Mathematical Literacy Grade 11 NSC Prelim 2018 Paper & Memo

| | education Department: Education PROVINCE OF KWAZULU-NATAL NATIONAL SENIOR CERTIFICATE GRADE I1 | MATHEMATICS COMMON TEST SEPTEMBER 2018 | MARKS: = 75 TIME: 14 hour This question paper consists of 9 pages and 2 diagram sheets.

Mathematics 2 Common Test September 2018 NSC - GRADE 11 INSTRUCTIONS AND INFORMATION Read the following instructions carefully before answering the questions: |. This question paper consists of 6 questions. 2. Answer ALL the questions. Clearly show ALL calculations, diagrams, graphs, et cetera, which you have used in determining the answers. we 4. Answers only will not necessarily be awarded full marks. 5. You may use an approved scientific calculator (non-programmable and non-graphical), unless stated otherwise. 6. Ifnecessary, round off answers to TWO decimal places, unless stated otherwise. 7. Diagrams are NOT necessarily drawn to scale. 8. Number the answers correctly according to the numbering system used in this question paper. 9. Write neatly and legibly. Copyright Reserved Please turn over

Mathematics 3 Common Test September 2018 NSC - GRADE 11 QUESTION 1 Ld For two events A and B, it is given that P(A) = 0,4 and P(B) = 0,3. Calculate P(A or B) if: 1.1.1 A and B are mutually exclusive. (2) 1.1.2) A and B are independent. (3) 1.2 A survey was carried out among 100 learners about which movies they have watched recently. The results are given below: ° 43 watched Braven (B) ¢ 41 watched Hereditary (H) ° 50 watched The Kissing Booth (K) * 6 watched all 3 movies ° 7 watched Braven and The Kissing Booth but not Hereditary e 18 watched Hereditary and The Kissing Booth * 15 watched only Hereditary The above information is represented in the Venn diagram below. (SLN e 1.2.1 Write down the values of a, b, c and d in the Venn diagram above. (4) 1.2.2. Calculate the probability that a learner selected at random from this group has not watched Braven or Hereditary or The Kissing Booth. (3) 1.2.3 Calculate the probability that a learner selected at random from this group has watched at least 2 of these movies. (3) {15} Copyright Reserved Please turn over

Mathematics 4 Common Test September 2018 NSC - GRADE 11 QUESTION 2 2.1 Jane bought a laptop for R9 600. Calculate the book value of the laptop after 3 years, if it depreciates at 20% p.a. on the reducing balance method. (2) 2.2 Sandra invests R20 000 in a savings account that pays interest at the rate of 12% p.a., compounded monthly. How much will she have accumulated in this account after 4 years? (3) 2.3. Thuli has two sons: Bongani and Sbu. They share the same birthday. On the day that Bongani had his 12" birthday and Sbu his 15" birthday, she opened an account for each of them, and invested some money in each account. The total amount that she invested in the two accounts was R150 000. Both accounts earn interest at 9,6% p.a., compounded quarterly. She divided the R150 000 between the two accounts in such a way that each of the boys will receive the same amount of money on his 25" birthday. How much money does she invest for Bongani? (5) [10] Copyright Reserved Please turn over

Mathematics 5 Common Test September 2018 NSC - GRADE 11 QUESTION 3 ABCD is a cyclic quadrilateral and CB is extended to E. DA = 155mm and DC = 136mm. ABE =81° and CE= 193mm. AD || CE. 193 mm 3.1 Give a reason why D=8l°. (1) eee) Calculate the length of AC. (3) 3.3 Calculate the size of DAC. (3) 3.4 Calculate the area of AACE (4) [11] Copyright Reserved Please turn over

Mathematics NSC - GRADE 11 QUESTION 4 The organiser of the regional school soccer league decided to have a trophy made to present to the winning team of the season. He had it made in the shape of a cylinder 6 Common Test September 2018 with a sphere on top of it. The radius of the cylinder is 5 cm and its height is 20 em. The total volume of metal used to make this trophy is 2000 cm?. \ Calculate the radius of the sphere. Copyright Reserved SOCCER TROPHY Ve | th 3 Vearh =—ar° 3 15] Please turn over

Mathematics 7 Common Test September 2018 NSC - GRADE 11 Give reasons for your statements and calculations in questions 5, 6 and 7. QUESTION 5 In the figure, O is the centre of the circle and A, B, D, E, and F are points on the circumference such that AB//FOC. D,=22° and C= 78°. Calculate the size of each of the following angles, giving reasons: 5.1 O, (2) 5.2 B, Q) 53. £ (2) 540 @) [9] Copyright Reserved Please turn over

Mathematics 8 Common Test September 2018 NSC - GRADE 11 QUESTION 6 6.1 In the figure below AB is a tangent at A to the circle with centre O. D and E are points on the circle Prove the theorem that states that DAB= E. (5) 6.2 Inthe figure PT is a tangent to the circle at T and TR is a chord. PR cuts the circle at Q. S is a point on TR so that PTSQ is a cyclic quadrilateral. PS, TQ and SQ are joined. p Q SS T 6.2.1 Prove that PS isa tangent to circle QSR at S. (5) 6.2.2 Prove that PT = PS. (6) [16] Copyright Reserved Please turn over

Mathematics 9 Common Test September 2018 NSC - GRADE 11 QUESTION 7 O is the centre of the circle. BOA is a diameter and cuts chord TC at N such that BONA LTC. Ne = a TC = 24 units. BN 9 Let NA = 4x. B T NC A 7A Write down, giving a reason, the length of TN. (2) 7.2 Write down, in terms of x, the lengths of: 7.2.1 BA () 7.2.2 ON (2) 7.3 Calculate the length of the radius of the circle, (4) 19] TOTAL: 75 Copyright Reserved Please turn over

Mathematics 10 Common Test September 2018 NSC - GRADE 11 DIAGRAM SHEETS: HAND IN WITH YOUR ANSWER BOOK NAME: GRADE: QUESTION 5 QUESTION 6.1 Copyright Reserved Please turn over

Mathematics 11 Common Test September 2018 NSC - GRADE 11 es QUESTION 7 Copyright Reserved Please turn over

MARKS: 75 education Department: Education PROVINCE OF KWAZULU-NATAL NATIONAL SENIOR CERTIFICATE = MATHEMATICS ‘ COMMON TEST SEPTEMBER 2018 MARKING GUIDELINE F This marking guideline consists of 7 pages.

Mathematics 2 Common Test September 2018 GRADE 11 Marking guideline QUESTION 1 1.1.1 | P(A or B) = P(A) + PB) = 0,4 + 0,3 ¥ subst. into correct formula =0,7 ~ answer (2) 1.1.2 | P(A or B) = P(A) + P(B) — P(A). P(B) Y correct formula = 0,4 + 0,3 — (0,4)(0,3) ¥ subst. into correct formula = 0,58 Y answer (3) 1.2.1 |a=12 Ya=12 b=8 vb=8 c=22 vo=22 vd =25 d=25 (4) 1.2.2. | Number that did not watch any of these movies = 100-(22+8+6+7+25+ 12+ 15) 100 — (22+8+ 6+7+25+12+15) =§ v5 5 P(did not watch any of these movies) = in 0,05 or 5% | “answer (3) 1.2.3 6 12 v¥ number that watched at least 2 of P(watched at least 2 of these movies) Se ese vomies waren BOESES ao or 0,33 or 33% 100 ¥ dividing by 100 ¥ answer (3) 15] Copyright Reserved Please tum over

3 Common Test September 2018 Mathematics GRADE 11 Marking guideline QUESTION 2 2.1 | 4=P-i)" Book value = 9 600(1—0,2)’ ¥ substitution into correct formula =R4915,20 ~ answer (2) 22 | 4=P(1+0)" 0,12 viz 8 A=20 ooo( 1-42)" 12 12 Y substitution into correct formula = R32 244,52 V answer (3) 2.3. | Let the amount that Thuli invested for Bongani be x -. the amount that Thuli invested for Sbu is150 000-—x axl3 4x10 0,096 ax13 x [' 42,096 a = (150 009-j(14 228) 2 eres } 4 4 4x10 0,096)" 0,096 \* 0,096) ¥ =(180000-9[ 1+ a 14+—— " =150.000 14+— —x|1l+— 4 4 0,096)" 0,096)" 0,096)" fs (122%) =150 000] 1+ ri Y simplification 150 al 0, ops)" 4 a= 40 0, 096)’ 0,096 1+ FT + ao Y answer x =R64 399,02 (5) OR OR 4x13 4x10 x (1228) = (150 no0-)[1+ a 7) 3,43239883x = (150 000—.x)(2,582249878) 3,43239883x = 387 337,4817 —2,582249878x 6,014648708x = 387 337,4817 _ 387337,4817 6,014648798 Y answer x = 64339,02 (5) NOTE: Answer only: full marks [10] 4x13 4x10 v= (1s0000—x)(1+ 908) Y simplification Copyright Reserved Please tum over

Mathematics 4 Common Test September 2018 GRADE 11 Marking guideline QUESTION 3 3.1 Exterior angle ofa cyclic quadrilateral = opposite Y correct reason interior angle. qd) 3.2 AC? = AD’ + DC? —2(AD)(DC)cosD Y applying cosine rule =155? +136° ~2(155)(136)(cos81’) ¥ substitution = 35925, 72... AC =189,54 mm v answer (3) 3.3 sinDAC _ sinD Y applying sine rule DC AC sin DAC _ sin81° = v ‘ati B6 189,54 substitution DAC =45,13° Y answer (3) 3.4 ACB= DAC = 45,13° [alt. 2’s; AD || BC] ¥ ACB= DAC Area of AACE = (AC)(CE)sin ACB ¥ applying area rule i : = 3 (189,54) (193) sin 45,13" Y substitution = 12962,73 mn? Y answer (4) [14] QUESTION 4 S Total V = V of cylinder + V of sphere 2000 = arh + Sa 2000 = x(5)’(20) + sm = 500r + 3m? 3 4 ,_ 2000—500z a r -(3}136.5.. 4 r* =102,4648.... r=4,68 cm V arh + Sa Y equating to 2000 Y substitution VY making +? the subject of the formula Y answer 65] Copyright Reserved Please turn over

Mathematics 5 GRADE 11 Marking guideline Common Test September 2018 QUESTION 5 5.1 O, = 100° —_[ext. angle of A] YSYR (2) 5.2 B, = 50° [Z at centre = 2 x Z at circumference] YSYR (2) 5.3 E = 130° [opp. angles of cyclic quad] vSYR (2) 5.4 ABC-C, [corresp. 2°s; AB |[FC] YSYR = 78° B, = 78°—50° * 28° Y answer 3) OR — f = 78°-50° [ext. angle of A] = 28° VS B, = 28° [alt. 2°s; AB ||FC] Vanswer YR (3) 1] Copyright Reserved Please turn over

Mathematics 6 Common Test September 2018 GRADE 11 Marking guideline QUESTION 6 6.1 Construction : Join OA and OD ¥ for construction Proof: Let A, =x wA, = 90° —x (tangent 1 to radius] ¥S/R AO =OD [radii] D, =A, =x [Z’s opposite = sides] v S/R AOD = 180° —2x (sum of 4’s of A] YS/R E = [AOD [4 at centre =2 x Z at YSYR circumference] = 90°-x 6) E = A, OR OR ; Construction: Draw diameter AC, and join C and D. ¥ for construction ae . . B A, +A, =90° [tangent 1 to radius] VSIR ADC=90° [Z in semicircle] c+A, =90° [sum of 2’s of A] VSIR nOnk, But C=E [Z’s in the same segment] YSVYR 6) 5 Copyright Reserved Please tum over

7 Common Test September 2018 Mathematics GRADE 11 Marking guideline Aue 6.2.1 t=8,= 2% [angles in the same segment] ¥ SYR T,=R =x [tan-chord-theorem] VY SYR §,=R PS is a tangent to circle QST [converse: tan- YR chord-theorem] (5) 622 | Let hy P, = i, =y {angles in the same segment] VY SYR PTR = T+T, =xt+y s, = B+R =yptx [ext. angle of A] VSYR §= PTR vS PT = PS [sides opp. equal angles] vR © [16] QUESTION 7 71 TN = 12 units [line from centre of circle 1 to chord] YSYR (2) 71 BA= 4x+9x = 13x v answer (1) 7.2.2 132 : OA = — [radius = * diameter] “length of radius ON = 0A — NA = B¥_4y = 2 ¥ answer (2) 7.3 TN’ +ON? =OT? [Pythagoras] 5x) (13x) “SR (8 2 2 Y substitution 2 2 1444 25x _ 169x 4 36x° =144 v=4 xeD 13(2 radius = ie BG) ) =13 Yvalue of x 2 ¥ answer (4) [9] TOTAL: 75 Copyright Reserved