Mathematics Grade 9 June Common 2011 Memo (English)

Downloaded from hlayiso.com Province of the EASTERN CAPE EDUCATION SENIOR PHASE GRADE 9 JUNE 2011 MATHEMATICS MARKING GUIDELINE MARKS: 100 This marking guideline consists of 9 pages.

Downloaded from hlayiso.com 2 MATHEMATICS (JUNE 2011) SECTION A 1. D√ (1) 2. C√ (1) 3. C√ (1) 4. C√ (1) 5. A√ (1) 6. A√ (1) 7. C√ (1) 8. B√ (1) 9. B√ (1) 10 A√ (1) 11. D √√ (2) 12. A√ (1) 13. C√ (1) 14. B√ (1) 15. A√ (1) 16. A√ (1) 17. A√ (1) 18. C√ (1) 19 B√ (1) 20. B√ (1) 21. B√ (1) 22 C√ (1) TOTAL SECTION A: 23

Downloaded from hlayiso.com (JUNE 2011) MATHEMATICS 3 SECTION B QUESTION NO SOLUTION MARK REASON ALLOCATION 1. 1.1 1.1.1 √25 ; 2,3 ; 16 ; 4 √ If all four numbers (1) mentioned. 1.1.2 If both numbers π ; √19 √ (1) mentioned. 1.2 Real numbers √ (1) Answer only. 1.3 0,005 629 √ (1) Answer only. 1.4 (2a2b4)3 x 4ab3 Product rule 8a-2b7 = 8a b x 4ab3 6 12 √ Simplification 8a-2b7 √ 9 8 = 4a b √ (3) Answer 1.5 Let the unknown distance be x 480 km : 4 hours x travelled km : 31/3 hours 480 4 = x3 √ Direct proportion x 10 12x = 4800 4800 X = 12 x = 400 √ Answer The family is left with 80 km to its destination √ OR Average speed for 4 hours: = 120 km/h √ Average speed Distance travel for 3 hours 20 minutes: 120 x 3 Distance travelled in 3 hours 20 minutes = 120 x = 400 km √ Distance left = 480 km – 400 km = 80 km √ (3) Difference [10]

Downloaded from hlayiso.com 4 MATHEMATICS (JUNE 2011) QUESTION 2 2.1 2.1.1 a 2  b2 x(3a  4b) X y ( a  b) 9a 2  16b 2 (a  b)(a  b) x (3a  4b) X y ( a  b) (3a  4b )(3a  4b ) √ ab x X Simplification y 3a  4b √ x(a  b) √ y (3a  4b ) (3) Answer 2.1.2 Removing of (5x – 1)(2x + 3) brackets = 10x 2 + 15x − 2x – 3 √ = 10x 2 + 12x − 3 √ (2) Answer OR 2.2 p 3 − p 2 r − 9p + 9r p 3 − 9p – p2r + 9r = (p 3 − p 2 r) – (9p − 9r) √ = p(p2 − 9) – r(p2 − 9) √ Correct grouping = p 2 ( p – r) - 9( p – r) √ = (p2 – 9)(p – r) √ Common factor = ( p – r)( p 2 − 9) = ( p – 3)(p + 3) )(p – 3 ) √ Answer = ( p – r )( p – 3 )( p + 3) √ (3) 2.3 2.3.1 6(x + 2) = 3( 3x – 4) – 3 6x + 12 = 9x – 12 – 3 √ 6x + 12 – 12 = 9x − 15 − 12 6x − 9x = - 27 √ - 3x = - 27 3x = 27 3x 27 = 3 3 x = 9 √ (3) 2.3.2 4x x7 = OR 5 3 4x x7 Cross X 15 = X 15 multiplication 5 3 12x = 5x − 35 12x = 5x – 35 √ 7x = -35 7x = -35 7x 35 7x 35 = - = - 7 7 7 7 x = -5 √ x = -5 (2) Answer

Downloaded from hlayiso.com (JUNE 2011) MATHEMATICS 5 2.3.3 3.5 x 1 = 75 3.5 x 1 = 75 3 3 x 1 Simplification 5 = 25 √ 5 x 1 = 5 2 Equal basis x+1 = 2 √ x+1–1=2–1  x = 1 √ (3) Answer 2.4 1 y = x − 4 2 To find the y-intercept, let x = 0 1 y = (0)− 4 2 Co-ordinate of y y = -4 intercept Co-ordinate of y intercept is (0; -4) √ To find the x-intercept , let y = 0 1 −4 = 0 2 1 = 4 2 x 2x = 2x4 2  = 8 Co-ordinate of x Co-ordinates of x-intercept (8;0 ) √ intercept y 4- 1 3- y= −4 2 2- 1- √ x-intercept in the -x | | | | | | | | | | | | | | | | | | | | | | x graph -2 -1- 2 4 6 8 10 12 -2- x-intercept in the -3- graph -4- √ -5- -6- -7- -y (4) [20]

Downloaded from hlayiso.com 6 MATHEMATICS (JUNE 2011) QUESTION 3 3.1 Structure 5 Correct drawing √ (1) 3.2 3.2.1 25 – 1 = 12 or any other relevant method 2 √ (2) Answer 3.2.2 Let the number of triangles be represented by n The difference in the matchsticks sequence is 2. √ Hence the number of matchsticks is 2 times number of triangles plus 1 i.e. 2n + 1 or any relevant method. √ (2) Answer 3.2.3 No. of n matchsticks = 2n + 1 n = 40 No. of matchsticks = (2 x 40) +1 = 80 + 1 = 81 √ There are 81 matchsticks (1) Answer 3.3 3.3.1 Let Linda’s age be x Then Thami’s age is x + 12 But x + (x + 12) = 82 √ Setting up the equation x + x + 12 = 82 2x = 82 − 12 2x = 70 2 2  x = 35 √ Linda’s age is 35 years (2) Answer 3.3.2 Thami’s age is x + 12 i.e. 35 + 12 = 47 Thami’s age is 47 years. (1) Answer [9]

Downloaded from hlayiso.com (JUNE 2011) MATHEMATICS 7 QUESTION 4 4.1 •B 4 ●A 3-- √ Position of C 2-- •C 1-- (-1;1) | | | | | | | | | | | | | | | -3 -2 -1 1 2 3 4 (1) 4.2 C (-1;1) (1) Answer 4.3 Translation (1) Answer 4.4 4.4.1 Trapezium (1) Answer 4.4.2 By rotating the shape all four angles meet a point, thus forming 360º. The sum of the angles of any quadrilateral is 360º or any other reasonable logical argument√√ (2) [6] QUESTION 5 5.1 5.1.1 R √ (1) Answer 5.1.2 Is similar √ (1) Answer 5.1.3 All angles of ∆ PQR = All angles of ∆ ABC √ (1) Answer 5.1.4 QR √ (1) Answer 5.1.5 RP √ (1) Answer 5.2 5.2.1 Statement Reason In ∆s ABO & PQO AB = PQ given √ Reason  ABO =  OPQ alternate angles √ Statement/Reas on  AOB =  POQ vertically opposite angles Statement/Reas √ on ∆ABC  ∆ RST  ,  ,S √ (4) Condition of congruency

Downloaded from hlayiso.com 8 MATHEMATICS (JUNE 2011) 5.2.2 Side, Side, Side (SSS) ;Side, Angle, Side (S,A,S); Right Angle, Hypotenuse and Side (RHS) (Any one of these) √ (1) Answer [10] QUESTION 6 6.1 6.1.1 Volume of can =  r2 h = x (32,5 mm)2 x 75 mm √ Substitution = x 1 056,25 mm3 x 75 mm Answer = 24 8973,21 mm 3 √ Convert 24 8973,21 mm 3 to cm 3 (1 000 mm 3 = 1 cm 3 ) √ Conversion 1 24 8973,21 mm 3 X = 248,97 cm 3 √ 1000 (4) Answer 6.1.2 Convert 248,97 cm 3 to mℓ (1 cm 3 = 1 mℓ) :. 248,97 cm 3 = 248,97 mℓ √ Conversion Space = Volume of can − liquid in the can = 248,97 mℓ – 200 mℓ :. Space = 48,97 mℓ √ (2) Answer [6] QUESTION 7 7.1 First arrange the numbers in ascending order 16; 18; 19; 23; 24; 26; 27; 29; 30; 33; 34; 35; 37; 38; 39; 40; 41; 42; 45; 49. Stem Leaf 1 mark per stem 1 6; 8; 9 √ 2 3; 4; 6; 7; 9 √ 3 0; 3; 4; 5; 7; 8; 9 √ 4 0; 1; 2; 5; 9 √ (4) 7.2 √ √ Interval Tally Frequency Tally 0 − 9 0 10 − 19 /// 3 frequency 20 − 29 //// 5 30 − 39 //// // 7 40 − 49 //// 5 Total 20 (2)

Downloaded from hlayiso.com (JUNE 2011) MATHEMATICS 9 7.3 7.3.1 Sum of marks obtained Average Mark = No. of Marks obtained Correct addition 645 = √ 20 Average Mark = 32,25 √ (2) Answer 7.3.2 Median − Middle Value 16; 18; 19; 23; 24; 26; 27; 29; 30; 33; 34; 35; 37; 38; 39; 40; 41; 42; 45; 49. 33  34 Median = 2 67 = 2 = 33,5 √ (1) Answer 7.3.3 Mode – value that appears most frequently/ often in a set of data. There is no mode √ (1) Answer 7.3.4 Range = Highest mark − Lowest mark = 49 – 16 √ = 33 (1) Answer 7.4 7.4.1 Labels (1) TEMPERATURES TAKEN ONE A DAY √ All 12 bars T (3) E 35 Max. 9 (2) Max.6 (1) P Less than 6 E 30 (NO R 25 MARKS) A T 20 U R 15 E 10 I 5 √√√ N 0 º c TIME IN HOURS (4) 7.4.2 16:00 (1) Answer [16] TOTAL SECTION B: 77 GRAND TOTAL: 100