer(a)(i) [1] (ii) What is the probability of choosing a card with a number which is smaller than at least three of the other numbers? Answer(a)(ii) [1] (b) Two of the six cards are chosen at random, without replacement. Find the probability that (i) they are both numbered 1, Answer(b)(i) [2] (ii) the total of the two numbers is 18, Answer(b)(ii) [3]

7 © UCLES 2009 0581/04/O/N/09 [Turn over For Examiner's Use (iii) the first number is not a 1 and the second number is a 1. Answer(b)(iii) [2] (c) Cards are chosen, without replacement, until a card numbered 1 is chosen. Find the probability that this happens before the third card is chosen. Answer(c) [2] (d) A seventh card is added to the six cards shown in the diagram. The mean value of the seven numbers on the cards is 6. Find the number on the seventh card. Answer(d) [2]

8 © UCLES 2009 0581/04/O/N/09 For Examiner's Use 4 A B C 140 m 180 m 240 m NOT TO SCALE The boundary of a park is in the shape of a triangle ABC. AB = 240 m, BC = 180 m and CA = 140 m. In part (a), show clearly all your construction arcs. (a) (i) Using a scale of 1 centimetre to represent 20 metres, construct an accurate scale drawing of triangle ABC. The line AB has already been drawn for you. A B [2] (ii) Using a straight edge and compasses only, construct the bisector of angle ACB. Label the point D, where this bisector meets AB. [2] (iii) Using a straight edge and compasses only, construct the locus of points, inside the triangle, which are equidistant from A and from D. [2] (iv) Flowers are planted in the park so that they are nearer to AC than to BC and nearer to D than to A. Shade the region inside your triangle which shows where the flowers are planted. [1]

9 © UCLES 2009 0581/04/O/N/09 [Turn over For Examiner's Use In part (b), use trigonometry. You must show your working and must NOT use any measurements from your construction in part (a). (b) (i) Show clearly that angle ACB is 96.4°. Answer(b)(i) [3] (ii) Calculate the area of the park. Answer(b)(ii) m2 [2] (iii) Use the sine rule to calculate angle ABC. Answer(b)(iii) Angle ABC = [3]

10 © UCLES 2009 0581/04/O/N/09 For Examiner's Use 5 D A C B P Q S R (x + 3) cm x cm (2x + 5) cm (x + 4) cm NOT TO SCALE The diagram shows two rectangles ABCD and PQRS. AB = (2x + 5) cm, AD = (x + 3) cm, PQ = (x + 4) cm and PS = x cm. (a) For one value of x, the area of rectangle ABCD is 59 cm2 more than the area of rectangle PQRS. (i) Show that x2 + 7x − 44 = 0. Answer(a)(i) [3] (ii) Factorise x2 + 7x − 44. Answer(a)(ii) [2] (iii) Solve the equation x2 + 7x − 44 = 0. Answer(a)(iii) x = or x = [1] (iv) Calculate the size of angle DBA. Answer(a)(iv) Angle DBA = [2]

11 © UCLES 2009 0581/04/O/N/09 [Turn over For Examiner's Use (b) For a different value of x, the rectangles ABCD and PQRS are similar. (i) Show that this value of x satisfies the equation x2 − 2x − 12 = 0. Answer(b)(i) [3] (ii) Solve the equation x2 − 2x − 12 = 0, giving your answers correct to 2 decimal places. Answer(b)(ii) x = or x = [4] (iii) Calculate the perimeter of the rectangle PQRS. Answer(b)(iii) cm [1]

12 © UCLES 2009 0581/04/O/N/09 For Examiner's Use 6 50 40 30 20 10 –10 –20 –30 0 –1 1 2 –2 –3 –4 –5 y x A B y = f(x) y = g(x) The graphs of y = f(x) and y = g(x) are shown above. (a) Find the value of (i) f(−2), Answer(a)(i) [1] (ii) g(0). Answer(a)(ii) [1]

13 © UCLES 2009 0581/04/O/N/09 [Turn over For Examiner's Use (b) Use the graphs to solve (i) the equation f(x) = 20, Answer(b)(i) x = or x = [2] (ii) the equation f(x) = g(x), Answer(b)(ii) x = or x = [2] (iii) the inequality f(x) < g(x). Answer(b)(iii) [1] (c) Use the points A and B to find the gradient of y = g(x) as an exact fraction. Answer(c) [2] (d) On the grid, draw the graph of y = g(x) − 10. [2] (e) (i) Draw the tangent to the graph of y = f(x) at ( −3, −27 ). [1] (ii) Write down the equation of this tangent. Answer(e)(ii) [1] (f) A region, R, contains points whose co-ordinates satisfy the inequalities −3 Y x Y −2, y Y 40 and y [ g(x). On the grid, draw suitable lines and label this region R. [2]

14 © UCLES 2009 0581/04/O/N/09 For Examiner's Use 7 60° A B O 24 cm NOT TO SCALE (a) The sector of a circle, centre O, radius 24 cm, has angle AOB = 60°. Calculate (i) the length of the arc AB, Answer(a)(i) cm [2] (ii) the area of the sector OAB. Answer(a)(ii) cm2 [2] (b) The points A and B of the sector are joined together to make a hollow cone as shown in the diagram. The arc AB of the sector becomes the circumference of the base of the cone. O A B NOT TO SCALE 24 cm

15 © UCLES 2009 0581/04/O/N/09 [Turn over For Examiner's Use Calculate (i) the radius of the base of the cone, Answer(b)(i) cm [2] (ii) the height of the cone, Answer(b)(ii) cm [2] (iii) the volume of the cone. [The volume, V, of a cone of radius r and height h is V = 1 3 πr2h.] Answer(b)(iii) cm3 [2] (c) A different cone, with radius x and height y, has a volume W. Find, in terms of W, the volume of (i) a similar cone, with both radius and height 3 times larger, Answer(c)(i) [1] (ii) a cone of radius 2x and height y. Answer(c)(ii) [1]

16 © UCLES 2009 0581/04/O/N/09 For Examiner's Use 8 Fifty students are timed when running one kilometre. The results are shown in the table. Time (t minutes) 4.0 < t Y 4.5 4.5 < t Y 5.0 5.0 < t Y 5.5 5.5 < t Y 6.0 6.0 < t Y 6.5 6.5 < t Y 7.0 Frequency 2 7 8 18 10 5 (a) Write down the modal time interval. Answer(a) min [1] (b) Calculate an estimate of the mean time. Answer(b) min [4] (c) A new frequency table is made from the results shown in the table above. Time (t minutes) 4.0 < t Y 5.5 5.5 < t Y 6.0 6.0 < t Y 7.0 Frequency 18 (i) Complete the table by filling in the two empty boxes. [1]

17 © UCLES 2009 0581/04/O/N/09 [Turn over For Examiner's Use (ii) On the grid below, complete an accurate histogram to show the information in this new table. 40 30 20 10 0 4 5 6 7 8 Time (minutes) Frequency density t [3] (iii) Find the number of students represented by 1 cm2 on the histogram. Answer(c)(iii) [1]

18 © UCLES 2009 0581/04/O/N/09 For Examiner's Use 9 (a) Solve the equation _ 3 + 4 _ + = 7 4 3 m m . Answer(a) m = [4] (b) (i) y = 3 2 _ _ 1 + 3 x x Find the value of y when x = 5. Answer(b)(i) [1] (ii) Write 3 2 _ _ 1 + 3 x x as a single fraction. Answer(b)(ii) [2]

19 © UCLES 2009 0581/04/O/N/09 [Turn over For Examiner's Use (iii) Solve the equation 3 2 1 _ = _ 1 + 3 x x x . Answer(b)(iii) x = [3] (c) p = _1 t q Find q in terms of p and t. Answer(c) q = [3]

20 © UCLES 2009 0581/04/O/N/09 For Examiner's Use 10 Total Row 1 1 = 1 Row 2 3 + 5 = 8 Row 3 7 + 9 + 11 = 27 Row 4 13 + 15 + 17 + 19 = 64 Row 5 Row 6 The rows above show sets of consecutive odd numbers and their totals. (a) Complete Row 5 and Row 6. [2] (b) What is the special name given to the numbers 1, 8, 27, 64…? Answer(b) [1] (c) Write down in terms of n, (i) how many consecutive odd numbers there are in Row n, Answer(c)(i) [1] (ii) the total of these numbers. Answer(c)(ii) [1] (d) The first number in Row n is given by n2 − n + 1. Show that this formula is true for Row 4. Answer(d) [1]

21 © UCLES 2009 0581/04/O/N/09 For Examiner's Use (e) The total of Row 3 is 27. This can be calculated by (3 × 7) + 2 + 4. The total of Row 4 is 64. This can be calculated by (4 × 13) + 2 + 4 + 6. The total of Row 7 is 343. Show how this can be calculated in the same way. Answer(e) [1] (f) The total of the first n even numbers is n(n + 1). Write down a formula for the total of the first (n – 1) even numbers. Answer(f) [1] (g) Use the results of parts (d), (e) and (f) to show clearly that the total of the numbers in Row n gives your answer to part (c)(ii). Answer(g) [2]

24 Permission to reproduce items where third-party owned material protected by copyright is included has been sought and cleared where possible. Every reasonable effort has been made by the publisher (UCLES) to trace copyright holders, but if any items requiring clearance have unwittingly been included, the publisher will be pleased to make amends at the earliest possible opportunity. University of Cambridge International Examinations is part of the Cambridge Assessment Group. Cambridge Assessment is the brand name of University of Cambridge Local Examinations Syndicate (UCLES), which is itself a department of the University of Cambridge. 0581/04/O/N/09 BLANK PAGE

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education MARK SCHEME for the October/November 2009 question paper for the guidance of teachers 0581 MATHEMATICS 0581/04 Paper 4 (Extended), maximum raw mark 130 This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the examination. It shows the basis on which Examiners were instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners’ meeting before marking began, which would have considered the acceptability of alternative answers. Mark schemes must be read in conjunction with the question papers and the report on the examination. • CIE will not enter into discussions or correspondence in connection with these mark schemes. CIE is publishing the mark schemes for the October/November 2009 question papers for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses. www.XtremePapers.com

Page 2 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – October/November 2009 0581 04 © UCLES 2009 Abbreviations cao correct answer only cso correct solution only dep dependent ft follow through isw ignore subsequent working oe or equivalent SC Special Case soi seen or implied www without wrong working 1 (a) (i) 8.4(0) B2 B1 for 1.2 or 3.6 seen or SC1 for figs 84 in answer (ii) 100 20 × (i) their oe 42 ft www2 M1 A1ft ft their 8.4 × 5 After 0 scored SC1 ft for 58% or 100 20 20 × − (i) their correctly given (b) 6 B2 M1 for 9 or 8 ÷ (1 + 8 + 3) soi (c) 3 2 4.2 × oe (= 3.6 seen) or their (a) (i) ÷ 7 × 3 9 12 3 × oe (= 2.25 seen) 1.6(0) cao www3 M1 M1 A1 (d) 25 .1 40 .2 oe 1.92 www2 M1 A1 Implied by figs 192 [11] 2 (a) (i) Reflection (M), x = 1 B1,B1 If extra transformations given in part (a) then zero scored (ii) Rotation (R) 180 (centre) (1, 0) B1 B1 B1 Must be “rotation”. Allow half turn for 180. Allow other clear forms of (1, 0) (iii) Enlargement (E) (centre) (6, 4) (scale factor) 3 B1 B1 B1 Must be “enlargement” Allow other clear forms of (6, 4) e.g. vector Accept 3 : 1 or 1 : 3 (iv) Shear (H) y-axis invariant oe (factor) –1 B1 B1 B1 Must be “shear” Allow other explanation for invariant but not “parallel to” isw after y-axis invariant seen

Page 3 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – October/November 2009 0581 04 © UCLES 2009 (b) (i)       −1 0 0 1 B2 B1 for correct right–hand column in 2 by 2 matrix (ii)       − 1 1 0 1 ft B2ft Ft only their factor in (a) (iv) provided not zero B1ft for left-hand column in 2 by 2 matrix provided shear factor is not zero or SC1 for       1 1 0 1 if not ft [15] 3 (a) (i) 1 B1 Penalty of –1 in question if any answers given as decimals or percentages (to 3sf) alone, but isw cancelling/conversion after correct answer (ii) 6 3 oe B1 (b) (i) 30 2 oe www2 B2 M1 for 5 1 6 2 × (ii) 6–12 and 12–6 and 7–11 and 11–7 soi k × 6 1 × 5 1 for k = integer 30 4 oe www3 M1 M1 A1 Evidence of all pairs adding up to 18 but no extras e.g. 4/6 × 1/6 Without seeing the first M, 6 4 × 5 1 oe scores M2, 6 2 × 5 1 oe scores M1 (iii) 5 2 6 4 × 30 8 oe www2 M1 A1 (c) 5 2 6 4 6 2 × + oe 30 18 oe cao www2 M1 A1 6 2 + their (b) (iii) (d) 4 B2 M1 for (1 + 1 + 6 + 7 + 11 + 12 + x) ÷ 7 = 6 or better [13]

Page 4 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – October/November 2009 0581 04 © UCLES 2009 4 (a) (i) Accurate triangle with 2 arcs seen, 2 mm accuracy for lines AC and BC B2 SC1 if accurate but no arcs or one arc or if AC and BC are wrong way round with arcs (ii) Accurate bisector of angle ACB, 2° accuracy and both pairs of arcs shown (accept equidistant marks on edges for 1st set of arcs) + must meet AB B2ft Ft their triangle SC1ft if accurate but no/one pair of arcs or short with arcs In both (ii) and (iii) isw (iii) Accurate perpendicular bisector of AD 2 mm accuracy at mid-point and 2° for right angle and shows both sets of arcs + must meet AC B2ft ft their D, which must be on AB SC1ft if accurate but no/one pair of arcs or short with arcs (iv) Correct region shaded cao B1 Dependent on correct triangle, accurate bisectors of angle ACB and side AD with correct D (b) (i) (cos C) = 180 140 2 240 180 140 2 2 2 × × − + oe – 0.111(1)…or better or 96.37 to 96.38 M2 E1 (–5600/50400 or –14/126) Allow use of 7, 9 and 12 M1 for correct implicit statement Verification using 96.4 scores M2 max Accept 9 1 − but not a non-reduced fraction (ii) 0.5 × 140 × 180 sin (their 96.4) oe 12521 to 12523 or 12 500 or 12520 cao www2 M1 A1 (s = 280), allow use of 7, 9 (31.3…) (iii) (Sin B =) 240 ) 4. 96 sin( 140 their oe 35.4 or 35.42 to 35.44 cao www3 M2 A1 Allow use of 7, 12 M1 for correct implicit statement SC2 for correct answer by other method [15] 5 (a) (i) (x + 3)(2x + 5) – x(x + 4) = 59 oe 2x2 + 6x + 5x + 15 –x2 – 4x = 59 oe x2 + 7x – 44 = 0 M1 A1 E1 Implies M1 (allow 11x for 6x + 5x) Correct conclusion – no errors or omissions (ii) (x + 11)(x – 4) B2 SC1 any other (x + a)(x + b) where a × b = – 44 or a + b = 7 (iii) –11, 4 www ft B1ft Strict ft dep on at least SC1 in (ii) allow recovery if new working seen (iv) tan = 5 ) ( 2 3 ) ( + + + + root ve their root ve their oe 28.3 (00…) ft www2 M1 A1ft Could be alt trig method oe M1 where trig function is explicit ft one of their positive roots (27.4° (27.40 – 27.41) from x = 11)

Page 5 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – October/November 2009 0581 04 © UCLES 2009 (b) (i) 4 5 2 + + x x = x x 3 + oe x2 + 4x + 3x + 12 = 2x2 + 5x x2 – 2x – 12 = 0 M1 A1 E1 Must be seen. Allow ratio or correct products Correct expansion of brackets seen (allow 7x for 4x + 3x) Correct conclusion – no errors or omissions M1 must be seen (ii) )1( 2 ) 12 )( 1( 4 ) 2 ( ) 2 ( 2 − − − ± − − or (x – 1)² –12 – 1 (B1) and x – 1 = ± 13 (B1) – 2.61, 4.61 final answers www4 B1,B1 B1,B1 In square root B1 for (–2)2 – 4(1)(–12) or better If in form r q p + or r q p − , B1 for – (–2) and 2(1) or better If B0, SC1 for –2.6 and 4.6 or both answers correct to 2 or more dps rot – 2.6055…, 4.6055…. (iii) 26.4 (26.42…. to 26.44….) ft B1ft ft 4 × a positive root + 8 [16] 6 (a) (i) –16 B1 (ii) 18 to 19 B1 (b) (i) –4.3 to –4.2, 1.5 to 1.6 B1,B1 (ii) –4.5 to –4.4 , 1.3 to 1.4 B1,B1 (iii) –4.5 to –4.4 < x < 1.3 to 1.4 ft B1ft Ft their (ii). Allow clear worded explanations and condone Y signs (c) 7 30 − oe isw conversion B2 Accept 7 2 4 − , 30/–7 M1 for 30/7 oe fracts, isw conversion or for –30/7 oe soi (d) Ruled line passing within 2 mm of (–5, 30) and (2, 0) B2 B1 for ruled line parallel to g(x). By eye (21° to 25° to horizontal if in doubt) allow broken line (e) (i) Ruled horizontal line through (–3, –27) B1 No daylight, not chord (allow broken) (ii) y = –27 B1 (f) Ruled lines x = –3, x = –2, y = 40 Region enclosed by lines x = –3, x = –2, y = 40 and y = g(x) B1 B1 Long enough to be boundary of region – allow broken or solid ruled lines Allow any clear indication [15]

Page 6 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – October/November 2009 0581 04 © UCLES 2009 7 (a) (i) 360 60 × π × 2 × 24 oe 25.1 (25.12 to 25.14) www2 M1 A1 Accept 8 π (ii) 360 60 × π × 242 oe 301 or 302 or 301.4 to 301.7 www2 M1 A1 Accept 96 π (b) (i) πd = their (a) (i) oe 4 (3.99 – 4.01) cao www2 M1 A1 (ii) 242 – (their radius)2 23.7 (23.66 to 23.67) cao www2 M1 A1 Alt trig method for h explicit Accept 35 4, 140 2, 560 (iii) 3 1 × π × (their r)2 × (their h) 394 – 398 cao www2 M1 A1 Not for h = 24 (c) (i) 27W B1 (ii) 4W B1 If B0, B0 in (c), SC1 for 27 and 4 alone [12] 8 (a) 5.5 < t Y 6 B1 Condone poor notation (b) 4.25, 4.75, 5.25, 5.75, 6.25, 6.75 (2 × 4.25 + 7 × 4.75 + 8 × 5.25 + 18 × 5.75 + 10 × 6.25 + 5 × 6.75) (= 283.5) ÷ 50 or their ∑f 5.67 www4 M1 M1 M1 A1 At least 5 correct mid-values seen ∑fx where x is in the correct interval allow one further slip Depend on second method After M3 allow 5.7 isw conversion to mins/secs and reference to classes (c) (i) 17, 15 B1 (ii) Rectangular bars of heights 11.3 and 15 Correct widths of 1.5 and 1 – no gaps B1ft B1ft B1 ft their 17 divided by 1.5 ft their 15 11.3 plot between 11 and 12 include lines and 15 to be touching the 15 line (iii) 2.5 cao B1 [10]

Page 7 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – October/November 2009 0581 04 © UCLES 2009 9 (a) 3(m – 3) + 4(m + 4) = –7 × 12 3m – 9 + 4m + 16 = –84 –13 www4 M2 A1 A1 Allow all over 12 at this stage M1 for 3(m – 3) + 4(m + 4) seen Allow all over 12 at this stage May be seen in stages (b) (i) 0.5 oe B1 (ii) ) 3 )( 1 ( )1 ( 2 ) 3 ( 3 + − − − + x x x x ) 3 )( 1 ( 11 + − + x x x final answer M1 A1 If brackets not seen allow 3x + 9 – 2x ± 2 as numerator with a correct denominator isw incorrect expansion of denominator if correct brackets seen (iii) 1 ) 3 )( 1 ( ) 11 ( = + − + x x x x ft or x + 11 = x 1 (x – 1)(x + 3) or better ft x2 + 11x = x2 + 3x – x – 3 3 1 − oe cso www3 M1 M1 A1 Must clear one denominator correctly Ft their (b)(ii) dep on fraction in (ii) with (x –1)(x +3) oe as denominator Depend on previous M1 – 0.33(33…) (c) p(q – 1) = t oe pq = t + p p p t + oe final answer www3 M1 M1 M1 Multiplying by (q – 1) Ft their first step e.g. pq only term on one side Ft their 2nd step e.g. dividing by p Note: q – 1 = p t is M2 and then q = p t + 1 is M1 [13] 10 (a) 21 + 23 + 25 + 27 + 29 = 125 31 + 33 + 35 + 37 + 39 + 41 = 216 B1 B1 (b) Cubes B1 (c) (i) n oe B1 (ii) n3 oe B1 (d) 42 – 4 + 1 = 13 www E1 Allow 16 for 42, otherwise all must be seen (e) 7 × 43 + 2 + 4 + 6 + 8 + 10 + 12 B1 All must be seen (f) n(n – 1) final answer oe B1 (g) n(n2 – n + 1) + their (f) n3 – n2 + n + n2 – n = n3 M1 E1 All must be seen, no errors or omissions [10]