Mathematics Grade 10 Paper 2 KZN November 2024 Paper & Memo

Downloaded from Stanmorephysics.com KWAZULU-NATAL PROVINCE EDUCATION REPUBLIC OF SOUTH AFRICA NATIONAL SENIOR. CERTIBICATE al VLA P anmorephisies.com MARKS: _ 100 TIME: 2 hours This question paper consists of 9 pages. Copyright Reserved Please turn over

Mathepsiwrilédded from Stanmorephysics.com KZN /Noyeraber 2024 INSTRUCTIONS AND INFORMATION Read the following instructions carefully before answering the questions. 1; This question paper consists of 10 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. Is If necessary, 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

Mathepsiwrilédded from Stanmorephysics.com aT ae QUESTION 1 Amahle recorded the amount of data (in MB) that she had used daily in the past 15 days. [3 [10 [12 [13 [15 [15 [16 [18 [24 [26 [34 [40 [42 | 43 [45 | 1.1 Calculate the range of the data. (2) 12 Calculate the mean data usage. (2) 1.3 Write down the five number summary of the data. (3) 1.4 Draw a box and whisker diagram to represent the data. (3) 1.5 Determine the interquartile range. (3) [13] QUESTION 2 The compan which a survey at a gym in KwaZulu — Natal, to find out health supplements. FREQUENCY 29 28 21 15 10 2 2.1 Identify the modal class for the data. (1) 2.2 Determine the interval in which the median lies. ql) 23 Estimate the mean age of the people that use the health supplements. (4) [6] Copyright Reserved Please turn over

Mathepsswilédded from Stanmorephysics.com eT ae QUESTION 3 In the cartesian plane below, A(-3;0), B(0;2), C(2;-1) and D(-1;-3) are points on a quadrilateral. A(-3 ; 0) 3.1 Determine the coordinates of R , the midpoint of AC. (3) 3.2. Calculate the gradient of the line AB. (2) 3.3. Calculate the length of AC. (3) 3.4 If ABCD is a rhombus, calculate the area of ABCD. (5) [13] Copyright Reserved Please turn over

Mathepsiwrilédded from Stanmorephysics.com KZN /Noyeraber 2024 QUESTION 4 If 4tan@=-3 and cos@>0, WITHOUT using a calculator, determine the value of: 41 sin (3) 42 10cos* 6 (2) 43 3cot6+5sind (3) [8] QUESTION 5 5.1 In each of the following equations, solve for x where 0°<x<90'. Give your answers correct to TWO decimal places. S211 cosx=2 (2) 3 $12, 4 +1=3 (3) cotx 5.2) Simplify fully, WITHOUT the use of a calculator. tan 45°.cosec 60° (5) cot 30°.cos’ 45 5.3 If x=56,3° and y=97,5°, evaluate the following correct to TWO decimal places. 5.3.1 cos(x+y) (2) 532 2sin” x-2 cos y (3) [15] Copyright Reserved Please turn over

Mathepsswilédded from Stanmorephysics.com eT ae QUESTION 6 The graphs of f(x)=psinx and g(x)=cosx+q for xe[0 ; 360°] are sketched below: & . x 270° 4 f 6.1 Determine the value of p and q. (2) 6.2 Write down the amplitude of g. (1) 6.3 Write down the range of /- (2) 64 For which value(s) of x will f(x)>02 (2) 6.5 The graph of g is reflected about the x-axis and then shifted 2 units upwards to obtain the graph h. Write down the equation of h. (2) 19] Copyright Reserved Please turn over

Mathepsuwidded from Stanmorephysics.com ET a QUESTION 7 In the diagram below ABCD is a rectangle. DC = 155cm and DE = 800cm. The angle of elevation from D to Eis 22° . ald 7A Calculate the length of BE. (4) 1 Determine the distance of BC. (3) 17] QUESTION 8 A metal marble has a diameter of 16 millimetres. Volume of sphere = Sar 8.1 Calculate the volume of metal used to make this marble. Give your answer correct to TWO decimal places. (3) 8.2 If the radius of the marble is doubled, write down the ratio of the new volume to the original volume. (3) [6] Copyright Reserved Please turn over

KZN/November 2024 Mathepsiwidded from Stanmorephysics.com QUESTION 9 9.1 In quadrilater Prove that WXYZ is a parallelogram. () 9.2 In the sketch below, ABCD is a parallelogram. DC=DE and A =120°. A s D 1 2 2 B C E 9.2.1 Calculate the size of: a) B (2) b) ¢, (2) 9.2.2 If D, =4x. Calculate the value of x. (4) [13] Copyright Reserved Please turn over

Mathepsiwrilédded from Stanmorephysics.com aT ae QUESTION 10 AXYZ is tight angled at Y. F and G are the midpoints of XZ and YZ respectively. H is the midpoint of XG. E lies on XY such that FHE is a straight line. x Y 10.1 Prove that E is the midpoint of XY. (3) 10.2 If EH=3,5cm and the area of AHEX =9,5cm’. Calculate the length of XY. (4) 10.3 Hence, calculate the area of AXYZ. (3) [10] TOTAL: 100 Copyright Reserved Please turn over

Downloaded from Stanmorgpiygics.com education Department: Education PROVINCE OF KWAZULU-NATAL Stanmorephysies.com sees 5 NATIONAL SENIOR CERTIFICATE (ane) MARKS: = 100 The marking guidelines consist of 10 pages. Copyright Reserved Please turn over

Mathemawnloaded from Stanmorephysics.com GRADE 10 QUESTION 1 Marking Guideline November 2024 1 Range = Max — Min =45-3 =42 Answer only, full marks YA __ substitution ¥CA answer (only if subst was shown) (2) Answer only, full marks VA 356 ¥CA_ answer (only if divided by 15) (2) YA Min and Max YAQ YA Q, and Q, (3) 1.4 Wdenree- _ ae-- Ze-- oe ¥CA Min and Max YCA Q, YCA Q,and Q, (3) 15 IQR = Q, -Q, =40-13 =27 Answer only, full marks YA formula YCA substitution ¥CA answer (3) [13] QUESTION 2 2.1 20<x<30 2.2 30<x<40 23 |< _ (25x29)+(35x28)+(45x21)+(55x15)+(6 _ 4275 105 = 40,71 105 (4) *Penalise for Rounding here* [6] Copyright Reserved Please turn over

Mathenewnloaded from Stanmorephysics.com November 2024 GRADE 10 Marking Guideline QUESTION 3 x, $x, + 31 Mack = 2 ; 1h) n( 22 130) YA substitution 2 2 R =l_-1 ¥CA_ x—value 2° 2 Answer only, full marks ¥CA_ y—value On condition the coordinates of A and C are used ONLY QB) 32 May = VeVi %,-% a= 2-0 YA substitution 0-(-3) 2 ma r only, full marks ¥CA answer Q) 43 AC= YA formula AC= VA substitution AC = 126 2 ? ¥CA answer i ; ‘ G3) 3.4 BR = (x, -x,)° +(V, -yy Area of ABCD = 2{ 4(40)(6R)] -(v26)( 22) 2 = 13units* YCA substitution YCA distance of BR YCA areaof Ax2 ¥CA substitution ¥CA answer Copyright Reserved Please turn over

Mathemewnloaded from Stanmorephysics.com GRADE 10 Marking Guideline November 2024 OR OR dy =\(-3+1) +(0+3) ¥CA length of AD = 13 YCA length of DC Ape = y(-1- 2y° +(-3+ ly YA formula for area =V13 Area of ABCD = base x height YCA substitution = diy Aye ¥CA answer =Vi3xVi3 = 13units* diagonal AC & BD = 26 units ¥CAYCA subst & length of DB 1 vA formula for area Area of ABCD = 5 (duc xdyp) 1 ¥CA substitution | (38%) = 13units* ¥CA answer (5) [13] QUESTION 4 4.1 vA tand= = 4 sin@=— vA value of r ¥CA answer G3) Copyright Reserved Please turn over

Mathemawnloaded from Stanmorephysics.com GRADE 10 Marking Guideline November 2024 4.2 10cos* =| of 4 ) ¥CA substitution 5 £92 Y¥CA answer 5 OR OR 10 sec’ 0 — 10 ¥CA substitution = = (3) 32 ¥CA answer 5 (2) 4.3 . 3cotO+5sind YCA =a 3 5 =-4-3 ¥CA simplification mmo Answer only, max 2/3 YCA answer See notes on last page OR OR 3 +5sin0 tan yl) = +5| — —3 = V¥CA —s 2) 5 4 4 =-4-3 Answer only, max 2/3 YCA __ simplification =7 See notes on last page YCA answer Q) [8] QUESTION 5 5.1.1 2 cosx=> 5 x=cos” (2) YA __ simplification using arc cos x=48.19° Answer only, full marks YA answer (2) Copyright Reserved Please turn over

Mathenawnloaded from Stanmorephysics.com November 2024 GRADE 10 Marking Guideline 5.1.2 4105 cotx cotx=2. vA cotx =2 1 ee tanx vA 1 _s4 1 tan x tanx = = 2 x =26,57° ¥CA answer GB) 5.2 tan 45°.cos ec60" cot 30’.cos” 45° 5 VA — 3B VA a/3) 2 va (2 2 See notes on last page YCA _ simplification YCA answer OR OR tan 45° tan 30° oe 1 sin 60° cos” 45° 3 1 5, 1) = BLES a 4 ale) (2) 2\ 2 2 _ NB ae 9 3 See notes on last page ¥YCA - simplification 8 4 ¥CA answer “3 (5) Copyright Reserved Please turn over

Mathemawnloaded from Stanmorephysics.com November 2024 GRADE 10 Marking Guideline 5.3.1 | coset) = c0s(56,3’ +97,5°) VA substitution =cos(153,6") vi ==0,90 CA answer (2) 5.3.2 | 2sin?x=2 cos y 2(sin 56,3" y -2 YA _ substitution in numerator a 97,5 vA _ substitution in denominator =4,72 ¥CA answer (3) [15] QUESTION 6 6.1 p=2 q=l 2) 6.2 amplitude =\ (1) 6.3 2sys2 interval notation OR ye [-2,2] A interval ¥A_ notation (2) 6.4 0 < x<180 YA interval YA notation sa OR ane (0' 5 180") YA. interval YA. notation Q) 6.5 h(x) =—cosx+1 YA =cosx ¥CA +1 (2) [9] Copyright Reserved Please turn over

Mathemewnloaded from Stanmorephysics.com November 2024 GRADE 10 Marking Guideline QUESTION 7 71 | fe oe Hor vA correct trig ratio AE = 800.sin 22° vA AE=299,69 AE=299, 69cm “BE = 299,69+155 ¥CA 299,69 + 155 BE = 454,69cem Y¥CA answer (4) 7.2 AD _ 5599 vA correct trig ratio AD = 800cos 22’ AD = 741,75 ¥CA AD=741,75 . AD = BC (opposite sides of a rectangle) BC = 741,75 cm YCA answer OR OR AD? + AE* = ED” ¥CA substitution AD* +(299,69)° =(800)" YCA — AD=741,75 AD=741,75 YCA answer BC =741,75cm G3) 17] Copyright Reserved Please turn over

Mathemawnloaded from Stanmorephysics.com GRADE 10 QUESTION 8 Marking Guideline November 2024 8.1 V = 2144,66 mm’ vA r=8 vA substitution YCA answer (3) 8.2 r=2x8mm =1l6mm OR 4 57 (16) :=2(8) (16)' :(8) 16:8 4096 : 512 8:1 VA - substitution with new r YCA. division of volumes ¥CA answer OR YA $2(16) YCA simplification ¥CA answer 3) [6] Copyright Reserved Please turn over

Mathemawnloaded from Stanmorephysics.com November 2024 GRADE 10 Marking Guideline QUESTION 9 9.1 In AWXY and AYZW 1) WY is common vA Ss 2) ¥,=W, (Alternate angles WZ // XY ) vA SIR 3) X=Z (Given) v “ AWXY = AYZW (Z,2,S) “ R XY = WZ (AWXY = AYZW) vA SIR «. WXYZis a parallelogram (one pair of sides equal and //)_ | VA R (5) 9.2.1(a) | B=180°-120° —_(co-interior angles AD // BC) vA SIR B= 60° Penalise for parallel lines vA answer here (2) 9.2.1(b) | A=C opposite sides of a parm VA S/R C, =120° % vA C, =60° angles on a str. line a OR OR C, =180°-120° (corr. angles AB//DC) 6, =60° YASIR vA answer —_ (2) 9:22 E=C, angles opposite equal sides VA S/R E+C,+ BD, =180° sumofA vA R 4x = 180° —120° vA substitution 4x = 60° x=15° vA answer (4) [13] Copyright Reserved Please turn over

Mathemawnloaded from Stanmorephysics.com GRADE 10 Marking Guideline QUESTION 10 November 2024 10.1 In AXGZ F and H are midpoints (Given) «. HE//GZ (midpoint theorem) F 5 vA S/R FE/YZ (same straight line) vA S In AXYG_ H is the midpoint EH//YG_ (Proved) .. E is the midpoint (line through midpt // to second side) VA R Q) 10.2 XEH =90° (corr. angles EF//YZ) vA R Area = 5 (EHYXE) 9,5 = 5 N5xXE ¥ A Sub Corr formula — 38_ te VA 38 5 43 76 “ XY = —=10,86 7 em YCA 2 =10,86 (4) 10.3 Y¥G=7 (midpt theorem) YZ=14 (YG=GzZ) VAS Area= EGG YCA sub 2 7 Area = 76cm’ ¥ CA answer GB) [10] TOTAL: 100 NOTES: QUESTION 4.3 Learners must show the simplification step to be awarded full marks. QUESTION 5.2 Learners must show the simplification step to be awarded full marks. Copyright Reserved Please turn over