Mathematics Grade 10 Paper 1 KZN November 2024 Paper & Memo

Downloaded from Stanmorephysics.com KWAZULU-NATAL PROVINCE EDUCATION REPUBLIC OF SOUTH AFRICA NAL IFICATE MARKS: 100 TIME: 2 hours This question paper consists of 7 pages. Copyright reserved Please turn over

Mathematics P1 ° N ber 2024 "Bownloaded from Stanmorephysics.com “me INSTRUCTIONS AND INFORMATION Read the following instructions carefully before answering the questions. 1. This question paper consists of 7 questions. 2. Answer ALL the questions. 3. Number the answers correctly according to the numbering system used in this question paper. 4. Clearly show ALL calculations, diagrams, graphs, etc. which you have used in determining your answers. 5: Answers only will NOT necessarily be awarded full marks. 6. You may use an approved scientific calculator (non-programmable and non-graphical), unless stated otherwise. ki Ifnecessary, round off answers correct to TWO decimal places, unless stated otherwise. 8. Diagrams are NOT necessarily drawn to scale. 9. Write neatly and legibly. Copyright Reserved Please turn over

Mathematics P1 ° November 2024 wnloaded from Stanmorephysics.com QUESTION 1 Il The value of 70 lies between two integers. Determine these two integers without finding the exact value of 70 ‘ (2) 1.2 Convert the following recurring decimal fraction 0, 23 to a common fraction in its simplest form. (2) (2) (4) (G3) 1.4 Factorise the following expressions fully: 14.1 x -13x4+42 Q) 1.4.2 m’ +m(4+n)+4n (3) 1.4.3 2y-250y4 @) [21] QUESTION 2 21 Solve for x, without using a calculator: 2.1.1 6x’ -x-1=0 (3) 1 zt ate 243 @) 22 Solve for x: -l< ra <5 2 (2) Copyright Reserved Please turn over

Mathematics PI fio November 2024 wnloaded from Stanmorephysics.com 2.3 Ata restaurant a glass of orange juice costs R4 more than a glass of ice tea. It is further given that five glasses of orange juice and three glasses of ice tea together cost R84. Let x be the cost of a glass of orange juice. Let y be the cost of a glass of ice tea. ye Set up a system of equations that represent the above information. (2) 2.3.2 Hence, determine the individual cost of orange juice and ice tea respectively. (4) [14] QUESTION 3 3.1 Consider the following linear number pattern: Be Ss-Qsws . 3.1.1 Write down the next term of the pattern. (1) 3.1.2 Determine the n' term of the pattern. (2) he value of the 25" term. (2) of the pattern is equal to —527? (2) ..are the first three terms of a linear number pattern. Determine (3) LCHAELMICHAELMICHAEL.....is continued in this way, what (2) [12] QUESTION 4 4.1 Given that f(x)=-2 44 x 4.1.1 Determine the equation of the horizontal asymptote of / . (1) 4.1.2 Determine the coordinates of the x intercept of f . (2) 4.1.3 Sketch the graph of / , clearly showing the intercept with the axes and the asymptotes. (3) 4.1.4 Determine the equation of g, if g is formed by shifting the graph of f 7 units down. qd) Copyright Reserved Please turn over

Mathematics P1 r N ber 2024 "Bownloaded from Stanmorephysics.com wee 4.2 The graph of h(x) = 2" —2 is sketched below. 4.2.1 Determine the coordinates of A . (2) 4.2.2 Write down the range of h. (2) 4.2.3 Determine the equation of / , if j is formed by reflecting the graph of h about the y-axis. (1) [12] Copyright Reserved Please turn over

Mathematics P1 r N ber 2024 "Bownloaded from Stanmorephysics.com wee QUESTION 5 5.1 Sketched below are the graphs of f (x)= ax’ +q and the straight line g. gis parallel to the x-axis and S(4;0) is the x-intercept of f. The graph of f passes through points T(2;3), PandR. Les 0 5.1.1 Write down the coordinates of: a) P (1) b) R (2) 5.1.2 Write down the equation of g. (2) 518 Determine the values of a and q. (4) 5.1.4 If f(x)=k, determine the value(s) of k for which f has no x—intercepts . 2) Sus. Determine the coordinates of F, if F is the reflection of T(2;3) about the x-axis. 2) 5.1.6 Using the graphs, determine for which value(s) of x: a) I (x)-g(x)>0 (2) b) J (x) .g(x)<0 (2) [17] Copyright Reserved Please turn over

Mathematics P1 r N ber 2024 "Bownloaded from Stanmorephysics.com wee aptop computer costing R11 500, on a hire-purchase the agreement are as follows: y a deposit of 20% of the purchase price. est is charged on the balance. ompulsory monthly insurance premium of R60,75. in 36 monthly instalments. 6.1.1 Calculate her balance after paying the deposit. (2) 6.1.2 Calculate Thembekile’s total monthly instalment. (4) 6.2 It costs £71,72 to fill a car with 55 litres of petrol in England. In South Africa, 1 litre of petrol costs R22,30. In which country is petrol more expensive, if the same car is used and the exchange rate between the countries is £1 = R23,23. GB) 6.3 Brent crude oil costs $77.36 a barrel. Calculate the cost in rands, of importing 100 barrels, if the exchange rate is R17,83 to the dollar. (2) [11] QUESTION 7 7.1 The letters of the word COORDINATES were put into a hat. A letter is chosen at random. Determine the probability that: TAs the letter N is chosen. qd) 7.1.2 the letter O is chosen. qd) Te Balls numbered from | to 10 were put into a container. 721 From the balls numbered | to 10, list the numbers that will be event A, the factors of 18. qd) U2i2 From the balls numbered | to 10, list the numbers that will be event B, all the odd numbers. qd) 72.3 Draw a Venn diagram to illustrate the above information. (3) 7.2.4 Determine: a) P(A and B) (2) b) P(AorB) (2) c) P((not A) or B) (2) [13] TOTAL: 100 Copyright Reserved

FINAL education Department: Education PROVINCE OF KWAZULU-NATAL NAHLONAS SENIGCk f MARKS: 100 Copyright Reserved This marking guideline consists of 7 pages. Please turn over

Mathematics Paper 1 2 KZN/November 2024 GRADE 10 Marking Guideline QUESTION 1 Ll 64 < 70 <V81 YA 8 vA 9 8< 70 <9 (2) 12 let 0,23 =k YA — 23,23 =100k 23,23 =100k YA answer 23 “99k Answer only: Full marks (2) YA 27x° 49x? y—9x? y—3xy? +3xy + y° ¥CA answer (2) YA (x+3)(x-4) YOK 3(x+3)+2(x-4)-21 (x-4)(x+3) _ 3x+9+2x-8-21 (x-4)(x+3) __ 5x-20 YCA simplification (x-4)(x+3) ae ¥CA answer x+3 (4) 2n a2 2n 2n a2 2n 13.3 | _5°5°-3.5°""5 VA = 3.5"5 54 5.4 5°" (25-15) YA 5°" (25-15) 5 ay ¥CA answer (3) 1.4.1 =(x-6)(x-7) YAYA each factor (2) 1.4.2 =m’ +4m+mn+4n vA m +4m+mn+4n =m(m+4)+n(m+4) YA m(m+4)+n(m+4) =(m+4)(m+n) YCA answer (3) 143° | =2y(1-125y') YA 2y(1-125y") =2y(1-Sy)(1+5y+25y’) YCAYCA answer (3) [21] Copyright Reserved Please turn over

Mathematics Paper 1 3 KZN/November 2024 GRADE 10 Marking Guideline QUESTION 2 2.1.1 (3x+1)(2x-1)=0 vA (3x+1)(2x-1)=0 1 1 FETT OTS YCAYCA each answer GB) 2.1.2 341 235 vA 3° 2x-1=-5 Y¥CA 2x-1=-5 x22 YCA answer - Lied GB) 2.2 —2<X+3<10 x+3<10 -5<x<7 x<7 (2) 2.3.1 x=4t+y ., vA 4+y 5x+3y =84 AS 3y =84 (2) 2.3.2 orep CS.CO plying (1) by —5 YCA 8y=64 YCA y-—value ¥CA — x—value OR OR x=4+y iC) Sx+3y =84 we Substitute (1) into (2) YCA substation 5(44+9)+3y =84 YCA 8y=64 8y = 64 y=8 YCA_ x—value x=19 ¥CA_ y—value (4) [14] Copyright Reserved Please turn over

Mathematics Paper 1 4 KZN/November 2024 GRADE 10 Marking Guideline QUESTION 3 3.1.1 T,=-7 vA answer () 3.1.2 T, =-5n+13 vA —Sn vA +13 (2) 3.1.3 Tes =-5(25)+13 vA substitution T.=-112 ¥CA Tf, ==112 25 ~~ Answer only: Full marks (2) 3.1.4 —5n+13=-527 ¥CA — -Sn+13=-527 n=108 YCA —_n=108 Answer only: Full marks (2) 3.2 2x—(3x+1)=3x-7-(2x) YA 2x—(3x+1)=3x-7-(2x) -2x=-6 YCA -2x=-6 ¥=3 ¥CA answer (3) ad an. 28 remainder is 5 vA = = 28 remainder is 5 <. letter number 5 = A VA answer Answer only: Full marks (2) [12] QUESTION 4 vA answer dQ) vA at +4=0 a4 vA answer (2) Copyright Reserved Please turn over

Mathematics Paper 1 5 KZN/November 2024 GRADE 10 Marking Guideline 4.1.3 y vA shape vA asymptote re sora deaaencnnss YCA x—intercept 1A ¢ a (3) 4.1.4 vA answer g(x)=-= dd) 4.2.1 h(0)=2"-2 vA x—value =<1 vA y-value A(0;-1) (2) 42.2 | y>2 YAVA answer OR ye(-2@) 2) 4.2.3 vA answer j(x)=2"-2 or i(s)=( a) [12] QUESTION 5 S.1.la | P(-4;0) vA answer (1) 5.1.1b | R(-2;3) vA x—value vA y-value (2) 5.19 g(x)=3 ¥AVA_ Answer only: full marks OR OR m=0 YA m=0 YA answer c=3 g(x)=3 (2) Copyright Reserved Please turn over

Mathematics Paper 1 6 KZN/November 2024 GRADE 10 Marking Guideline 5.1.3 | f(x)=a(x-4)(x+4) through T(2;3) v = a 3=a(2-4)(2+4) A 3=a(2-4)(2+4) ¥CA value of a on condition a<0 YCA f (x) a-j44 ¥CA value of g (4) 5.1.4 k>4 YCA value of q vA k> (2) 5.1.5 F(2;-3) vA x—value vA y—value (2) 5.1.6a | xe(-2;2) OR -2<x<2 YCAYA -2<x<2 OR V¥CAYA xe (-2;2) (2) 5.1.6b | x<—4orx>4 OR ¥CAYA — answer x €(—v0;-4] or x €[4;00) (2) [17] QUESTION 6 vAv 6.1.1 Loan = 90 Ri 500 AYCA answer 100 =R9 200 OR OR Deposit =aaxRI 1500 YA Deposit =R2 300 = R2 300 Loan = R11 500-— R2 300 YCA answer = R9 200 (2) 6.1.2 | A=P(1+in) vA A=P(I+in) A=9 200(1+0,11x3) vA substitution A=RI12 236 ¥CA A=RI12 236 Monthly instalment = 2.255 + R60,75 ¥CA answer =R400,64 If compound interest formula used max | out of 4 for monthly instalment (4) Copyright Reserved Please turn over

Mathematics Paper 1 7 KZN/November 2024 GRADE 10 Marking Guideline 6.2 In South Africa, vA R1 226,50 price of full tank = R22,30 x 55L=R1 226,50 In England, price of full tank = £71,72 = Rx £1 = R23,23 vA RI 666,06 ¥CA conclusion @) 63 83 x100 YA _ $77,36 x R17,83 x100 ¥CA answer Answer only: Full marks (2) (11) QUESTION 7 v TAA p(N)=_ =0,09 A answer ll d) v 72 P(0)=~=0,18 A answer ll dd) 7.2.1 | A= {1;2;3;6;9} vA answer () 7.2.2 | B= {1;3;5;7;9} vA answer (1) G23 1;3;9 4:8;10 2&6 ; 5&7 (3) 7.2.4a) _ so CA numerator P(A and B)=——=0,3 A answer Answer only: Full marks = (2) 7.2.4b) 24+34+2 = 7 CA numerator P(A or B) = 10 = 79 70? YA answer Answer only: Full marks (2) 7.2.4c) _—34+24+3 8 _ ¥CA numerator P((not A) or B) = ia “7 e Ph aneRE Answer only: Full marks (2) [13] TOTAL: 100 Copyright Reserved Please turn over