Mathematics Grade 11 Paper 1 KZN November 2025 Paper & Memo

Downloaded from Stanmorephysics.com KWAZULU-NATAL PROVINCE EDUCATION REPUBLIC OF SOUTH AFRICA NATIONAL SENIOR CERTIFICATE GRADE 11 MARKS: 150 TIME: 3 hours This question paper consists of 9 pages and | information sheet.

Matbsmaitioaded from Stanmerephysigs.com KZN/November 2025 INSTRUCTIONS AND INFORMATION Read the following instructions carefully before answering the questions. 1. 10. This question paper consists of 10 questions. Answer ALL the questions in the SPECIAL ANSWER BOOK provided. Clearly show numbering if additional space has been used to answer some question. If NOT, cross out all rough work done in the additional space. Clearly show ALL calculations, diagrams, graphs, etc. which you have used in determining your answers. Answers only will NOT necessarily be awarded full marks. You may use an approved scientific calculator (non-programmable and non-graphical), unless stated otherwise. , =) If necessary, round off ans} O decimal places, unless stated otherwise. Diagrams are NOT necessaril An information sheet with fe d at the end of the question paper. Write neatly and legibly. Copyright Reserved Please Turn Over

““"SBWiloaded from Stanmerephysigs.com QUESTION 1 1.1 Solve for x: 11 4x°-9=0 1.2 2x? +6x-3=0 (answers correct to TWO decimal places) 2 _ 113 pega lt te 6 x+2 1.14 x? -2x-1529 1.2 Two equations are given below: 1.3 1.4 Copyright Reserved 2” 1.2.1 Show that —=4" written as y=2x-2. 1722 Hence, or of e for both x and y. The roots of a quadratic equation are given as: x=J/2—k , where ke (-9 : 0). Determine the integral values of k for which the roots are rational. where b <0 and —bt The roots of the equation ax” + bx +c =0 are cel c=b+18. Calculate the values of a, band c. KZN/November 2025 (3) (3) (4) (4) (2) (4) (2) (5) [27] Please Turn Over

““"S8Wiloaded from Stanmerephysigs.com QUESTION 2 2.1 Simplify fully, WITHOUT using a calculator: 0 2 2x(27x° ) { = } Die es -2 ai (6-288) iB m+4 4 5m+tl 2.2; It is given that a8 5.243 22.1 Simplify the above expression. m+3 2.2.2 Hence, solve for x if a : ’ amass 4am! 125° 2.3 Solve for x: QUESTION 3 KZN/November 2025 WhatsApp messages were sent using face emojis with two popping hearts per face (shown below). (3) (4) (3) (3) (5) [18] MESSAGE 1 MESSAGE 2 MESSAGE 3 These faces and popping hearts in the different messages forms a linear number pattern. 3.1 3.2 3:3) 3.4 EB How many popping hearts will appear in message 4?. Determine the n” term (general term) of the popping hearts in the messages. Calculate the number of popping hearts in the tenth (T,, ) message. In which message will there be 710 popping hearts? A new series of messages using the same emojis was created by: e keeping the number of faces in message 1 and message 2 the same as the original messages. e adding 2 more faces to the original message 3. e adding 6 more faces to the original message 4. The new series of messages forms a quadratic number pattern. (1) (2) (2) (2)

Matbsmaitioaded from Stanmerephygigs.com KZN/November 2025 QUESTION 4 The first four terms of a quadratic number pattern are —7 ; m ; 5 ; n. The constant second difference is 4. 4.1 Prove, by calculations, that m=-3 and n=17. (3) 4.2 Which two consecutive terms of the quadratic number pattern has a first difference of 292? (3) 4.3 It is given that the general term of a quadratic number pattern is T, =2n* —2n—7. Calculate the value of n if 7, =1 005. (4) [10] QUESTION 5 Consider the function f defined by f(x) - +5 . Graph f passes through (4 ; 0). r xz 5.1 Write down the y-inter Se a (1) 52 Write down the equations ¢ ptotes of / (2) 5.3 Calculate the x-intercept (2) 5.4 Draw a neat sketch of f Clear ly ‘abel all the intercepts with the axes and asymptotes. (3) 55 Determine the values of y for which f lies x<-—2 or 0<x<4. (2) 5.6 Graph / is reflected about the y-axis and shifted : units up to form p. Write down the equation of p in the form of p(x) =.... (2) [12] QUESTION 6 6.1 The function h(x) =ax’ +bx +c has the following properties: ° h(2)=0 e the value of od is 5 2a e therangeof his y< 10;yeR. Draw a neat sketch of f. Clearly label the x-intercepts and turning point. (3) Copyright Reserved Please Turn Over

Matbsmaitioaded from Stanmerephygigs.com KZN/November 2025 6.2 The graphs of g(x)=—-x°+9 and h(x) =ax? +bx-22 are sketched below. 6.2.1 6.2.2 6.2.3 6.2.4 6.2.5 6.2.6 Copyright Reserved Ais an x-intercept of g and / and lies on the negative x-axis. A and D are the points of intersection of graphs g and h. C is the turning point of h. B lies on the horizontal line that passes through the turning point of g such that BC is parallel to the y-axis and passes through the other intercept of g. BC = 18 units ¥ h x Calculate the coordinates of A. (3) 1 3 Show that a=— and b=-~. (4) 4 2 Write down the range of h. (2) It is given that m(x)= x? —4. Explain, in words, the transformation from the graphs of g to m. (2) Calculate the coordinates of D. (5) It is given that p(x)=k. Determine the integral value of k such that the distance between the points of intersection of h and p is 8 units. (2) [21] Please Turn Over

Matbsmaitioaded from Stanmerephysigs.com KZN/November 2025 QUESTION 7 > The graj (x)=2"+q is drawn below. . Graph f passes through the point (= : -13). a Inn a f (2) 72 If g=—8, calculate the coordinates of A. (3) 73 Determine the average gradient between A and B. (3) x 7.4 Determine the values of x where <0 2 7) “ TS: Determine the values of k for which 8—2*—k =0 will have negative roots. (2) [12] Copyright Reserved Please Turn Over

“"SBWiloaded from Stanmerephysigs.com QUESTION 8 8.1 8.2 8.3 8.4 8.5 Copyright Reserved interest at the rate of 10% per annum. pine invests R15 000 for a period of 6 years in an account that pays ulate the interest his investment earned at the end of 6 years. Miss Coetzee bought a brand new car. The value of the car depreciates annually at the rate of 8,25% p.a. on the reducing balance method. If the book value of the car was R277 216-after 3 years, calculate the purchase price of the car. An advertisement reads “XY Bank offers loans at an effective interest rate of 12% per annum”. Calculate the nominal rate of interest if interest on the loan is compounded quarterly. Mr Dlezi invested R25 000 in an account that offers 9,25% per annum compounded half-yearly. Calculate the accumulated amount in this account at the end of nineteen half-years: Mr Masango invested 13% p.a., compounded rate changed to 10,5% initial deposit, Mr M: years, he had R928,30 moi value of x, to the nearest rand ings account that paid interest at the rate of first two years. Thereafter, the interest d quarterly. Three years after making the from the account. At the end of 5 nt than his initial deposit. Calculate the KZN/November 2025 (3) (2) (4) (3) (5) [17] Please Turn Over

Mi i 5 KZN/N ber 2025 “SoWiloaded from Stanmerephysigs.com vember QUESTION 9 In the Venn diagram below, events A, B and C are shown. S A B iS 9.1 It is given that P(A)=0,47 and P(c)=3 9.1.1 (1) 9.1.2 (2) "4 5 9.2 It is given that events A coon fil and P(A) = 5 PB). 9.2.1 Determine the P(A and B) in terms of P(B). (2) 9.2.2 It also given that P(A or B)=70%. Calculate P(B). (5) [10] QUESTION 10 A survey was conducted amongst 45 males and 85 females who were randomly selected. This survey was about the brand of clothes they would love to buy between Gucci (G) and Louis Vuitton (LV). The results are summarised below. e 3y females said they would love to buy Louis Vuitton. ey males said they would love to buy Gucci. e The total number of males who said they would love to buy Louis Vuitton is double the total number of females who said they would love to buy Gucci. 10.1 Calculate the total number of females who said they would love to buy Gucci. 10.2 Are the events of being a ‘male’ and having said that they would love to buy one of the brands independent? Justify your answer with calculations. Copyright Reserved TOTAL MARKS: (6) (4) [10] 150

KZN/November 2025 “"SoWiloaded from Stanmerephysigs.com INFORMATION SHEET Sab EAD’ -4a0 2a A= P(1+ni) A=P(\-ni) A=P(1-i)” A= P(i+i)" T.=a+(n-l)d 8, =F Pa+(n—1d] T, =ar™"! 5 _ alr” 1) sr#l S,=-—4;-l<r<l % r-1 l=? pass" = paslal+i" i i "(x)= lim f(xt+h)- fo) h>0 h x, +x. + d= (-—™)? +029) M22 Ae%) y=mxt+e m=22 JL m= tan 0 X5—X, (x-a) +(y-by =r? In apc: 4 = sn A sinB a =b’ +c? —2be.cos A area AABC = 5 ab.sin C sin(a+f)=sina.cos B +cosa@.sin B cos(a@ +f) =cosa@.cos B-sina.sin B cos’ @—sin®> a cos2a =41—2sin? a 2cos*a-1 Copyright Reserved sin(a@— £)=sina.cos B —cosa.sin B cos(a@—f) =cosa.cos # +sina.sin B sin 2@ = 2sina@.cosa Yo, —x/y n o P(A or B) = P(A) + P(B) — P(A and B) pa =D) Le-H?

Downloaded from Stanmorephysics.com KWAZULU-NATAL PROVINCE EDUCATION REPUBLIC OF SOUTH AFRICA NATIONAL SENIOR CERTIFICATE GRADE 11 Full Name & Surname For Educator / Moderator Use Only ( NOT for learners to complete) QUESTION Ql Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 | Q10 Total Per Question 27 18 13 10 12 21 12 17 10 10 Mark Scored Moderated Mark GRAND TOTAL 150 This answer book consists of 18 pages.

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Downloaded from Stanmorephysics.com ‘ FINAL & CLOSED education MARKING Department: GUIDELINES Education PROVINCE OF KWAZULU-NATAL NATIONAL SENIOR CERTIFICATE (“ann MARKS: 150 These marking guidelines consist of 19 pages.

MathemRawnloaded from Stanmorephysics.com QUESTION 1 CAPS — Grade 11 Marking Guidelines KZN —November 2025 Ld 4x7 +920 (2x+3)(2x-3)=0 Accept subst. to quadratic formula YA correct factors pa laYe) lor x=— YCA answer Y¥CA answer 2 2 (3) 4x -9=0 Ve _./9 r ‘ 2 x =t 4 VA isolating x i] 3 x=-> or x== YCA answer YCA answer 2 G3) 1.1.2 | -2x7+6x-3=0 ye —b +b? —4ac 2a Ha . _ (6) + 4(6)° = 4(-2)(-3 _ + (6) -4(2)(3) | VA correct subst. into x=. ORs x= : 2(-2) 2(2) quadratic formula x=2,37 or x=0,63 Penalise 1 mark for incorrect rounding ¥CA answer ¥CA on) 1:1:3 _ _ 2x +x-6 x+2 (x-3)(x+2)=2x" +x-6 x? -x-6=2x? +x-6 YA multiplication 2 = at + 2-0 YCA standard form x(x+2)=0 x=0 or x¥-2 YCA answer VA rejection (4) 1.1.4 | (x-5)(x+3)29 x? =2x-15-920 x’ —2x-2420 Final answer must be given 2 or 0. vA standard form YCA critical value (accept if shown on a sketch) ¥YCA both answers (4) Copyright Reserved Please turn over

MathemRawnloaded from Stanmorephysics.com CAPS — Grade 11 KZN —November 2025 Marking Guidelines 12.1 | 1 gee 2Y j 2 LAA 1.2.1 is a proof, all SK 77 —98) : marks are accuracy. Var -y=2-2x VA -y=2-2x SPS 2x 2 (2) 1.2.2 | x+6=x7-y x+6=x° -(2x-2) vA substitution x’ -3x-4=0 (x-4)(x+1)=0 ¥CA factors x=4 or x=—l ¥CA both values of x y=2(4)-2 or y=2(- y=6 or p=-4 YCA both values of y (4) OR x+6=x-y y=x-x-6 (2x-2)=x' -x-6 vA substitution x’ -3x-4=0 (x-4)(x+1)=0 YCA factors x=4 or x=-l ¥CA both values of x y=2(4)-2 or y=2(-1)-2 y=6 or p=-4 YCA both values of y (4) 1.3 A=2-(2)=4 & A=2-C7)=9 k=-2 or k=-7 YA k=-7 VA k=-2 (2) YA value of a YCA subst. a & c into A=72 Y¥CA standard form YCA value of 5 with rejection YCA value of c (5) [27] Copyright Reserved Please turn over

MathemRawnloaded from Stanmorephysics.com CAPS -Grade 11 Marking Guidelines KZN — November 2025 QUESTION 2 2 ) 2.1.1 3); (2% 2x.(27x? )3 | —— m(2if)) (3) 2 =2x.(3°x')> x1 VAI a2 =2x.(3x) 3 = 2x(3x)’ =2x(3°x") YA Px =18x! ¥CA answer 3) 2.12 | (6-2v3) iB — 36- 243 +12 vA removing the brackets in 3 numerator b= 243 3 vA rationalising denom. i.e. a3 BOS va _ 483-243 3 3(16V3 = 24) YCA simplifying the numerator —_—— = 16V3 24 ¥CA answer (4) m+4 m+ 994 Dede 5.2+3 ~ 2” edo") = Oi ee2”.2! vA expansion using exponent law 25 23 ans 4 , 93 (51 = 2" (2 ~4x2) a 2 (2 -1) vA factorising the numerator 2"5.8 225 ee or - or 0,2 ¥CA answer 40 eS) Q) OR OR ¥¥A factorising the numerator Copyright Reserved Please turn over

MathemRawnloaded from Stanmorephysics.com CAPS -Grade 11 Marking Guidelines KZN — November 2025 qinr. = 42m som ¥CA answer 4 3/51 @) 2" (24-4x2) 2m.25(2'-1) OR saa OR = 275.8 oP 275 Ler or as or 0,2 vA simplifying terms in numerator 40 5) OR gmt4 4 pmtl vA adding like terms in numerator 5.2m+3 2” 16-8." ¥CA answer 5.28 (3) _ 278 OR 5.2.8 =o. or Ee or 0,2 ~ 40 5 > vA splitting to two fractions gmt _ 4gmtl 5.2mt3 _ gms = ggmtl YA simplification 2758 ISB gmt4-m gy gmtl-m YCA answer 40 40 (3) 16 8 40 40 =5 or iL or 0,2 40 5 1 222 “os YA equating ow to 5 1 5= 5 5:51 5 ise = 5° VA gas oy 5 1+3x=0 1 x= 3 ¥CA answer GB) OR OR YA equating oe to 5 Copyright Reserved Please turn over

MathemRawnloaded from Stanmorephysics.com CAPS — Grade 11 Marking Guidelines KZN —November 2025 i YA 5=5* 125* all 53 ¥CA answer at @) 1=-3x. 1 x=—— 3 vA isolating surds vA squaring on both sides ¥CA standard form x=11 or x#-3 YCA answers with selection. (4) [17] QUESTION 3 Bl; 14 vA answer Q) 3.2 2;6;10;14 T, =4n-2 YA 4n VA -2 (2) 3.3 Tj. =4(10)-2 or 7, =[2(10)-1]x2 YCA subst. of n=10 Ty =38 Answer only full marks vCA answer Q) 3.4 T, =4n-2 710 =4n-2 “CA, T= 710 4n=712 Answer only full marks OR n=178 If n is a decimal, it must be YCA answer rounded to whole number (2) OR OR ¥GCA. T,=355 Copyright Reserved Please turn over

MathemRawnloaded from Stanmorephysics.com CAPS — Grade 11 Marking Guidelines KZN —November 2025 ¥CA answer Q) 2n~1-=355 2n=356 Answer only full marks OR n=178 If 1 is a decimal, it must be 35 2p 614; 26 ccc hearts qi rounded to whole number CS a) 0 first difference MA, essscervssesveneoud second difference 2a=4 3(2)+b=4 2-2+c=2 a=2 b=-2 c=2 T, =2n? -2n+2 266 = 2n -2n+2 2n? —2n+2-266=0 n’—n-132=0 (n-12)(n+11)=0 n=12 or n#-I11 whole number Only the 12" mes: pemnorephaemeom OR 133 fF BEB eecccceee’ faces quad seq . first difference n must be a positive whole number Only the 12" message YA 14 and 26 in NEW pattern VA ALL values of a, b&c YCA general term YCA T, =266 ¥CA factors ¥CA both answers with rejection (6) OR YA 7 and 13 in NEW pattern YA ALL values of a, b&c YCA general term YCA T, =133 ¥CA factors ¥CA both answers with rejection (6) [13] Copyright Reserved Please turn over

MathemRawnloaded from Stanmorephysics.com QUESTION 4 CAPS -Grade 11 Marking Guidelines KZN — November 2025 4.1. Dlilfm; 5357 (m+7)-3 (5—m) ; (n—5).. —2m-2 ; m+n-10.. first difference ..Second difference vA sec diff. in terms of m and n m2 = 4 “.m+n-10=4 2m = = -IO= " un 4.1 is a proof, all VA -2m=6 VA -34+n-10=4 —2m=6 —3+n-l0=4 marks are accuracy. (3) m=-3 n=17 4.2 FG RG LD: sssesoanas 4n=292 VA 4n=292 n=73 YCA value of n -.T,, and T,, ina qu YCA Ty & Try G3) 43 T, =2n?-2n-7 1005 = 2n? —2n—7 2n? —2n-1012=0 n’ —n-506=0 (n—23)(n+22)=0 n=23 or n#-22 VA T,=1005 vA standard form ¥CA factors ¥CA both answers with rejection T,, =1005 (4) [10] QUESTION 5 Sel . 3 1 pe: oe (92 =a 1 0)= ie f= 3*2 y=-l vA answer Q) 5:2 yea? vA answer armies vA answer 2 (2) Copyright Reserved Please turn over

MathemRawnloaded from Stanmorephysics.com CAPS — Grade 11 KZN —November 2025 Marking Guidelines 53 ‘ 3 4641 ye=—_ + — A ) Mi2 2 Ant 0=— 4+ — VA I = see 2 et y=0 seg | Alle 2 Px=2=26 = Answer only full marks, ae and ALL marks are x=4 accuracy since it is given vA answer (2) 5.4 YA shape ¥CA both asymptotes ¥CA both intercepts GB) 5.5 vv A answer (2) OR OR 1 ve(-»:5] or ye[-1; 0] ¥vA answer 2 (2) 5.6 | p(xy=—_42 VA VA x-2 z-2 (2) OR OR 3 -3 P(x)= +2 vA vA +2 ged —x+2 (2) [12] Copyright Reserved Please turn over

MathemRawnloaded from Stanmorephysics.com CAPS -Grade 11 Marking Guidelines KZN —November 2025 VA TP (5510) YA both x-intercepts. vA shape DO NOT penalise if the y-intercept is shown (3) 6.2.1 | -x2+49=0 YA let y=0 x -9=0 (x+3)(x-3)=0 YA correct factors x=-3 or x=3 Answer only full marks ¥CA answer A(-3;0) @) OR OR -x°+9=0 YA let y=0 x =9 cee YA isolating the square x=-3 or x=3 Answer only full marks A(-3; 0) ¥CA answer @) Copyright Reserved Please turn over

MathemRawnloaded from Stanmore ph ysics .com KZN—November 2025 CAPS — Grade 11 Marking Guidelines 6.2.2 | Otherx—intercept inh is (9 0) YA identifying(9 ;0) ysa(x—x,)(x-x,) oe os is a proof, i -9 all marks are accuracy. WAAL —9) VA substitution of »x—intercepts = 2a G40+3,0-9) OR -9=a(3+3)(3-9) 27 VA subst.of (0 77) or 27 f= a(-27 OR —9=-36a = 4 al ) (3; 9) 1 a 4 i a 7 DR 2) yeni 9 vA simplification of A. 4 xp 2 ot 8 Pe 2 me 2 and b=->5 (4) 4 5 OR OR C(3;-9) YA identifying C(3;—-9) Since 6.2.2 is a proof, _ 2 yaa(x+ py +q all marks are accuracy. y=a(x-3)?-9 YA substitution of C(3;—9) 27 . 0=a(-3-3)-9 OR a = a(0—3)?—9 VA subst. (0 2) or (-3;0) 9 9 =a(36) OR 47% 1 a=— 4 =1(-37-9 74 1 y= at *—6x+9)-9 A simplification to h. 123,27 ar a ear 4 nase and ba 4) 4 623 | C(3;-9) VE od yeR:y2-9 Answer only full marks VA answer -- --9) om , Answer only full marks YA Yo=W9 ye[-9; 0) vA answer (2) Copyright Reserved Please turn over

MathemRawnloaded from Stanmorephysics.com CAPS — Grade 11 Marking Guidelines KZN — November 2025 Reflection about x-axis, and shift-of 5 units up 6.2.4 OR shift of 5 units down , and reflection about the x —axis Must be in this order Must be in this order YA reflection about the x—axis YA Sunits up (2) OR VA 5 units down YA reflection about the x—axis (2) 6.2.5 1, 3 27 =x? -x-—=-x7 49 4 2 Sepa: 63g 4 2; 4 5x° -6x-63=0 (x+3)(5x-21)=0 x=—3 or el - 4, 25 5 vA equating both equations ¥CA standard form ¥CA correct factors ee D(4,2 ; -8,64) YCA value of x YCA value of y : (5) 6.2.6 | h(-l)=-5 & h(7)=-5 VA x-value of —1 or 7 “k=—-5 Answer only full marks “A answer (2) [21] QUESTION 7 7A h (x) =2*+q as 8 3 _ YA substitut 2; -7T— “15227 +4 substitting { $ ;) vA answer =-8 2 Q) 72 2°-8=0 VA let y=0 2 =23 YA 2 =23 x=3 A(3;0) Answer only full marks || VA answer GB) Copyright Reserved Please turn over

MathemRawnloaded from Stanmorephysics.com CAPS —-Grade 11 Marking Guidelines KZN — November 2025 73 fOy=2°-8 f@)=-7 B(0 >-7) YA coordinate of B m=. Jn 7 VN eX 0-(-7 m= 7) YCA substitution 3-0 —— Z ¥CA answer 3 G3) 4 O<x<3 vv AA answer OR (2) OR xe[0;3) v¥VvAA answer Q) 75 T<k<8 ¥¥CACA answer (2) OR OR ke(7;8) ¥¥CACA answer (2) [12] QUESTION 8 8.1. A=P(1+i.n) A=R 15000(1+0,10x6) vA subst. into correct formula A=R 24000 VA value of A Interest = R24000 — R15000 =R 9000 ¥ CA answer GB) OR OR Interest = pxixn vA correct formula =15000x 0,16 YA subst. into correct formula =R 9000 ¥CA answer QB) 8.2 A=P(1-7)" — ae 3 R29) vA subst. to correct formula 455625 (1—0,0825)° = ¥CA answer “. P= R358 922 (2) Copyright Reserved Please turn over

MathemRawnloaded from Sta inmorephysics .com KZN—November 2025 CAPS — Grade 11 Marking Guidelines 8.3 - \m Lg fey fom YA correct formula os m ‘ 4 140,12= ( + i) vA subst into correct formula F =4(¥,12 = 1)x 100% YCA simplification r=11,49% ¥CA answer (4) 8.4 A= P(l +i)’ A=R250000{ 1+ vA D,0225) v n=19 A=R59022 ¥CA answer GB) 8.5 12x2 Fi ) 4x2 0. 1B 24 6soo(1+- 933 [+ } =7428,30 | VA 6s00{1+%2) 12 ; 12 24 ; 0,105)” v , 6s00{1+%9 (i+° | —7428,30 “ [1.3 } x= 8 ( caer YA «(14 23] 1+ 4 YA all equated to 7428,30 x =R3300 ¥CA answer (5) OR OR 4 4 8 0,13)" 6500 —_ {1 ot) = (i) =7428,30 | “A 6s00{ +28) 12 4 4 ; 0,105 «os vA (re ) ox -x= ae j -500{1+ 51> (6 } 4 Bs 8 Coe va (148385 —x=-R 3300 VA ALL equated to 7428,30 “x = R3300 ¥CA answer (5) OR OR Copyright Reserved Please turn over

MathemRawnloaded from Stanmorephysics.com CAPS —-Grade 11 Marking Guidelines KZN — November 2025 A,= e501 + | 7428,30 of mos [fp¥.t0> 4 pS wy 3) ( nies) .| L+—— ] -x. 12 4 ...reverse equation forward equation 7428, 30 6500{1+ 0,13)" a} (It 1+ 0,105 } mes) 4 “( 24 vA es0o( 1+ 93?) 12 4 vA (122) —x a108) 4 vA [+ VA ALL equated to 7428,30 4 4 oy = 148.30 659 ) ( 0,10. } 1+ — 4 —x =—-R 3300 “x =R3300 Y CA answer (5) OR un 24 4 A, =6500( 1+ 923) (1-22°] soovenves forward equation 5 0,13 v , rane A 6s0o{ + 3 f= aT t+ Xeseesseed reverse equation ; (1) En [1+ 2426) . 4 4 A,=P. 5 37 F3 . , VA (1-222) 6500 14.2:13 1 4.2:105 _ _7428,30 4 1D) js | 4) ¢ 0.05% YA ALL equated to 7428,30 1+ — Gre 24 4 x=6500{1+ 3} (10808) ——7428,30_ 12 4 ( a ara ¥CA answer [17] Copyright Reserved Please turn over

MathemRawnloaded from Stanmorephysics.com CAPS — Grade 11 Marking Guidelines KZN — November 2025 QUESTION 9 9.1.1 | P(AandC)=0 vA answer : 9.1.2 P(AorC)=P(A)+P(C)—P(A and C) | P(A orc > Kore) =0,47+ 3-0 7 P(A or C)=0,90 P(A or C)=P(A) + P( P(A or€)=0.47 +5 vA substitution to a correct rule ¥ A answer (2) OR YA substitution to a correct rule P(A orC)=0,90 vA answer - 9.2.1 | P(A and B)=P(A)xP(B vA P(A and B)=P(A)xP(B) P(A and B)==P(B)xP(B) P(A and B)==P(B)° ° vA answer (2) 9.2.2, | P(A or B)= P(A) + P(B)— P(A and B) 0,7= =P) +P(B) -2P(By = 1 xp) -2 P(By? 0.7= <P(B) ~=P(B) 206) ~ ; P(B) st =0 OR 4P(B)’-14P(B) +7 =0 T-Oe P(B) = ‘) 5 2 op) = 2 eV) 40907) 2(4) If answer is CA, P(B)=0,60 or P(B) # 2,90 0< P(B) <1 OR YCA substitution to correct rule ¥CA equation in terms P(B) YCA substitution into quadratic formula ¥CA answer (5) OR Copyright Reserved Please turn over

MathemRawnloaded from Stanmorephysics.com CAPS — Grade 11 Marking Guidelines KZN — November 2025 P(A or B) = P(A) + P(B) — P(A and B) 0,75 P(A) + SPA) = m(A).> P(A) 7 5 0,7= —P(A) -=P(A)? pF) = MA) 5A)" ~7P(A) + =0 OR 25P(A)’ —35P(A) +7 =0 ‘TO P(A) = ¥CA substitution to correct rule ¥CA equation in terms of P(A) YCA substitution into quadratic formula P(A) = P(A) =0,24 or P(A) 4 YCA P(A) P(A) = 2PCB) 0,24 = =P) . _ 3 If answer is CA -. P(B) = 0,60 or z 0<P(B) <1 ° ¥CA answer [10] QUESTION 10 10.1.1 Let the number of females who will love to buy Gucci be x Gucci Louis Vuitton Males y 2x 45 Females a 3y 85 2x+3y 130 : x=85-3y x+3(45-2x)=85 OR y+2(85-3y)=45 x+135-6x =85 OR y+170-6y=45 —5x =-50 OR —S5y=-125 x=10 y=25 x=10 x must be a positive whole number VA males : y+2x=45 VA females: x+3y=85 YA y=45-2x OR x=85-3y ¥CA substitution YCA simplification ¥CA answer (6) Copyright Reserved Please turn over

MathemRawnloaded from Stanmorephysics.com CAPS — Grade 11 Marking Guidelines KZN — November 2025 OR Gucci Louis Vuitton Males y 45-y 45 Females 85—3y 3y 85 85-2y 45+2y 130 45—y =2(85-3y) OR VA Mand LV = 45-y VA Fand Gucci= 85-3 VAVA 45-y=2(85-3y) 45—y=170-6y Sy =125 ” 95 MCA values of y _ , CA answer x=10 females who would (6) OR OR Allow n to represent total number n(M)=85 & n(F)=45 n(M and G)=y n(M and LV)=2(85-3y)=170-6y VA 2(85-3y) or 170-6y n(F and G)=85-3y VA 85-3y n(F and LV)=3y -.2(85-3y)+y=45 YAVA 2(85—3y)+y=45 170-6y+y=45 Sy =-125 yeas CA value of y x=10 females who would love Gucci items CA. answer. (6) 10.2 25 25 5 P(M == VGA. — — 0,19 (M and G) 130 mag ge =0,19 45 35 45. 35 og 76K 22 2 P(M)*P(G)= 739" 739 130 * 130 - ¥CA pica or — or 0,09 675 675 75 = 0,09 P(M and G) # P(M)xP(G) Both events are NOT independent ¥CA NOT independent (4) Copyright Reserved Please turn over

MathemRawnloaded from Stanmorephysics.com CAPS — Grade 11 Marking Guidelines KZN —November 2025 OR OR P(Mand Lv) = 72 2 2 130 ¥CA — or — or 0,19 bas B03 P(M)xP(LV) => x22 4s 95 130° 130 vca 4545 i71 130 130 676 ¥CA 1 oo 0,25 =0,25 676 P(M and LV) # P(M)xP(LV) Both events are NOT indeper YCA NOT independent (4) [10] TOTAL: 150 Copyright Reserved Please turn over