GRADE 9 NOVEMBER 2012 MATHEMATICS MARKING GUIDELINES

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Godwin Shon Mathews
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1 Province of the EASTERN CAPE EDUCATION SENIOR PHASE GRADE 9 NOVEMBER 01 MATHEMATICS MARKING GUIDELINES MARKS: 100 This marking guideline consists of 1 pages.

2 MATHEMATICS (NOVEMBER 01) QUESTION D (1) 1. D (1) 1.3 B (1) 1.4 D (1) 1.5 A (1) 1.6 C (1) 1.7 C (1) 1.8 D (1) 1.9 C (1) 1.10 C (1) [10] QUESTION.1 Let the amount she had be = x Amount she spent = Amount left = 5 3 x = 30 3 x = 30 5 x OR 3 : 5 = 30 : x 3/5 = 30/x 3x = 5 (30) Forming equation x ( ) = 30 ( ) x = 50 x = 150/3 3x = 150 x = 50 Hence Angela had R50 at first. () (0,0898) = mm (1)

3 (NOVEMBER 01) MATHEMATICS = 1,7856 x 10 4 (1) Correct answer Principal is R1 500 R 150 = R1 350 S I = P x R x T 18 = R1 350 x x = R79 Calculating the principal Formula Interest Amount paid = Principal + Interest + Deposit = R R79 + R150 = R 9 (4).3. Monthly instalment = + insurance = R insurance = R57,75 + R10,50 = R68,5 (1) [9] QUESTION Structure 5 (1) Correct drawing

4 4 MATHEMATICS (NOVEMBER 01) st term : 1() = 1 nd term : (3) = 3 3 rd term : 3(4) = 6 4 th term : 4(5) = 10 Therefore n ( n 1) () 3.. n b = ( n 1) If n = 6 6 b = (6 1) = 3 x 7 b = 1 1 blocks can be used to form structure 6 (1) 3.3 Input Output Process x (3) Minus 1 for any wrong output value The y-intercept is 3 i.e. c Take points (0 ; 3) and (- ; 0) If x = - ; y = 0 mx + c = y -m + 3 = 0 substituting into formula and calculating m -m = -3 m = 3 Hence y = 3 x + 3 ()

5 (NOVEMBER 01) MATHEMATICS If x = = y Multiplying by LCM + 3 = y 9 6 = y Removing the brackets 15 = y 1 y = 7 (4) Grouping like terms 3.5 Let the first year s admission be represented by x (any letter can be used) Representing the unknown x + x + 4x + 8x = x = Forming the equation Simplifying left hand side x = 100 Therefore 100 learners were admitted in the first year. (4) [14] QUESTION p q 81p q 3 = 9p q (1 9q ) = 9p q [(1 3q) (1 + 3q)] (4) (3x ) (5x + 1) = 15 x + 3x 10x = 15 x 7x () Common factor and difference of squares Correct factors of difference of squares Removing brackets

6 6 MATHEMATICS (NOVEMBER 01) x y z 3 8x y z 4 x 8x y 16xy 3 = x y z x y z 4 Product of numerator and Product of denominator = 3y 4 3 z (4) x 6 + 3( x 8) 4 = x + 3 4( x 6) 1( x 8) + = 4(x + 3) 4 (x 6) + 3(x + 8) = 4(x + 3) x 1 + 3x + 4 = 4x + 1 Simplifying the left hand side and the right hand side Grouping like terms 5x + 1 = 4x + 1 5x 4x = x = 64 x = 0 (4) x = 6 Converting 64 into power x = 6 Equating exponents x = 3 (3) [17] QUESTION º(n ) = 1 60º 180( n ) = 180 n = Simplification n = 7 + n = 9 Hence, if the sum of the angles of a polygon is 1 60º, it has 9 sides. ()

7 (NOVEMBER 01) MATHEMATICS 7 5. In SPQ and QRS PQ = RS (opposite sides of a parm.) SQP = QSR (alt s, PQ // RS ) SQ = SQ (common) SPR = QRS (S S) OR PQ = RS opposite sides of a parm. SP = QR opposite sides of a parm. SQ = SQ common Hence SPQ QRS (SSS) (4) Statements with reason x 6x = 10 6x 10 = 6 6 x = 0 cm Setting up the proportional sides The length of the longest side is 0 cm. () L (3 ; -5 ) M (5 ; -6 ) N (6 ; -5 ) O (5 ; -3 ) () marks for 4 coordinates correct 1 mark for 1or wrong coordinate No mark for 1 correct co-ordinate

8 8 MATHEMATICS (NOVEMBER 01) () Award for correct position of image on diagram above with the following coordinates L'' (-1; -5) M'' ( 1; -6) N''(; -5) O'' (1; -3) () [14] Correct gliding on ANNEXURE is to be awarded

9 (NOVEMBER 01) MATHEMATICS 9 QUESTION P R + = 180º sum of angles on a straight. line Hence = 180º 95º = 85º but P R = corresponding angles, PQ//ST Therefore P R = 85º (3) Obtaining angle QTS Equating with reason angles PQR and QTS x 1 (1) QUESTION x 4 (1) [5] S = D t 0km 1,5 = 16 km/h Speed of train B is 16 km/h (1) 7.1. Speed of train A D 30 = t 0, 75 = 40 km/ h Therefore Train A is faster than Train B because it runs at a speed of 40km /h whilst Train B runs at a speed of 16 km/h. () Reason In ADB OR ADC DB = DC =,1 cm AD = AB DB Pythagoras Theorem = (3 cm) (,1 cm) = 9 4,41 = 4,59 AD = 4, 59 =,14 cm () Theorem

10 10 MATHEMATICS (NOVEMBER 01) 7.. Total Surface Area = (triangular base area) + area rectangle +(Area of other rectangular face) = ( 1 b x h) + (l x b) + (l x b) = ( 1 x 4, x,14) + (4, x 15) + (3 x 15) Formula Correct substitution = 8, = 161,99 cm (4) [9] QUESTION No. of learners that will fail = = Minimum total marks she can obtain = 80 x 8 = 640 (1) (1) [] Addition and correct answer Multiplication and answer

11 (NOVEMBER 01) MATHEMATICS 11 QUESTION Hand length and Shoe Size Correlation Graph Labelling x-axis 1 mark; Labelling y-axis 1 mark; Title of graph 1 mark Plotting points marks (5) 9.1. The bigger the shoe size the longer the length of the hand and vice versa. (1) Correct reason for relation.

12 1 MATHEMATICS (NOVEMBER 01) ; 4; 5; 5; 6; 7; 7; 8; 9; 9 Median = 6 7 = 13 1 = 6 () Identifying middle numbers and dividing Mode is 15 (1) Mean = = 130 / 10 = 13 Therefore the mean is 13. () 1 mark for correct sum Range = 9 = 7 (1) 9. Win (W) Draw (D) Loss (L) Win (W) WW WD WL Draw (D) DW DD DL Loss (L) LW LD LL (3) 1 mark per row / column Correctly completed table (1) 9.3. (1) (1) Graph 1 (1) 9.4. In graph 1 the scale units on y-axis are small i.e. 10 units This results in graph being stretched (enlarged). On the other hand the scale units on y-axis of graph are bigger i.e. 50 units this results in the graph being compressed. (1) [0] TOTAL: 100

GRADE 9 NOVEMBER 2012 MATHEMATICS

Province of the EASTERN CAPE EDUCATION SENIOR PHASE GRADE 9 NOVEMBER 2012 MATHEMATICS MARKS: 100 TIME: 2 hours This question paper consists of 16 pages. 2 MATHEMATICS (NOVEMBER 2012) INSTRUCTIONS AND INFORMATION