# Mathematics Grade 11 September Test 2016 Paper & Memo > Grade 11 Mathematics question paper and memo from the September Test 2016, 12 pages. Read online or get the free PDF download. - Page: https://www.hlayiso.com/library/maths-sept-qp-memo/debccdc8-5a2d-463d-a5a5-7e9a2ef36b71 - Updated: 2026-09-27 - Subject: Mathematics - Grade: Grade 11 - Document type: Question paper and memo - Year: 2016 - Exam period: September Test - Curriculum: CAPS - Pages: 12 Breadcrumbs: [Home](https://www.hlayiso.com/) › [Library](https://www.hlayiso.com/library) › [Grade 11](https://www.hlayiso.com/library/grade-11) › [Mathematics](https://www.hlayiso.com/library/grade-11/mathematics) › [Mathematics Grade 11 September Test 2016 Paper & Memo](https://www.hlayiso.com/library/maths-sept-qp-memo/debccdc8-5a2d-463d-a5a5-7e9a2ef36b71) ## About this document Mathematics · Grade 11 · September Test · 2016. Question paper and memorandum, 12 pages. Read online or download the PDF. Read it free online on the page above; download the PDF from there. ## Document text Text extracted from the PDF. Layout, diagrams and tables may not survive extraction. ### Page 1 Downloaded from hlayiso.com testpapers.co.za ### Page 2 Downloaded from hlayiso.com Mathematics 2 September 2016 Common Test NSC INSTRUCTIONS AND INFORMATION Read the following instructions carefully before answering the questions. lL. This question paper consists of FIVE questions. 2. Answer ALL the questions. 3. Clearly show ALL calculations and diagrams that you have used in determining your answer. 4. You may use an approved scientific calculator (non-programmable and non-graphical). 5. Ifnecessary round off answers to TWO decimal places, unless stated otherwise. ( ; 6. Answers only will not be awarded full marks. 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 ### Page 3 Downloaded from hlayiso.com Mathematics 3 September 2016 Common Test NSC QUESTION 1 1.1 A travel agent did a survey amongst his clients as to which type of holiday they prefer. The table below shows the results of this survey. Game Sea Drakensberg Total reserve Male 250 70 a 500 Female 150 b 160 c Total d 170 e 910 1.1.1 Write down the values ofa, b, ¢, d and e. (2) 1.1.2 Ifa client is selected at random, what is the probability that this client } (a) Will prefer visiting a game reserve? (1) (b) Will be a female? qd) (c) Will be a female and prefer visiting a game reserve? dd) (d) Will be a female or prefer visiting a game reserve? (2) 1.1.3. Consider the following two events: Event A: a client is a female Event B: a client prefers visiting a game reserve when going on holiday. Are events A and B independent? Motivate your answer with the necessary calculations. (3) 1.2 Figures obtained from a city’s police department indicate that of all the vehicles stolen, 70% were stolen by syndicates (gangs) to be sold off, and 30% were stolen by individual Co} persons for their own use. Of the vehicles stolen by syndicates: e 10% were recovered (found back) within 24 hours; *® 30% were recovered after 24 hours; and © 60% were never recovered. Of the vehicles stolen by individual persons: « 30% were recovered within 24 hours; e 40% were recovered after 24 hours; and * 30% were never recovered. 1.2.1 Draw a tree diagram to represent the above information. @) 1.2.2 Calculate the probability that if a vehicle was stolen in this city, it would be stolen by a syndicate and recovered within 24 hours. (2) 1.2.3 Calculate the probability that a vehicle stolen in this city will not be recovered. (3) [18] Copyright Reserved Please Turn Over ### Page 4 Downloaded from hlayiso.com Mathematics 4 September 2016 Common Test NSC QUESTION 2 A frustum is the portion of a cone which remains after a smaller cone has been cut off its upper part. FRUSTUM 13,5 em arem Gaba Smaller cone that has been cut off. The height of the small cone is 13,5 cm and the height of the large cone is 27om. C i The diameter of the base of the large cone is 24cm and the diameter of the base of the small cone is 12cm. 2.1 Find the volume of the frustum, giving your answer in terms of 7. (4) 2.2 The frustum has the same volume as a certain sphere. Calculate the radius of this sphere. List of formulas to choose from when answering Question 2: Vata? ( Vearh : Vetmn (3) [7] Copyright Reserved Please Turn Over ### Page 5 Mathematics 5 September 2016 Common Test NSC : Downloaded from hlayiso.com | QUESTION 3 3.1 Inthe diagram A(-15; p) is a point such that OA=17 and DOA =6, where @ isa reflex angle. y °C eL) Se BO; y) } A(-15; p) 3.1.1 Calculate the value of p. (2) 3.1.2 Determine the value of each of the following without the use of a calculator: (a) cos 0 qd) (b) tan (180° — 8) (2) (c) — sin (@ — 360°) . cos (90° — 0) (3) 3.1.3 B isapointon OA, such that OB = 10, Calculate the values of x and y without calculating the size of angle @. (4) (5 3.2 Simplify the following and express your answer as a single trigonometric term. _ o cos ( x). tan 225 + fan 6 60° — x) (1) sin 90° ~ sinx 3.3. Prove the following identity: 2 sin’ x cosx =< (4) 2tanx — 2 sinx cosx sinx 3.4 Determine the general solution for x if: sin (2x + 15°) = cos (x ~ 15°) (5) [28] Copyright Reserved Please Turn Over ### Page 6 Downloaded from hlayiso.com Mathematies 6 September 2016 Common Test NSC QUESTION 4 The following sketch represents the graphs of: SR) = aeos (x+ 6) and. a) = esinx for x @[-180° ; 180°] : : - Bd 9, a — [}- Ne aati a To a a re OT ED C AL { /- sew an — S Rane Ga N WA 41 From the sketch‘above write down the-values of @, b and ¢. (3) 42. A(-14,64°; p) and B(q;-0,51) are the coordinates of the points of intersection of f and g. Write down, rounded off to:2:deeimal places, the values of: 4.2.1 p O% 422 ¢ (dy 43 Fort which values of x will f(x) . g(x) 2 0? (2) 44 Ifthe.graph of f is shifted 45° to the left, write down the equation of the new graph, a (8) Copyright Reserved Please Tarn Over ### Page 7 Mathematics QUESTION 5 Downloaded from hlayiso.com 7 NSC September 2016 Common Test 5.1 In the figure below, AB = 107 mm, BC =68mm and ABC = 79°. AC L DC. DAC=22°. 5.2 (a) (b) 5.1.1 Calculate the length of AC. Q) 5.1.2 Calculate the length of DC. QQ) 5.2.1 Complete: Areaof A ABC = sab ees () 5.2.2 In the figure below AB=AD with DC=%BD and DC=r. It is also given that ABD =o, and sin 20 = 2 sina cosa. A a Cc B D Show that AD = —* (4) cos a Now determine the area of A ADC, in terms of o and 7. A [14] TOTAL MARKS: 75 Copyright Reserved ### Page 8 Downloaded from hlayiso.com ### Page 9 48NQ WIN Bsea|q poasay iy6uAdo “saSed 1 jo sysisuoo wWnpuBIoMeU SIG], “gq"N fe) oO © Surpnyouoo 10j YT // Taspusdsput iON, {o) (q pue va #(@adx(W)d 103 VOI (a pur Wa # (Max) n (@ax(y)d Smemozeo 105 YT Bblo / oz'o= 8 _ 016, 016 = (@d x (Vd | OFT OOF OlP > aio | oro= US _ Cqpeyya & mm ST 0ST vl & 16 _ 016 = Jomsue 10} VOT Ela | 997099 = SLVOIILYSD YOINSS ro} vpMLULOY yoaTIO9 Futsn JOJ YD T (p2i6._016 , O16 “WNOILLVN Y Ost 00r Oly | PCT e oD 7 16 _ 016 ro} sonsue 10f-V5 1 aifo | Ao ost loc SOU rnnnangassenenne nes 9) () : - 9T0t MADNALAAS So 1 16 _ 016 . © Josue 107 YO I sto] +yray acti pope. : » LSELNOWNWOD ' oO {8 vomaue th fo / pile Hi — 10) a c 2vot Ooh [CT : WOAGNVYOWR @ a ey s 016 [Ove [OLT| oor | roy . - SOLLVNGAILVIA re) oy | oot [oor] ost | emarg ent waweeeecy. eee ey en sampeA joo y 30 ¢ 10} vI | |_00S ost _[o| ose | aw] | | | “S#88RSSa==esee5=55 Seeeeneeesss Qa SONJEA JOHIOD ¢ JOJ WT suedoxe ‘DATISOE Tey, | Aq pavay | vag | suey ITT a TNOLLSANO 910z nquadeg yoy wouMeT) ox ™~ sonEMOREA bite. shew ### Page 10 Downloaded from hlayiso.com Mathematics 3 Common’ ~~ ptember 2016 . NSC . ww 12.1 y ‘within 24h) 10% 2 7a 24 LA for Birst branching: into S and IP 1A for branching from S into R(within 24h), R(after 24h) and on R(within 24h) NR 1A for branching from IP into wT Rests 24h) R(within 24h), R(after 24h) and. NR S: syndicates; IP: individual persons; | R: recovered; NR: not recovered (3) 1.2.2 ag x 10% = 0,07 1A for 70% x 10% OR 0,7x0,1 0,7x0,1=0,07 1A foi angwer 2 70% x 60% + 30% x 30% 9 123 Jost LA for 70% x 60% OR 0,7x0,6 OR 1A for 30% x 30% OR 0;3x0,3 0,7x0,640,3x0,3 1A for correct answer =051 (3) [18]. QUESTION 2 2.1 | Volume of large cone = 477k =ix(2Peny 1A for substitution = 12960 ¥ 1CA for answer Volume of small cone = 17h = 426)'03,5) =162 2v 1A for answer Volume of frustum = 1296 7-162 7 =11340 ¥ 1CA for answer (4) 2.2 | Volume of sphere= $777 $m? = 11340 ¥ 1 CA for equating r=850,5 ¥ 1CA for simplification r=94AT om ¥ 1CA for answer (3) 7 Copyright Reserved Please Tum Over Mathematics t : nec ‘Common Test September 2016 QUESTION 3 | 3.1 | 3.1.1 |p? =1PF -Cisev 1A for using Pythagoras | = 8 1A for correct answer Q) 3.12 15 : a) cos “TT v 1A for answer ay 3.12 | tan (180° — @) =—tane ¥ 1A for correct reduction ) _8 = iofv 1€A for answer 15 2) 3.12 sin © — 360°) . cos (90°— 6) © ; | 1A for sin(@—360°) =sin@ = sind sind 77 1A for cos(90" -6)=sin8 = six? = (<8) 17 4 oa ~ 339 1CA for answer _ @) 313 |g w 82 ICA for equating the rato 17-10 ly = -80 i 80 v i v= a or —4,71 1CA for answer -15_ x - . cos@ = ———-=— ¥ 1CA for equating the ratio 1710 17x = -150 ' 150 =p oF -89L 1CA for answer (4) is-t _ le “FF £ = 61% f. m= - 8,82 \ oO A = ie & 7 DA = 4ytl = -%7l Copyright Reserved ### Page 11 Downloaded from hlayiso.com JONO wun @Sediq Penteseyy 1yBuXdog JOAQ WN] esedig peniesey yBuAdog © somsue JO} YT wal 66°97 = uonrorydmis roy YO] A CT WITIESIL = OG ecm = 2Y ones 10) VT an “oa | crs (3) TOMSUE TOF VOT 7 wm TESTE = OV PF SLYEOEZEL = PMULOY UY MoNMINSGNS IOFVI | —_p 64 $09 E9 “LOI ‘Z— 789 + LOL = g[1 onIs0o SurAydde soy WT A FS00(QG AW) - d+ AV = OV] rs rs SNOLLSIN0 Isl () JSMSUR JO} YT AXxX800 = £ rr @ [.08T ‘oS€t] 203 VT [.0 £6$b-]393 VI / [ost ‘set 9 -LO%st]| er W _ JomasuB OF WT GEz'sor_ 0 .99'S91 10) » o¢'s91 = b| Tp @ sous JOY YT siso=4| rer aa (3) 2 par q Jo sonfea 1oy WE ZO = 9 ASH =F AT=P| Th 9107 1eqmardas yay wou » NOLLSANO SOMEMIDMBEY Iszt ©) Z 4‘ 0vC%+ OF=* VOI PY ZY OVCA+ OF=X 09EY+ 09=XFT OIA EXE SL S142T 0964 +(*F— S01) 081= S147 VOI fp A9EY+(XF— SOI) — O8T= $1427 10 Z24 ‘welt 9€=* VOI yp ZY ‘ppvy+og=x 09EF+ .06=XFZ O9EF+XE— .SOL= ST+*Z WOT 4 O9EY+XE— SOL= SI+XT (x4— soijus=(.s1+z)ms Uys Jo SUD} UI BUNA 305 YT ZA [(st-*4)— oguis=(.st+xz)ms |e @) x WIS x SOO = x UIs“ xUIS Z xX UIS=X S0O-] OY V1 4 Xe FUT = x S09 ¢ 800 — ]) xUISZ BUIZLO}OBy 103-1 * x MIS Z “ XSCO xowuyousp Burxyduns 10 yy 4 ase xXxS00 800 800 X UIST — See ag 7 2 VE 4 mT gg (2) 800 ae SASUE IO} YOT I (xuts — ]) xs00 xUIS— 1A [=x ,8004% (UIs 107 VOT (xups— }) xsoo x, UIS+x WIS — x $00 (201s — 1) xs00 suonoeyy SuIppe Jog YD | xsoOo xs —] 800 203 VI ,eUS ~ [x80 = xuls xuej—=(x— ,09¢)ue soy YT T= .06UIS PUB T= .C7Z UE) JO} YT A xus—pe 800 =(x—)soo 10y YJ (#8) + 50,4 re rm OSN 9107 eqmaydes so] wou ¢ sonemampepy ### Page 12 Common? ptember 2016 = - _& [52 Copyright Reserved 521 |*%absincv 1M for correct formula @ 322 @ BD = 2r BAD = 180° - 20v 1A for finding BAD = AD _ _BD_y 1A for applying sine rule sinB- ‘sinBAD oO AD == 2 v 1A for correct substitution sina sin(L80°- 2a) 4 {o) Ap = 2rsing ” sin 2@ —_ or si > = ome 1CA for simplification/substitution oO 2sing cos a@ cd or = cos & (4) £ va ADC= 180° -a ¥ 1A for finding ADC re) s.Area of AADC = ¥% AD.x. sin(180°—a@) | 1A for applying area rule ‘* . r= = 4 r jaaY 4a me LA for substituting AD=—7— and xe) cos & (