Cambridge IGCSE Mathematics - International 0607 — 2013 May/June Paper 4 · Variant 2

0607/42/M/J/13 · 120 marks · ≈135 min

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Question paper20 pages

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Mark scheme7 pages

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Question paper, page 1

This document consists of 19 printed pages and 1 blank page. IB13 06_0607_42/2RP © UCLES 2013 [Turn over *1604463664* For Examiner's Use UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education CAMBRIDGE INTERNATIONAL MATHEMATICS 0607/42 Paper 4 (Extended) May/June 2013 2 hours 15 minutes Candidates answer on the Question Paper. Additional Materials: Geometrical Instruments Graphics Calculator 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. Do not use staples, paper clips, highlighters, glue or correction fluid. You may use a pencil for any diagrams or graphs. DO NOT WRITE IN ANY BARCODES. Answer all the questions. Unless instructed otherwise, give your answers exactly or correct to three significant figures as appropriate. Answers in degrees should be given to one decimal place. For π, use your calculator value. You must show all the relevant working to gain full marks and you will be given marks for correct methods, including sketches, even if your answer is incorrect. The number of marks is given in brackets [ ] at the end of each question or part question. The total number of marks for this paper is 120. www.XtremePapers.com

Question paper, page 2

2 © UCLES 2013 0607/42/M/J/13 Formula List For the equation ax2 + bx + c = 0 x = 2 _ ± 4 2 _ b b ac a Curved surface area, A, of cylinder of radius r, height h. A = 2πrh Curved surface area, A, of cone of radius r, sloping edge l. A = πrl Curved surface area, A, of sphere of radius r. A = 4πr2 Volume, V, of pyramid, base area A, height h. V= 1 3 Ah Volume, V, of cylinder of radius r, height h. V = πr2h Volume, V, of cone of radius r, height h. V = 1 3 πr2h Volume, V, of sphere of radius r. V = 4 3 πr3 = = sin sin sin a b c A B C a2 = b2 + c2 – 2bc cos A Area = 1 2 bc sin A A a B C b c

Question paper, page 3

3 © UCLES 2013 0607/42/M/J/13 [Turn over For Examiner's Use Answer all the questions. 1 (a) (i) Kim’s wage is $720 each month. She spends $196 each month on food. Calculate $196 as a percentage of $720. Answer(a)(i) % [1] (ii) She pays 25% of the $720 in taxes. Find the ratio money spent on food : money paid in taxes. Give your answer in its simplest form. Answer(a)(ii) : [2] (iii) The $720 is an increase of 44% on Kim’s previous wage. Calculate her previous wage. Answer(a)(iii) $ [3] (iv) Next year the $720 will increase by 4%. Calculate next year’s monthly wage. Answer(a)(iv) $ [2] (b) Jay’s monthly wage is $650. Each year Jay’s monthly wage increases by 5%. Calculate the number of years it will take for Jay’s monthly wage to exceed $1000. Answer(b) [3]

Question paper, page 4

4 © UCLES 2013 0607/42/M/J/13 For Examiner's Use 2 (a) 2x 3x 2x + 1 x + 3 NOT TO SCALE The areas of the rectangles are equal. Find the value of x. Show all your working. Answer(a) x = [4] (b) θ y + 1 2y NOT TO SCALE Find the value of y when tan θ = 3 1 . Show all your working. Answer(b) y = [3]

Question paper, page 5

5 © UCLES 2013 0607/42/M/J/13 [Turn over For Examiner's Use (c) Jo walks 10 km at w kilometres per hour. Sam cycles 10 km at (w + 9) kilometres per hour. The difference between the times taken by Jo and Sam is 2 1 2 hours. (i) Show that w2 + 9w – 36 = 0. [4] (ii) Find the time, in hours and minutes, taken by Jo to walk the 10 km. Answer(c)(ii) h min [4]

Question paper, page 6

6 © UCLES 2013 0607/42/M/J/13 For Examiner's Use 3 y x 5 4 3 2 1 –1 –2 –3 –4 –5 0 –1 –2 –3 –4 –5 5 4 3 2 1 L (a) Find the equation of the line L. Answer(a) [2] (b) (i) On the grid, draw the line y = 2x + 4. [2] (ii) On the grid, shade the region where y [ 0 and y [ 2x + 4. [2] (c) P is the point (1, – 4) and Q is the point (3, 2). Find the equation of the line passing through P and Q. Answer(c) [3]

Question paper, page 7

7 © UCLES 2013 0607/42/M/J/13 [Turn over For Examiner's Use 4 The masses of 100 apples are measured. The results are shown in the table. Mass (m grams) 20 < m Y 100 100 < m Y 150 150 < m Y 240 Frequency 28 45 27 (a) Calculate an estimate of the mean mass. Answer(a) g [2] (b) Use the information in the table to complete the histogram. 1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 40 80 120 Mass (grams) 160 200 240 20 60 100 140 180 220 Frequency density m [3]

Question paper, page 8

8 © UCLES 2013 0607/42/M/J/13 For Examiner's Use 5 B P H K 998 km 1060 km 1185 km 988 km 30° NOT TO SCALE The diagram shows some straight line distances between Bangkok (B), Hanoi (H), Phnom Penh (P) and Kuala Lumpur (K). Angle BHP = 30°. (a) Calculate BP and show that it rounds to 535 km, correct to the nearest kilometre. [3]

Question paper, page 9

9 © UCLES 2013 0607/42/M/J/13 [Turn over For Examiner's Use (b) Calculate angle BKP. Answer(b) [3] (c) The bearing of P from K is 020°. Find the bearing of B from K. Answer(c) [1]

Question paper, page 10

10 © UCLES 2013 0607/42/M/J/13 For Examiner's Use 6 R C A B Q P 12 cm 20 cm 6 cm NOT TO SCALE The diagram shows a triangular prism of length 20 cm. The cross-section of the prism is triangle ABC with angle BAC = 90°, AC = 6 cm and AB = 12 cm. (a) Calculate the volume of the prism. Answer(a) cm3 [2]

Question paper, page 11

11 © UCLES 2013 0607/42/M/J/13 [Turn over For Examiner's Use (b) (i) Calculate the total surface area of the prism. Answer(b)(i) cm2 [4] (ii) The surface of the prism is painted at a cost of $0.005 per square centimetre. Calculate the cost of painting the surface of the prism. Answer(b)(ii) $ [1] (c) Calculate the angle between the diagonal line CQ and the base ABQP. Answer(c) [3]

Question paper, page 12

12 © UCLES 2013 0607/42/M/J/13 For Examiner's Use 7 A flight from London, England to Auckland, New Zealand departs at 14 00 on February 7th. The journey takes 2 1 27 hours and the distance is 18 400 km. The time in New Zealand is 13 hours ahead of the time in England. (a) Find the time and the date that the flight arrives in Auckland. Answer(a) Time Date [3] (b) Calculate the average speed of the journey. Answer(b) km/h [1] (c) The cost of a ticket for the flight is 3600 pounds (£). £1 = 2.09 New Zealand dollars (NZD). (i) Calculate the cost of the ticket in NZD. Answer(c)(i) NZD [1] (ii) Calculate the cost of the journey, in NZD per kilometre. Give your answer correct to 2 decimal places. Answer(c)(ii) NZD/km [2]

Question paper, page 13

13 © UCLES 2013 0607/42/M/J/13 [Turn over For Examiner's Use 8 (a) Solve the equation 2 2 3 + = x x . Answer(a) x = or x = [4] (b) Solve the inequality x 2 [ x3 + 2 . Answer(b) [3]

Question paper, page 14

14 © UCLES 2013 0607/42/M/J/13 For Examiner's Use 9 70° 8 cm NOT TO SCALE B A O AB is a chord of the circle centre O. Calculate (a) the length of the chord AB, Answer(a) cm [3] (b) the length of the arc AB, Answer(b) cm [2] (c) the area of the shaded region. Answer(c) cm2 [4]

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15 © UCLES 2013 0607/42/M/J/13 [Turn over For Examiner's Use 10 2 –2 0 180° 360° x y f(x) = cos x g(x) = 2sin     2 x (a) On the diagram, sketch the following graphs. (i) y = f(x) [2] (ii) y = g(x) [2] (b) Write down the equation of the line of symmetry of the graphs. Answer(b) [1] (c) Write down the co-ordinates of the local minimum point on the graph of y = f(x) for 0° Y x Y 360°. Answer(c) ( , ) [2] (d) Write down the period and amplitude of g(x). Answer(d) period = amplitude = [2] (e) Write down the range of g(x) for the following domains. (i) 0° Y x Y 360° Answer(e)(i) [1] (ii) o Answer(e)(ii) [1] (f) Solve the equation f(x) = g(x) for 0° Y x Y 360°. Answer(f) x = or x = [2] (g) Shade the regions on the diagram where y Y f(x) and y [ g(x). [1]

Question paper, page 16

16 © UCLES 2013 0607/42/M/J/13 For Examiner's Use 11 2 5 1 1 1 10 The diagram shows a disc, with six equal sectors, and an arrow. When the disc is spun, each sector is equally likely to stop next to the arrow. (a) The disc is spun. Write down the probability that the sector next to the arrow is labelled with (i) 1 or 2, Answer(a)(i) [1] (ii) an even number, Answer(a)(ii) [1] (iii) a number which is a factor of 10. Answer(a)(iii) [1] (b) The disc is spun twice. (i) Complete the tree diagram by writing the missing probabilities on each branch. 10 first number second number not 10 10 not 10 … … … 10 not 10 … … 1 6 [2]

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17 © UCLES 2013 0607/42/M/J/13 [Turn over For Examiner's Use (ii) Find the probability that the arrow is next to the number 10 twice. Answer(b)(ii) [2] (iii) Find the probability that the arrow is next to the number 10 at least once. Answer(b)(iii) [2] (c) The disc is spun n times until it stops with the number 10 next to the arrow. Find n when the probability that this happens is 7776 625 . Answer(c) n = [2]

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18 © UCLES 2013 0607/42/M/J/13 For Examiner's Use 12 Month Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Temperature (t ˚C) 13 13 15 16 19 23 25 26 24 20 18 13 Rainfall (r mm) 59 49 62 46 25 6 1 3 28 62 63 66 The table shows the average monthly temperature, t, and rainfall, r, in Malaga, Spain. (a) Find the mean, median, upper quartile and range of the average monthly temperatures. Answer(a) mean = °C median = °C upper quartile = °C range = °C [4] (b) (i) Find the equation of the line of regression for this data, giving r in terms of t. Answer(b)(i) r = [2] (ii) Describe the type of correlation between r and t. Answer(b)(ii) [1] (iii) Calculate an estimate of the rainfall when the temperature is 22°C. Answer(b)(iii) [1]

Question paper, page 19

19 © UCLES 2013 0607/42/M/J/13 For Examiner's Use 13 P O Q Y p q X NOT TO SCALE The diagram shows a triangle OPQ. The point X is on PQ so that PX : XQ = 1 : 2. = p and = q. (a) Find in terms of p and q. Give your answer in its simplest form. Answer(a) [2] (b) OQY is a straight line and OY = 2OQ. Find in terms of p and q. Give your answer in its simplest form. Answer(b) [3] (c) p =       k 3 and |p| = 5. Find the two possible values of k. Answer(c) k = or k = [2]

Question paper, page 20

20 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. © UCLES 2013 0607/42/M/J/13 BLANK PAGE

Mark scheme, page 1

CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education MARK SCHEME for the May/June 2013 series 0607 CAMBRIDGE INTERNATIONAL MATHEMATICS 0607/42 Paper 4 (Extended), maximum raw mark 120 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 should be read in conjunction with the question paper and the Principal Examiner Report for Teachers. Cambridge will not enter into discussions about these mark schemes. Cambridge is publishing the mark schemes for the May/June 2013 series for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level components and some Ordinary Level components. www.XtremePapers.com

Mark scheme, page 2

1 2 (a (b (a (b Pa a) ( (i (ii (iv b) a) b) age (i) ii) ii) v) e 2 2 4 5 7 9 2 7 y 3 2 27.2 49 : 500 748 9 o ( 2x 7 2 + y y 3 × 5 1 2 ( : 45 0 8.8[ or 8 2 ( x 1 = + y 2y o. (27 5 f [0] 8.83 + x 3 1 = = y .e. .22 fina ca 3 ( )1 3 1 y + 2… al a ao f 8.8 3 = + 1 ) answ fina 829 ( 3x o.e wer al a to ( + x e. IG © r answ 8.8 3 + or GC Ca wer 830 ) 3 bet M SE amb r 0) o.e tter Mar E – brid e. r rk S M dge Sc ay/ e In che /Ju nter eme une rnat e e 2 tion 201 nal 13 Ex 1 2 3 2 3 M2 B2 M1 B1 B1 xam 2 2 1 1 1 mina S o a o M o M ÷ S M A o o M o If m B o i. y atio SC1 or B answ or 2 M2 or M M1 ÷ 12 SC2 M2 Allo or M or 6 M2 or M f M mis B1 or fi .e c y = ons 1 fo B1 f wer 27.2 for M1 for 2 d 2 fo for ow M1 650 for M1 M0, sin for fina corr 5 1 s 20 or a for r) 2 (2 r 72 for r 72 oes or a r lo 1.5 for mu r gr for , B1 ng b r x2 al an rec 013 ans r 18 27.2 20 r 72 20 s no ans og ( 54 r 65 ulti rap r gr 1 fo brac 2 = nsw ct eq S 3 we 80 s 22. ÷ 1 20 × 1 ot s we (10 for 50 ipli phs rap or cke x 7 = wer qua Syll 06 er 4 see ..) : 1.4 = 1 1.0 spo er 8 000 r 10 × 1 ied tha ph o 2x ets x r 0, atio lab 607 45:4 n ( : 25 4 o 144 4 o oil m 8.85 0/65 000 1.0 d by at i of y 2 ( x x if c o.e , 7 on w bus 7 49 o ma 5 o o.e. 4% o.e. met 5 50) 0/65 5n = y 1. inte y = + x corr e. o wit s or ay b o.e. se im tho ÷ l 50 = 1 .05 erse = 1. )1 + rec or 2 tho 1.0 be i en mp od log in m 100 at ect .05x or ct ex 2(2 ut f 09 ( imp plie g(1. me 00 lea as x 3 r x xpa 2x + fra P (1.0 plie d b .05 etho ast sho (x x ans + 1) ctio Pa 4 088 ed b by 7 ) ods twi ow + x sion ) = ons pe 42 8 to by 749 . ice wn ) 3 ns f 3(x s le er o 1. the 9. co Co foll x + eadi 089 eir r orre ond low + 3) ing 9) rati ectl done w ) o. g to io y e e.

Mark scheme, page 3

3 (c (a (b Pa ) ( (i a) b) ( (i age (i) ii) (i) ii) e 3 1 w 2 b o w w 3 y R 3 w 10 10 + w 2× bett or (w w 2 w 3 h = y Rul so 9 0 + ( 10 ter 90 + w 9 + 20 − = led oi s (w 9 0 + 9w mi − x lin soi 9 + ) = 3 − w in 1 − ne th ) 9 − 2 = 36 ww o. hro 2× − 2 1 = ww .e. f oug 10 × o. 0 w fin gh ( IG © 0w .e. al a (– 2 GC Ca 5 = ans 2, 0 M SE amb 5w swe 0) a Mar E – brid (w w er and rk S M dge 9 + d (0 Sc ay/ e In )) 9 , 4) che /Ju nter o.e ) eme une rnat e. o e e 2 tion or 201 nal 2 13 Ex B1 B1 M1 E1 4 2 2 2FT xam 1 1 1 1 T mina If w i. c a a i. E li B M q c o a s s T tr (u B h B B B o o F S atio f no wx = .e. corr allo also .e. Esta ine B3 M2 qua com or M a + ubs imp Tria rial unl B1 F hou B1 B1 B1 or or ti FT SC1 ons ot s = 1 cor rect w o al cor abli e wi for for dra mple M1 b = stit plif al a ls f less FT urs a for for for iny on 1 fo s 20 see 0 i rrec t fo 10( llow rrec ish ith r w r (w atic etin for = 9 tuti fyin and for s co T fo and r y r ru r y pa ly i or c 013 en o is n ct m orm (w + w o ct c hed no = 3 + w c sh ng r (w or ons ng im com orre or 1 d m − = uled art n if p corr S 3 only not mul mat + 9 over coll wi err 3 or 12 + how sq w + sk s in or mpro mp ect 0 ÷ minu x − d lin not pos rec Syll 06 y im suf ltip 9) – r co lect ith ror r fo )( 2 w wing quar ) a + ketc n qu (w ove pari t an ÷ th ute c + ne t t sh itiv ct a lab 607 mp ffic plic – 10 om tion at l rs o or w − w g z re r )(w ch o uad + w em son nsw heir s c o. thr had ve g rea bus 7 lied cien cati 0w mmo n o leas or o w = )3 − zero reac w + of q drat ) 2 2 9 ent n o wer r po e. oug ded gra a un s d b nt. on = 2 on d of tw st o omi = 3 or os o chi ) b qua tic 2 − t – f di fou osi o gh die nsh by f of 2.5 den wo one issi or go or q ing wi adra for 4 81 − allo ista und tiv r y (0, ent hade furt f an w(w nom ter e m ons – 1 ood qua g − ith atic rmu 3 = ow anc d w e ro = y , 4) and ed P ther eq w + min rms more s 12 sk adra 9 ± ab c or ula 36 o w M ces with oot kx or d c Pa 4 r w quat + 9 nato s e in w ketc atic 2 ± = – r co be o.e M2 f and h 1 t co 1 − r gr uts pe 42 work tion ) o or nter ww ch o c fo 22 – 3 orre efor e. for d ti or 2 orre 1 o adi s x- er kin n in .e. rme w of orm 5 36 o ect re at ime 2 tr ectl o.e. ient axi ng. n th edia mula or lea es rial ly i . se t of is e.g he ate a or ast 3 ls) into een f 2 g. r 3 o n

Mark scheme, page 4

Page 4 Mark Scheme Syllabus Paper IGCSE – May/June 2013 0607 42 © Cambridge International Examinations 2013 (c) 7 3 − = x y o.e. 3 B2 for c x y + = 3 o.e. or 7 − = kx y o.e. or M1 for rise/run = 1 3 4 2 − − − o.e. and M1 for correct method for finding c Answer 3x – 7 implies M1 M1 4 (a) 125.7 or 126 2 M1 if at least 2 mid-values seen (60, 125 and 195) or 2 from 1680, 5625, 5265 or sum of 12570 (b) Columns from 100 to 150 and 150 to 240 Heights 0.9 and 0.3 1 1,1 Accept freehand 5 (a) 534.6… www 3 3 M1 for 9882 + 10602 – 2 × 988 × 1060 cos 30 A1 for 285800 to 285802 or 286000 (b) 998 1185 2 535 998 1185 ] [cos 2 2 2 × × − + = 26.6 (26.62 to 26.65) M2 A1 Allow use of 534.6… for 535 M1 for correct implicit statement Strictly dependent on at least M1 SC2 if correct without working (c) 353 (353.3 to353.4) 1FT FT 380 – their (b) only if answer between 270 and 360. 6 (a) 720 2 M1 for 0.5 × 12 × 6 × 20 o.e. (b) (i) 700 (700.2 to 700.4) 4 Allow 432 + 120 5 as final answer for full marks M1 for [BC2]= 122 + 62 M2 for BC × 20 + 12 × 20 + 6 × 20 + 2 × area triangle ABC or M1 if one of the five areas missing or is incorrect (ii) 3.5[0] (3.501 to 3.502) 1FT FT their (i) × 0.005 (c) 14.4 (14.42 to 14.43) 3 M1 for 202 + 122 (544) or 202 + 122 + 62 (580) (Square roots 23.323…, 24.08…) M1 for tan = 2 2 12 20 6 + their or sin = 2 2 2 6 12 20 6 + + their or cos = 2 2 2 2 2 6 12 20 12 20 + + + their their o.e.

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7 8 9 (a (b (c (a (b (a (b (c Pa a) b) ) ( (i a) b) a) b) ) age (i) ii) e 5 [ 6 7 0 – 0 x x x 9 9 9 5 0]6 669 752 0.41 – 1. 0.7 x Y x > x Y 9.18 9.77 9.02 6 30 9 (6 24 1 .49 798 Y –1 0 Y=0. 8 (9 7 or 2 to 0 o 669 cao (– (0 1.4 798 9.1 r 9 o 9. o.e., 9.0 o fi 1.4 0.79 9 o 8 o 77… .78 .03 , [F to inal 496 976 or – or 0 …) 8 (9 3 Feb 66 l an 6 to 6… –1.4 0.79 ) 9.77 b] 9 9.1 nsw o – …) 496 976 73 t IG © 9th 1) wer 1.4 6 to 6… to 9 GC Ca 494 o –1 9.7 M SE amb 4) 1.4 75 w Mar E – brid 94 …. ww rk S M dge .) w 4 Sc ay/ e In che /Ju nter eme une rnat e e 2 tion 201 nal 2 1 1 13 Ex 3 1 1 2FT B2 B1 B1 1FT 1 1FT 3 2 4 xam T 2 1 1 T T mina B B o s F 0 B S o C F F M o w M A e M o M atio B1 B2 or M een FT 0.40 B1 Ske or a Con FT FT M2 82 or M with M1 Allo exac M3 or M M1 ons for for M1 n the 089 for etch a di ndo the the for 2 + M1 hou for ow ct for M1 for (3 s 20 r [F r [0 for eir 9… r po h co ffe one eir eir r 2 82 + for ut s r 3 7 9 28 r 3 7 for r 0 9.0 013 Feb] ]6 r 27 (d) i.e oor ould ren < f (a) (a) × 8 − r A squa 360 70 9 8π 360 70 r 3 7 .5 × 09 t S 3 ] 9t 30 7 2 1 )(i) no qu d b nt fu for if if 8 × 8.2 8 2 AB are 0 ×π o.e 0 ×π 360 70 × 8 to 3 Syll 06 th. o.e + ÷ ot 2 uali be w func r Y on on × si 8.8 B = e ro × π e. a × π 0 ×π 8 × 39. lab 607 e. 13 184 2 dp ty s with ctio etc e n e p n35 cos sin oot 16 × as f 8 × × π 8 s 10… bus 7 so 400 p ske h d on e c nega posi 5 o 70 s n 35 6o.e fina – 8 × a sin7 …) s i or 0 B etch diffe e.g ativ itiv o.e. 0 ( 5 o. e. al a 0.5 and 70 ) r B 1F h fere g. y ve r ve r . e ( 8 e. ans 5 × d o.e B1 f FT f ent w = y roo root .g. . 84 or we 8 × e. ( P for for win 2 x ot t 22 . r ab er b × 8 (30 Pa 4 17 0.4 ndo −x … bov but sin .1 o pe 42 30 409 ow. 3 − x ) ve e mu n70 or 3 er 0 o 9 or . 2 − exp ust 0 o 30. or 0 r 2 [1 pres be o.e .07 3 0 14] ssio . …. 00 on ..)

Mark scheme, page 6

10 11 0 (a (b (c (d (e (f) (g (a (b Pa a) ( (i b) ) d) ) ( (i ) g) a) ( (i (ii b) ( (i age (i) ii) (i) ii) (i) ii) ii) (i) ii) e 6 x ( 7 2 0 – 4 C 1 3 6 x = 180 720 2 0 Y –2 Y 42.9 Cor 6 4 6 2 1 o 6 5 , 36 1 18 0, – 0 Y g( Y g 9 (4 rrec o. o. o.e 6 1 cao 0 – 1 (x) g(x 42. ct a .e. .e. . an o ) Y ) Y 94… area nd 2 Y 2 …) a sh 6 5 , ), 3 had 6 1 17 ded an IG © (3 . nd GC Ca 17. 6 5 M SE amb .0 t Mar E – brid to 3 rk S M dge 317 Sc ay/ e In 7.1) che /Ju nter ) eme une rnat e e 2 tion 201 nal 13 Exam minatio 1 1 ons 2 2 1 1,1 1 1 1 1 1,1 1 1 1 1 2 2 s 2013 B1 B1 Co Co Co Co or in Fo pe fo Do ca Do SC B1 po M (0 0. S 3 1 fo 1 fo ond ond ond ond r y f equ or a erc or 3 o n anc o n C1 1 fo osit M1 f 0.02 027 Syll 06 or i or i don don don don for ual all cent 3 sf not cell not for or a tion for 278 778 lab 607 ina ina ne π ne π ne 4 ne s g(x itie pa tag f. pe ing ac r 0. any n the 8 or 8) bus 7 ccu ccu π f π f 4 π stri x) a es in arts ges ena g or cep .33 y on eir r 0 s urat urat for for for ct i and n p s ac wi lise r co pt r if ne 6 1 × 02 . te s te s 180 180 r 72 ine d al part cce ith e in onv rat no of t th × 7 2 ske ske 0 0 20 qua llow ts ( ept the nco ver tios fra the heir or P etch etch alit w s i) a de e u orre rtin s or acti e 6 5 6 1 r 0.0 Pa 4 h of h of ties sepa and cim usua ect ng. r w ion in 027 pe 42 f co f 2 s, al ara d (ii mal al r t word n se a c 777 er osx 2sin llow ate i) ls o rul ds. en cor 7 to x n( 2 x w x or les . rrec 2 x ) x ct

Mark scheme, page 7

Page 7 Mark Scheme Syllabus Paper IGCSE – May/June 2013 0607 42 © Cambridge International Examinations 2013 (iii) 36 11 cao 2 M1 for 1 – 6 5 6 5 their their × (0.306 or 5 30 .0 or 0.3055 to 0.3056) or 6 5 6 1 6 1 × + or (ii) + 2 × 6 5 6 1 their their × or 6 1 6 1 their their × + 2 × 6 5 6 1 their their × (c) 5 2 SC1 for answer of 4 or 65 = 7776 seen or 54 = 625 seen or M1 for attempted products of 1 , 6 1 6 5 > ×       k their their k 12 (a) 18.75 (18.7 or 18.8) 18.5 23.5 13 1 1 1 1 (b) (i) r = –4.31t + 120 2 – 4.313…., 120.0…. B1 for r = –4.31t + c or r = kt + 120 Allow x for t (ii) Negative 1 (iii) 25 (25.1 to 25.4) 1FT FT their equation only if linear 13 (a) 3 2 p + 3 1 q o.e. 2 M1 for correct route from O to X or PQ = q – p o.e. or correct unsimplified answer (b) – 3 2 p + 3 5 q o.e. 3 B2 for kp + 3 5 q or – 3 2 p + kq, k ≠0 or correct unsimplified expression or M1 for correct route from X to Y or – 3 2 p + mq + nq, m 0 ,0 ≠ ≠ n (c) 4 ± 1,1 If 0 scored , M1 for 2 2 2 5 3 = + k o.e.

What you needed in this session

Cambridge’s own grade thresholds for 2013 May/June, Paper 4 · Variant 2. A higher threshold means an easier paper — the bar moves with how the cohort did.

A81/120
C46/120
E20/120