# Mathematics Grade 9 November EC 2011 Memo (English) > Grade 9 Mathematics memorandum from the November EC 2011 (English), 13 pages. Read online or get the free PDF download, question paper linked when available. - Page: https://www.hlayiso.com/library/grade-9-mathematics-2011-november-ec-memorandum/7758c2b7-fdc7-48fa-9be5-eef378cad671 - Updated: 2026-10-29 - Subject: Mathematics - Grade: Grade 9 - Document type: Memorandum - Year: 2011 - Exam period: November EC - Curriculum: CAPS - Pages: 13 Breadcrumbs: [Home](https://www.hlayiso.com/) › [Library](https://www.hlayiso.com/library) › [Grade 9](https://www.hlayiso.com/library/grade-9) › [Mathematics](https://www.hlayiso.com/library/grade-9/mathematics) › [Mathematics Grade 9 November EC 2011 Memo (English)](https://www.hlayiso.com/library/grade-9-mathematics-2011-november-ec-memorandum/7758c2b7-fdc7-48fa-9be5-eef378cad671) ## About this document Mathematics · Grade 9 · November EC · 2011. Memorandum, 13 pages. Read online or download the PDF. Read it free online on the page above; download the PDF from there. ## Document text Text extracted from the PDF. Layout, diagrams and tables may not survive extraction. ### Page 1 Province of the EASTERN CAPE EDUCATION SENIOR PHASE GRADE 9 NOVEMBER 2011 MATHEMATICS MARKING GUIDELINE MARKS: 100 This marking guideline consists of 12 pages. ### Page 2 2 MATHEMATICS (Memo) (NOVEMBER 2011) QUESTION 1 1.1 A √ 1 1.2 B √ 1 1.3 B √ 1 1.4 A √ 1 1.5 C √ 1 1.6 C √ 1 1.7 C √ 1 1.8 D √ 1 1.9 B √ 1 1.10 D √ 1 [10] QUESTION 2 2.1 2.1.1 CI =? Principal for the 1st year : R 12 000 Interest for the 1st year (8,8% of R12 000) : R 1056,00 √ Correct calculation of Principal for the 2nd year : R 13 056,00 interest Interest for the 2nd year (8,8% of R13 056) : R 1148,93 √ Correct amount after 2 Amount Ceolo will pay for Compound Interest after 2 years Years is R12 000+ R1 056 + R1 148,93 =R 14204,93 SI=? For 2 years SI = P . r . t = 12 000 x 9% x 2 = R 2 160,00 The amount Ceolo will pay for SI is R2 160 + R12 000 Correct SI amount = R 14 160,00 √ The best option Ceolo should choose is Simple Interest √ (4) Correct choice 2.1.2 Amount to be paid by Ceolo’s friend in rand is R12 000 Dollars($ ) to be sent by friend = = $ 1 685, 39 √ (1) Correct answer ### Page 3 (NOVEMBER 2011) MATHEMATICS (Memo) 3 2.2 150 litres : 1 min 81 000 litres : x mins. = 150 x = 81 000 x = = 540 mins. Time taken in mins. 1 hr. : 60 mins x hrs. : 540 mins. √ = 60x = 540 x = Answer in hours = 9 hrs. √ Therefore it will take 9 hrs to fill the tank with a volume of 81 000 litres (2) [7] QUESTION 3 3.1 3.1.1 Changing of division sign = √ Simplification = √ Answer = √ (3) ### Page 4 4 MATHEMATICS (Memo) (NOVEMBER 2011) (4 – 3x)(3x + 2) + 7x - (1 – 2x ) Removing of the = 12x + 8 – 9x - 6x + 7x - 1 + 2x √√ brackets = – 9x + 7x + 2x + 12x - 6x + 8 -1 √ Grouping of like terms Answer = 6x + 7 √ (4) 3.2 3.2.1 36a b c + 16a b - 4a bc = 4a b (9abc + 4b - c ) √√ (2) Answer 3.2.2 25a - 49b = (5a – 7b)(5a + 7b) √√ (2) Answer 3.3 Factorisation 5 -3 = (5 – 3)(5 + 3) √ = 2x8 = 16 √ (2) Answer 3.4 3.4.1 5 = Changing to exponential 5 = 5 √ form x =-3 √ (2) Answer 3.4.2 Multiplying by LCM √ 4(x +3) – 2(x- 2 ) = x + 4 4x + 12 – 2x + 4 = x + 4 √ Removing the brackets 2x + 16 = x + 4 Grouping like terms 2x - x = - 16 + 4 √ Answer x = - 12 √ OR (4) ### Page 5 (NOVEMBER 2011) MATHEMATICS (Memo) 5 OR = 3.5 3.5.1 0 = -1 3 = 2 -1 Answer 8 = 3 -1 the rule is, the square of the input number minus 1 i.e. x -1 √√ (2) 3.5.2 y = x -1 = (17) -1 = 289 – 1 = 288 Answer A i.e. y = 288 when x = 17 √ (1) 3.5.3 x -1 = y x - 1 = 528 Simplification x = 529 √ x = = 23 Answer B i.e. x = 23 when y = 528 √ (2) ### Page 6 6 MATHEMATICS (Memo) (NOVEMBER 2011) 3.6 3.6.1 For 2x + 3y = 6 If x = -3 x=0 x= 3 2(-3) + 3y =6 2(0) + 3y = 6 2(3) + 3y = 6 -6 + 3y = 6 3y = 6 6 + 3y = 6 3y = 12 y =2 3y = 0 y = 4 y=0 pt (-3; 4) pt (0;2) pt (3;0) for y = x + 2 if x=-3 x=0 x=3 y = -3 + 2 y = 0+ 2 y = 3 + 2 y = -1 y = 2 y = 5 pt (-3;-1) pt (0;2) pt (3;5) OR 2x + 3y = 6 ……… equation 1 3y = -2x + 6 y = √ correct standard form for equation 1 y = x +2 ………equation 2 x -3 -2 -1 0 1 2 3 Correct table for 4 2 0 equation 1 y= 3 2 1 √ x -3 -2 -1 0 1 2 3 Correct table for equation 2 y = x+2 -1 0 1 2 3 4 5 √ ### Page 7 (NOVEMBER 2011) MATHEMATICS (Memo) 7 cont. 3.6.1 6_ y=x+2 5_ √√ correct drawing 4_ of graphs 3_ 2 1_ √ correct labelling both graphs | | | | | | | | | | | X -5 -4 -3 -2 -1 0 1 2 3 4 5 6 -1_ 2x + 3y = 6 -2_ (6) 3.6.2 (0;2) √ (1) [31] QUESTION 4 i.e. √ 10p = 180 Cross multiplication Answer p = 18 cm √ (2) 4.1.2 The corresponding sides of the two triangles are proportional. √ (1) 4.2 4.2.1 LM = LO + OM = 6 cm + 3 cm √ Length of LM LM = 9cm Similarly QO= QM + MO = 6cm + 3 cm √ Length of QO QO = 9 cm LM = QO (2) ### Page 8 8 MATHEMATICS (Memo) (NOVEMBER 2011) 4.2.2 In POQ and NML POQ = N ML = 90° (PO LQ & Statement MN LQ ) √ PQ = LN (given) √ Statement OQ = ML proved in 4.2.1 POQ NML (RHS) √ (3) Conclusion 4.3 Reflected image 10_ 9_ enlarged image 8 Y'(0;8) √√ 7_ √√ 6_ 5_ . . 4 Y(0;4) . . Z''(-6;4) X''(-2;4) 3 _ X'(2;4) Z'(6;4) 2_ . . 1 _ x(1;2) z(3;2) | | | | | | | | | | | | | | | | | | | -8 -6 -4 -2 -1 _ 2 4 6 8 10 -2 _ -3 _ 4.3.1 See enlarged image in diagram above √√ (2) Correct enlargement 4.3.2 Refer to reflected image in diagram above √√ Correct reflection 4.3.3 X(-2; 4) Y(0; 8) & Z (-6 ; 4) √√ (2) All points to be 2/2 4.4 4.4.1 Statement/reason KMN = 3x (alt. s LK// MN ) √ Statement/reason 2x + 3x + x = 180 ( s on straight line)√ 6x = 180 Answer x = 30 √ (3) 4.4.2 LKM = 3x = 3(30) = 90 √ (1) Answer 4.4.3 Triangle MKL is a right-angled triangle √ (1) Answer [19] ### Page 9 (NOVEMBER 2011) MATHEMATICS (Memo) 9 ### Page 10 10 MATHEMATICS (Memo) (NOVEMBER 2011) QUESTION 5 5.1 5.1.1 2 (l + b ) = 48 2 (l + b ) = 48 2 2 l+ b = 24 But l = (48) l = 16 m √ Correct calculation of length l+b = 24 16 + b = 24 Correct calculation b = 24 – 16 of breadth b = 8m √ (2) 5.1.2 2 ( l+ b) = P [P is perimeter of Unshaded portion] 2( x16 + 8) = P √ Substitution 2(12 + 4 ) = P 2 (16) = P 32 = P √ The Perimeter of the unshaded portion is 32 m (2) Answer 5.2 (VA) = (VO) +(OA) (Pythagoras Theorem) √ Statement/reason (VA) = (12) + (5) = 144 + 25 = 169 VA = Answer = 13 cm √ (2) [6] ### Page 11 (NOVEMBER 2011) MATHEMATICS (Memo) 11 QUESTION 6 6.1 6.1.1 School A received 44% of R94 000 = x = R 413 600 √ (1) Answer 6.1.2 x Knowing that a full circle º = x √ has 360º Answer = 79,2 √ (2) 6.1.3 School C had 12% of R94 000 = x R94 000 Amount received by = R11 280 √ School C School C received R11 280 + R 5 300 Answer = R 16 580,00 √ (2) 6.2 6.2.1 Range = 73 – 17 = 56 √ (1) Answer 6.2.2 Median = = Answer = 43,5 √ (1) 6.2.3 Frequency table of 50 parents who attended parents meeting Interval √ Tally √ Frequency√ 0 - 10 - 0 11 - 20 //// 4 Interval 21 - 30 //// / 6 1 mark 31 - 40 //// //// 10 41 - 50 //// //// / 11 Tally 51 - 60 //// //// 10 1 mark 61 - 70 //// / 6 71 - 80 /// 3 Frequency 1 mark Total 50 (3) ### Page 12 12 MATHEMATICS (Memo) (NOVEMBER 2011) 6.3 The total body mass of 5 parents is Total body mass 51 + 56 + 54 + 59 + 53 = 273 √ of 5 parents The average body mass of 6 parents is 56 kg The total body mass of 6 parents is 6 x 56 = 336 kg √ Total body mass of 6 parents The body mass of the 6th parent = 336 – 273 = 63 kg √ (3) Answer 6.4.1 % Marks obtained by Alive in 6 Tests Y 24_ 22_ x √ Heading 20_ x M A 18_ √ Correct scale on each R x axis K 16_ x √ Labelling axes 14_ I x N 12_ √ correct graph 10_ P E 8_ (4) R C 6_ E N 4_ x T 2_ | | | | | | | | | 0 1 2 3 4 5 6 7 8 9 X TEST NUMBER 6.4.2 Average % of test marks = Calculating sum √ = = 15,33 % √ (2) Answer ### Page 13 (NOVEMBER 2011) MATHEMATICS (Memo) 13 6.5 1st Choice 2nd Choice Outcomes G GG G GG G W GW B GB G GG √ 1st Choice S G G GG W GW B GB √ 2nd Choice G WG W G WG W WW B WB √√ All correct possible B G BG Outcomes G BG Missing one outcome 1, Missing more than 1 no W BW mark B BB (4) 6.5 6.5.1 Answer P (White counter ) = √ (1) 6.5.2 Answer P (White and Black counters) = = √ (1) 6.5.3 Answer P (at least one Black counter ) = √√ (2) [27] TOTAL: 100 ## Related documents - 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