Mark Scheme (Results) January 2011 - Bland

Mark Scheme (Results)January 2011O LevelGCE O Level Mathematics B (7361/02)Edexcel Limited. Registered in England and Wales No. 4496750Registered Office: One90 High Holborn, London WC1V 7BH

Edexcel is one of the leading examining and awarding bodies in the UK andthroughout the world. We provide a wide range of qualifications includingacademic, vocational, occupational and specific programmes for employers.Through a network of UK and overseas offices, Edexcel’s centres receive thesupport they need to help them deliver their education and training programmes tolearners.For further information please call our Customer Services on + 44 1204 770696, or visit our website at www.edexcel.com.If you have any subject specific questions about the content of this MarkScheme that require the help of a subject specialist, you may find our AskThe Expert email service helpful.Ask The Expert can be accessed online at the following link:http://www.edexcel.com/Aboutus/contact-us/January 2011All the material in this publication is copyright© Edexcel Ltd 2011

Mathematics B, Mark SchemePaper21. ∠ AOD = 30 oB1∴∠OCDOR ∠ ODC = 15 oB1∠ CDE = 75 oB1 3NB: Award Bs for angles seen on diagramTotal 3 marks2. (a) 20-x, x, 16-x, 7 shown on Venn diagram B3 (-1eeoo) 3(b) “(20-x)” + x + “(16-x)” + 7 = 35 (no slips) M1x=8 A1 2 5Total 5 marks3. l =2 26 + 2.5 (= 6.5) M12π× 2.5 × "6.5" + 2π× 2.5× 9 + π×2.5(Sum of 2 correct areas)M1(Sum of 3 correct areas)M1 DEP212 cm 3 A1 4Total 4 marks4. (a) 1.5 x 20000 /100 M1300 m A1 2(b) 1.2 x 100 000/20000 cm M16 cm A1 2(c) 60 000 x (100) 2 x (1/20000) 2presence of one of (100) 2 or (1/20 000) 2in above formulaM1bothM1 DEP1.5 cm 2 A1 3 7Total 7 marksGCE O Level Mathematics Syllabus B (7361) Paper 2 January 2011 1

⎛ 2 13 ⎞5. B.C = ⎜ ⎟⎝2y y+3x⎠)⎛ 5 15 ⎞(A + B.C = ⎜⎟⎝2y+ x 2y+3x⎠ 2y + x = 42y + 3x = 6Elimin. x or y from 2 linear “SEs” in x and ySubst x or yNB: allow 1 sign slip only for both MsB2 (-1 eeoo)B1B1M1M1 (DEP)x = 1, y = 3/2 A1, A1 8Total 8 marks6. (a) 1/3 , 120/360, 0.333, 33.3% B1 1(b) 1/3 + 1/4 or120 + 90360M1210/360, 7/12 , 0.583, 58.3% A1 2(c)1 1× or 60/360 x 90/3606 4M11/24 , 0.0417, 4.17% A1 2(d) 1/3 x ¼, ¼ x 1/3, 1/6 x ¼, ¼ x 1/6 (2 off) B1allB11/3 x ¼ + ¼ x 1/3 + 1/6 x ¼ + ¼ x 1/6 M1¼ , 0.25 , 25% A1 4 9Total 9 marksGCE O Level Mathematics Syllabus B (7361) Paper 2 January 2011 2

47 (a)(i) PA =5 a B1(a)(ii) AB = b – a B1 2(b)(i) AQ = 4 9“(b – a)”B1 ft(b)(ii) PQ = “ 4 5 a” + “ 4 9(b – a)”M1OR PQ = - 1 5 a + b - 5 (b – a) (no errors) M19 16∴ PQ =45 a + 4 9 b A1 4(b)(iii) QC = b - “9(b – a)” M1OROR 5QC = (b – a) + a (no errors) M19 16QC = -“45 a + 4 9 b” - 1 5 a + b + a M1 4∴ QC =9 a + 5 9 b A1 5(c) 4PC = " a” + b 5and attempting (but NOT using vector division) to show that PQ = nPC and QC = mPCM1Either 4⎛4⎞PQ = ⎜ a+b⎟= 4 9⎝5⎠ 9 PCOR 5⎛4⎞QC = ⎜ a+b⎟= 5 9⎝5⎠ 9 PCA2[OR Attempting (but NOT using vector division) to show that 4 5 PQ = " QC " or QC = " PQ"5 4M1GCE O Level Mathematics Syllabus B (7361) Paper 2 January 2011 3

4⎛4 5 ⎞PQ = ⎜ a+b⎟5⎝9 9 ⎠OR 5⎛16 4 ⎞QC = ⎜ a+b⎟4⎝45 9 ⎠A2]c.c A1 4 11Total 11 marks_____________________________________________________________________8. Penalise labelling ONCE only(a) ∆ ABC drawn and labelled B1 1(b) (i) A 1 = (3, 1), B 1 = (7, 3), C 1 = (8, 2)B2 (-1 eeoo)(ii) ∆ A 1 B 1 C 1 drawn and labelled B1 ft 3(c) A 2 = (2, -2), B 2 = (6, -4), C 2 = (4, -6)B2 (-1 eeoo)∆ A 2 B 2 C 2 drawn and labelled B1 ft 3(d) (i) enlargement factor 2(ii) 270 oantclockwiseNB: Last B1 is DEP on previous B and the B in (b) (i)B1B1B1 (DEP)(OR 90 o , clockwise B1, B1 (DEP) ) 3OR d(i) enlargement factor -2 B1d(ii) 270 o clockwise B1, B1 (DEP)OR 90 o anticlock. B1, B1 (DEP)NB: The 3 rd B of (d) is DEP on the 1 st and 2 nd Bs of (d)(e)⎛ 0 2⎞⎜ ⎟⎝−2 0⎠B2 (-1 eeoo) 2 12Total 12 marksGCE O Level Mathematics Syllabus B (7361) Paper 2 January 2011 4

9. (a) (10 – t) (3 + t) = 0 (solving trinomial quadratic) M1t = 10 secs. A1 2(b) 38.3, 42, 41.3, 38.3 B3 (-1 eeoo) 3(c) curve-1 mark for incorrect/non-uniform scalestraight line segmentseach point missedeach missed segmenteach point not plottedeach point incorrectly plottedtramlinesvery poor curve B3 3(d) 3.5 secs ( ± 0.1 secs) (1 ss = 0.1sec) B1 ft 1(e) Attempt to measure gradient at t = 2M1Answer rounding to 3 m/ sec A1 2(f) 2 < t < 5 ( ± 0.1 secs)B1 ft, B1 ft 2 13Total 13 marks_____________________________________________________________________2 210. (a) L = (3 x) + (4 x)M15x (o.e) A1 2(b)112× × 3x× 4x+ 3x× 10xOR × (10 x + 18 x ) × 3 x M12242x 2 (o.e) A1 22(c) 2 × "42 x " + 2 × "5 x" × y+ 10xy+ 18xy= 1008 M12“ (10x + 10x+ 18 x) y = 1008 − 84x” M1 (DEP)y21008 −84x=38x(c.c) A1 322 1008 − 84x(d) V = "42 x " ×38xM121 x2V = (1008 − 84 x )19(c.c) A1 221× 1008 3× 21×84 2(e) One of or - x19 19(o.e) M121× 1008 3× 21×84 2- x19 19(o.e) A121× 1008 3× 21×84 2“ - x ”=019 19(o.e) M1 (DEP)x =1008252(o.e) (isolating x from quadratic in x 2 only)M1 (DEP)x = 2 cm A1 5 14Total 14 marksGCE O Level Mathematics Syllabus B (7361) Paper 2 January 2011 5

_____________________________________________________________________11. Penalise ncc ONCE onlyEC(a) sin 35 =2M11.15 cm A1 23(b) tan 35 =BEM14.28 cm A1 2(c)(d)"4.28" "1.15"4.43 cm , 4.44 cm A1 2"4.28"tan ∠ BCE = (o.e) M1"1.15"75 o , 75.0 o A1 22 2BC = + M1(e) ∆ABP : ∠ BAP = 55 and ∠ ABP = 105B1OR ∆CPD : ∠ CDP = 145 and ∠ DCP = 15 B1OR ∆CEP: ∠ CEP = 55 and ∠ ECP=105B1OR ∆BEP : ∠ EBP = 15 and ∠ BEP = 145B1∠ BPA = 20 o , 20.0 o B1 2(f)X is a pt on AE st BX is perpendicular to AEAXcos55 = (AX=1.721)3M1BXsin 55 = (BX = 2.457)3AND"2.456"tan"20" = (PX = 6.748)PX=> M1AP = “AX” + “PX” = “1.721” + “6.748”M1 (DEP)ORAXcos55 = (AX=1.721) M13XEcos35 = ( XE = 3.5092)"4.284"ANDM1( DC = 2× cos35 = 1.6383)"1.6383" × sin15DP =(=1.2893)sin"20"AP = “AX” + “XE” + 2 + “DP” = “1.721”+”3.5092”+2+”1.2398”M1 (DEP)GCE O Level Mathematics Syllabus B (7361) Paper 2 January 2011 6

OR3sin 35 = ( AE = 5.2303)M1AECDcos35 = (CD = 1.6383)2ANDM1"1.6383" × sin15DP= (DP = 1.2398)sin"20"AP = “AE” + ED + “DP” = “5.2303” + 2 + “1.2398” M1 (DEP)OR3sin 35 = ( AE = 5.2303)AEM1sin"105" × "1.15"EP = (EP = 3.2478)sin"20"M1AP = “AE” + “EP” = “5.2303” + “3.2478”M1 (DEP)ORSpecial Case : Sine Rule∠ ABP = 105 oM1AP 3=sin"105" sin"20"M1 (DEP)3×sin"105"AP =sin"20"M1 (DEP)AP = 8.46, 8.47 cm, 8.48 cm A1 4 14Total 14 marks_____________________________________________________________________TOTAL 100 MARKSGCE O Level Mathematics Syllabus B (7361) Paper 2 January 2011 7

Further copies of this publication are available fromInternational Regional Offices at www.edexcel.com/internationalFor more information on Edexcel qualifications, please visit www.edexcel.comAlternatively, you can contact Customer Services at www.edexcel.com/ask or on + 44 1204 770 696Edexcel Limited. Registered in England and Wales no.4496750Registered Office: One90 High Holborn, London, WC1V 7BH