Mathematics Grade 10 Paper 2 KZN June 2024 Memo

Downloaded from hlayiso.com Downloaded f rom St anmorephysics. com NATIONAL SENIOR CERTIFICATES GRADE 10 MATHEMATICS P2 COMMON TEST JUNE 2024 MARKS: 50 TIME:1 Hour Copyright Reserved Please turn over

Downloaded from hlayiso.com Downloaded f rom St anmorephysics. com

Downloaded from hlayiso.com Downloaded f rom St anmorephysics. com

Downloaded from hlayiso.com Downloaded f rom St anmorephysics. com

Downloaded from hlayiso.com Downloaded f rom St anmorephysics. com (2) (2) (4) [9] TOTAL: [50]

Downloaded from hlayiso.com Downloaded f rom St anmorephysics. com FINAL MATHEMATICS P2 COMMON TEST JUNE 2024 MARKING GUIDELINES NATIONAL SENIOR CERTIFICATE GRADE 10 MARKS: 50 These marking guidelines consist of 6 pages. Copyright Reserved Please turn over

Downloaded from hlayiso.com Downloaded f rom St anmorephysics. Mathematics 2 com Common Test June 2024 GRADE 10 Marking Guideline QUESTION 1 1.1  x + x y + y2  M 1 2 ; 1  2 2  A Formula  6 + (−5) 6 + 3  ✓A Substitution M ;   2 2  1 9 CA Answer M ;  2 2 (3) 1.2 y2 − y1 mWT = x2 − x1 ✓A Formula 6−0 A Substitution = 6 − (−3) 2 CA Answer = 3 (3) 1.3 y2 − y1 mRW = x2 − x1 3−0 = −5 − (−3) A substitution −3 = −3 2 A 2 2 −3 mWT  mRW =  = −1 3 2 2 −3 A  = −1 ∴ RW is perpendicular to WT 3 2 (3) 1.4 x1 + x2 y1 + y2 x= y= 1 a −3 2 2 CA = 2 2 1 a −3 9 b+0 = = CA a = 4 2 2 2 2 a=4 b=9 9 b+0 ✓CA = 2 2 OR Sliding method ✓CA b=9 a = 6−2 b = 6+3 ANSWER ONLY MAX: 2/4 a=4 b=9 (4) [13] Copyright Reserved Please turn over

Downloaded from hlayiso.com Downloaded f rom St anmorephysics. Mathematics 3 com Common Test June 2024 GRADE 10 Marking Guideline QUESTION 2 2.1.1 x2 + y 2 = r 2 x 2 + 52 = 132 A substitution x = 12 x = −12 CA x = −12 (CA provided x is negative) (2) 2.1.2 cos −12 CA answer = (1) 13 2.1.3  sin   12    cos    5    = 12  13  CA substitution  −12   13  = −5 CA answer (2) 2.2.1 cos 2 ( 21 ) = cos 42 NOTE: Penalise only here for rounding A answer = 0, 74 (1) 2.2.2 sin ( 20 + 21 ) A (20 + 21) = sin 41 = 0, 66 A answer ANSWER ONLY: FULL MARKS (2) 2.2.3 sec ( 20 ) 1 = 1 cos ( 20 ) A cos20 = 1, 06 A answer ANSWER ONLY: FULL MARKS (2) [10] Copyright Reserved Please turn over

Downloaded from hlayiso.com Downloaded f rom St anmorephysics. Mathematics 4 com Common Test June 2024 GRADE 10 Marking Guideline QUESTION 3 2 2  1  1 1  1  2  2 3.1   − 2 + tan10 tan10 A   accept    2  2  2  1 1 1 = − +1 ✓A 2 2 2 =1 1 A tan10 CA answer (4) 6sin x = 1 A 6sin x = 1 3.2 A simplification 1 sin x = 6 CA answer x = 9, 6 (3) NOTE: do not penalise for rounding [7] QUESTION 4 4.1 In MRC and XCR 𝑅̂1 = 𝐶̂ 2 Alt  s. MR//CX. A S/R 𝐶̂ 1 = 𝑅̂2 Alt  s. MC//RX. A S/R CR = CR Common side. MRC  XCR (AAS) A MRC  XCR …(AAS) MR=XC A conclusion MC=XR (4) RC=CR NOTE: Penalise ONCE if parallel lines are not stated 4.2.1 8𝑥 − 16° = 5𝑥 + 5° opp  s of parm A S/R 3𝑥 = 21° A 3𝑥 = 21° 𝑥 = 7° CA x-value (CA provided x is positive) (3) 4.2.2 Dˆ = Bˆ opp  s of parm A S/R Bˆ + Dˆ + Aˆ + Cˆ = 360 Sum  s of quad A S/R 2Bˆ + 80 = 360 CA S 2 Bˆ = 280 Bˆ = 140 CA answer Copyright Reserved Please turn over

Downloaded from hlayiso.com Downloaded f rom St anmorephysics. Mathematics 5 com Common Test June 2024 GRADE 10 Marking Guideline OR Cˆ = 8 ( 7 ) − 16 CA Cˆ = 40 40° SR Bˆ + Cˆ = 180 Co-int  s. AB DC CA answer Bˆ = 180 − 40 (4) Bˆ = 140 [11] QUESTION 5 5.1.1 Vˆ1 + 30 = 100 Ext  of  A S/R Vˆ1 = 70 A answer (2) OR R̂1 = 80 s on a str. line A S/R V̂1 = 70 sum of  A answer 5.1.2 Sˆ1 = Vˆ2  s opp = sides A S/R Sˆ + Vˆ + VRS 1 2 ˆ = 180 Sum  s  A S 2 Sˆ1 + 100 = 180 2 Sˆ = 80 1 Sˆ1 = 40 A answer OR (3) Rˆ1 + VRS ˆ = 180  s on a str line Rˆ1 = 80 A Rˆ1 = 80 Sˆ1 = Vˆ2  s opp = sides Sˆ1 + Vˆ2 = Rˆ1 Ext  of  A S/R 2 Sˆ = 80 1 Sˆ1 = 40 A answer Copyright Reserved Please turn over

Downloaded from hlayiso.com Downloaded f rom St anmorephysics. Mathematics 6 com Common Test June 2024 GRADE 10 Marking Guideline 5.2 ES = ER diagonals of rectangle A S/R R̂2 = t  s opp =sides ˆ = 180 Eˆ + Rˆ + ESR Sum  of  2 2 Eˆ 2 + t + t = 180 Eˆ = 180 − 2t 2 A Ê2 Eˆ 4 = Eˆ 2 Vert opp  s A S/R Eˆ = 180 − 2t 4 A answer (4) OR Qˆ1 = t Alt ∠s PS∥QR ✓A S/R ∠s opp = sides ✓A Pˆ = t S/R 1 Eˆ 4 + Qˆ1 + Pˆ1 = 180 Sum ∠s ∆ ✓A S ✓A Eˆ = 180 − 2t answer 4 [9] TOTAL: 50 Copyright Reserved Please turn over