Mathematics Grade 9 June Common 2010 Memo (English)

Province of the EASTERN CAPE EDUCATION SENIOR PHASE GRADE 9 JUNE 2010 MATHEMATICS MARKING GUIDELINE MARKS: 100 TIME: 2 hours This Marking Guideline consists of 05 pages.

2 MATHEMATICS (JUNE 2010) SECTION A 1. D√

(JUNE 2010) MATHEMATICS 3 2. C√ 3. B√ 4. C √√ 5. A √√ 6. A√ 7. C√ 8. B√ 9. A √√ 10 A√ 11. D√ 12. A √√ 13. C√ 14. D √√ 15. B√ 16. C√ 17. C√ 18. D √√ 19 D √√ [26] SECTION B QUESTION NO SOLUTION MARK ALLOCATION 1.1.1 2 1 1 mark 3 = 3k m √√ -4 specifically for negative index 1.1.2 4 2 2 =-[(a -4a+4) +(a +a-6)] +8√ = -[2 a2 – 3a – 2]+8√ =-2 a2+3a+2+8√ ==-2a2 + 3a + 10√ 1.2 4(a + b) – x2(a+ b) 3 = (a+b)(4 – x2) √ =(a+b) (2 – x)(2+ x) √√ 1.3 1112-992 2 = (111 – 99)(111 + 99) √ =12 x 210 = 2 520 √ 3 1.4.1 3x + 15 – 6 = -2x – 2√ 5x = -11√ x = -11/5√ 1.4.2 3x + 1 = 27 3 3x + 1 = 33√ x + 1 = 3√ x=2 √ 2 2.1.1 Pattern 5 =32 since the common difference is 7√ 1

4 MATHEMATICS (JUNE 2010) 2.1.2 7x1–3 =7-3=4 or any logical method 3 7x2–3 = 14–3 =11 1 mark for 7x3–3 = 21–3 =18√ exploration 1 mark for 7n 7xn -3= 7n-3√√ and 1mark for -3 2.1.3 7x200 -3 =1397 1 2.2.1 √ 1 2.2.2 √√ Since parallel lines have the same gradient 2 Give full marks even without reason 2.2.3 At x-intercept, the value of y is zero. 3 √ √ √ 2.2.4 √√ 2 3 3.1.1 Sum of angles of a hexagon =4x 180º 1 = 720º 3.1.2 Number of triangles in any polygon = n-2√ 2 Therefore sum of angles in any polygon =n–2 x 180º√ 3.1.3 1080= 180º x 6√ 2 =180º x (8–2) So this is an octagon√ 3.2.1 2 1st Quadrant image

(JUNE 2010) MATHEMATICS 5 3. 3.2.2 See 3.1.1 4th quadrant√ 3 A  (2;-1), B  (3;0), C  (2;0), D  (1;-1) √√ 1 mark for correct image 2 marks for correct co- ordinates 4 4.1 S = d/ t 4 20 km/h = 1√ t t = 0,05 h√ = 3 min√ The rear of the train will emerge at 1:06 pm or 13:06√ 4.2 8 - 5 =135 ℓ √ 4 11 8 64-55 =135ℓ 88 9 = 135ℓ √ 88 So full capacity =135ℓ x88√ 9 =1320ℓ√ 4.3 Volume of a cylinder = πr2h√ 4 =22x 64cm2x17cm√√ 1 mark 7 specifically for the correct = 3 419,43 cm3√ height 5 5.1.1 72º =1/5 √ 1 360º 5.1.2 Number of learners for English= 126º x 120 = 42 learners√√ 2 360º 5.1.3 180º -72º =108º (angles on a straight line) √ 3 Number of learners for Maths=108º x120 360º = 36√ Percentage for Maths =36 x100 = 30%√ 120 5.2.1 Arranged data in ascending order: 2 15; 18; 20; 20; 23; 23; 25√ Median is 20√ 5.2.3 Range is 25-15=10 1 5.2.4 Mean = 15+18+20+20+23+23+25√ 2 7 = 20,57√ 5.2.5 2 1 5 8√ 2 0 0 3 3 5√

6 MATHEMATICS (JUNE 2010) 5.3.1 4 1 mark for correct scaling 1 mark for labelling axes 2 marks for correct plotting 5.3.2 This indicates a weak correlation between the results in the 2 two subjects since the points are fairly scattered. √√ 1 mark for conclusion 1 mark for the reason