Mathematics Grade 12 Paper 3 Eastern Cape Mock 2012 Memo

Province of the EASTERN CAPE EDUCATION NATIONAL SENIOR CERTIFICATE GRADE 12 SEPTEMBER 2012 MATHEMATICS P3 MEMORANDUM MARKS: 100 This memorandum consists of 13 pages.

2 MATHEMATICS P3 (SEPTEMBER 2012) QUESTION 1 1.1 2; 4; 10; 20 (3) 4; 10; 20 1.2 2a = 4 a=2 a=2 3a + b = 2 3(2) + b = 2 6 +b=2 b = -4 (OR OTHERWISE) b = -4 a+b+c=2 2–4+c=2 c=4 c=4 Tn = an2 + bn + c Tn = 2n2 – 4n + 4 (4) Tn = 2n2–4n+4 [7] QUESTION 2 2.1 × 36 600 OR 36 600 – 12,5% (36 600) 87,5 or 12,5% = 32 025 = 32 025 (2) answer 2.2 Increase = 41 109 – 36 600 = 4509 4509 Percentage increase = × 100 = 12,32% (2) answer 2.3 Average monthly sales = 36 578 = 36 578 No, because 36 578 ˂ 37 000 (2) No/reason [6]

(SEPTEMBER 2012) MATHEMATICS P3 3 QUESTION 3 3.1 IQ score ˃ 115 × 48 000 000 = 7 680 000 people (2) answer 3.2 IQ score ˂ 130 = × 48 000 000 = 47 040 000 people (2) answer 3.3 48 000 000 – 47 040 000 960 000 = 960 000 × 100 answer only (full marks) answer = 2% OR (2) 3.4 Sample = × 100 10% = 10% Yes, sample is 10% of the population. (2) conclusion [8] QUESTION 4 4.1 4.1.1 A only/ B only A B A B (A B)' 0,2 0,2 0,3 0,3 (3) 4.1.2 For mutually exclusive events: probability rule P(A) + P(B) = P (A B) 0,4 + 0,5 = 0,7 substitution 0,9 0,7 P(A) + P(B) P(A B) NOT mutually exclusive events (3) conclusion

4 MATHEMATICS P3 (SEPTEMBER 2012) 4.1.3 For independent events: probability rule P(A) × P(B) = P(A∩B) P(A) + P(B) – P(A∩B) = P(A B) expansion 0,4 + 0,5 – P(A∩B) = 0,7 - P(A∩B) = -0,2 P(A∩B) = 0,2 P(A) × P(B) = 0,4 × 0,5 = 0,2 answer P(A) × P(B) = P(A∩B) independent events (4) conclusion 4.2 4.2.1 Ages ˂ 30 30 – 39 40 – 49 50 TOTALS ages qualification Qualifications 3 year, professional 15 152 102 221 490 4 year, professional 43 337 311 166 857 Degree, professional 211 578 298 145 1232 Higher degree, professional 12 121 127 37 297 Degree, no professional 13 55 45 15 128 TOTALS 294 1243 883 584 3004 (2) 4.2.2 P (teacher ˂ 40) = addition = = 0,51 (2) answer 4.2.3 P (teacher ˂ 40 and professional degree) addition = = = 0,31 (2) answer

(SEPTEMBER 2012) MATHEMATICS P3 5 4.2.4 P (teacher, no degree) = addition = = 0,45 (2) answer 4.2.5 P (teacher ˃ 40, no degree) addition = = = 0,27 (2) answer 4.2.6 For mutually exclusive events: addition P(A) + P(Q) = P (A Q) + = answer 0,383 0,070 not mutually exclusive events (3) conclusion [23]

6 MATHEMATICS P3 (SEPTEMBER 2012) QUESTION 5 5.1 Number of letters available: 26 26 and 9 Number of digits available: 9 Must have a total 75 000 inequality 1 digit: 9 × 26 = 234 2 digits: 9 × 9 × 26 = 2106 3 digits: 9 × 9 × 9 × 26 = 18 954 multiplication 4 digits: 9 × 9 × 9 × 9 × 26 = 170 586 The inventory system must have FOUR digits. (4) conclusion 5.2 5.2.1 5 × 3 = 15 (1) answer 8 5.2.2 C2 = ways counting OR principle 7 + 6 + 5 + 4 + 3 + 2 + 1 = 28 (2) answer [7]

(SEPTEMBER 2012) MATHEMATICS P3 7 QUESTION 6 NOTE: According to the NCS the solutions to data-handling problems should be done with the use of a calculator. The alternative is to use the pen and paper method. 6.1 160 140 120 Waistline after 6 weeks 100 80 60 40 20 0 0 20 40 60 80 100 120 140 160 Weight after 6 weeks 9 points (2) 6.2 weak, positive relationship weak, (1) positive 6.3 r = 0,597 (2) answer 6.4 ŷ = a + bx a = 74,8 + 0,3x (3) b 6.5 y = 92,8 cm from the graph answer line (2) 6.6 ŷ = 74,8 + 0,3x = 74,8 + (0,3 × 200) substitution = 134,8 cm (2) answer [12

8 MATHEMATICS P3 (SEPTEMBER 2012) * FOR QUESTIONS 7 TO 10 FOLLOW CANDIDATES REASONING * QUESTION 7 C B . O1 2 D A Construction: Make O the centre of the circle. Draw radii BO and DO construction To prove: ̂ + ̂ = 180° Proof: ̂2 = 2 ̂ (angle at centre = 2 × angle at circumf.) statement/reason ̂1 = 2 ̂ (angle at centre = 2 × angle at circumf.) statement/reason ̂ 2 + ̂ 1 = 360° (sum of angles around a point = 360°) statement/reason 2 ̂ + 2 ̂ = 360° 2( ̂ + ̂ ) = 360° ̂ +̂ = = 180° deduction [5]

(SEPTEMBER 2012) MATHEMATICS P3 9 QUESTION 8 8.1 C B 1 1200 3 2 1 1 O 55 0 D 2 1 A 8.1.1 ̂ 1 = 2A ̂ B (angle at centre = 2 × angle at circumf.) statement = 2 (55°) reason = 110° (2) 8.1.2 ̂ 2 = 35° (∆AOB is isosceles, OB = OA radii) statement reason B ̂ D = 60° (sum of opposite. s of cyclic quad. = 180°) O ̂ D = 60° – 35° statement/reason = 25° (4) answer

10 MATHEMATICS P3 (SEPTEMBER 2012) 8.2 P 1 x S 1 2 3 2 1 T 2 R 1 3 Q 8.2.1 ̂ 3 = x (two tangents joining from the same point) answer/reason ̂ = x (tangent/chord theorem) answer/reason ̂ 1 = x (corresponding angles, RQ ST answer/reason (3) 8.2.2 ̂ 3 = ̂ 1 = x (see above) statement TPSQ is a cyclic quadrilateral; angles subtended by conclusion the same chord are equal. (2) 8.2.3 T ̂ Q = ̂2 ( on same chord: TPSQ is a cyclic statement quadrilateral) T ̂ Q = ̂ 1 = x (from QUESTION 8.2.1) ̂1 = ̂2 conclusion/ TS bisects P ̂ Q (2) reason 8.2.4 ̂ 1 = ̂ 2 = x (alternate angles, RQ ST) statement/reason P ̂ Q = ̂ 1 = x (proven in QUESTION 8.2.1) ∆ RQS is an isosceles triangle, base angles are statement/ equal. (2) conclusion [15]

(SEPTEMBER 2012) MATHEMATICS P3 11 QUESTION 9 E 2 1 G 1 2 1 2 F 2 D C 2 1 1 A 2 1 B 9.1 In ∆ABC and ∆ADB: ̂ 1 is common statement/reason ̂ 1 = ̂ 1 (tangent/chord theorem) statement/reason ∆ABC///∆ADB ( ) = 2 AB = AC.AD (3) conclusion 9.2 In ∆ACE and ∆AED: ̂ 2 is common statement/reason ̂ 1 = ̂ 1 (alternate angles, AE GF) statement/reason ̂ 2 = ̂ 1 (ext. angle of cyclic quadrilateral GDCF) statement/reason ̂ 1= ̂ 2 ∆ACE///∆AED ( ) (3)

12 MATHEMATICS P3 (SEPTEMBER 2012) 9.3 = (∆ACE///∆AED) statement/reason 2 AE = AC.AD and deduction AB2 = AC.AD (proved in QUESTION 9.1) AB2 = AE2 AB = AE (3) conclusion [9] QUESTION 10 A D F E C B 10.1 = (given) answer (1) 10.2 = = (line to one side of ∆) statement (1)

(SEPTEMBER 2012) MATHEMATICS P3 13 10.3 = ratio = AB = 6x AB = 6x = = (3) answer 10.4 Area of ∆CFB = FB.h formula Area of ∆CFA = AF.h AB = FB + AF FB = AF = 3x 6x = 3x + 3x (AB = 6x from QUESTION 10.3) = = Area of ∆CFB = Area of ∆CFA (3) answer [8] TOTAL: 100