Mathematics Grade 10 KZN Prelim 2025 Paper & Memo

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Downloaded from Stanmorephysics.com FINAL NATIONAL SENIOR CERTIFICATE GRADE 10 MATHEMATICS SEPTEMBER 2025 MARKING GUIDELINE MARKS: 75 This marking guideline consists of 7 pages. Copyright reserved Please turn over

Downloaded from Stanmorephysics.com Mathematics 2 Common Test September 2025 CAPS – Grade 10 Marking Guideline QUESTION 1 Consider the sequence: 17 ; 15 ; 13 ; ... 1.1.1 T4 = 11 A answer (1) 1.1.2 Tn = −2n + 19 OR Tn = 19 − 2n A −2n A +19 (2) 1.1.3 −71 = −2n + 19 CA equating −90 = −2n n = 45  T45 = −71  CA answer ( n  ) (2) 1.2 4 x + 7 − (3 x − 2) = 2 x − 5 − (4 x + 7)  A equating 1st diff. x + 9 = −2 x − 12 3x = −21  CA simplification  x = −7  CA answer (3) 1.3.1 T67 = 9  A answer (1) 1.3.2 Tn = 3n + 2  A 3n + 2 T250 = 3(250) + 2 CA subst n = 250  T500 = 752 CA answer (3) 1.4 Numerator: Tk = 4k + 6 A numerator Denominator: Tp = 5 p + 3 A denominator 4n + 6 Tn = CA answer 5n + 3 Answer Only: Full Marks. (3) [15] Copyright reserved Please turn over

Downloaded from Stanmorephysics.com Mathematics 3 Common Test September 2025 CAPS – Grade 10 Marking Guideline QUESTION 2 A = P (1 + i ) n 2.1.1 A formula A = 25000 (1 + 0, 0825 ) 9 A substitution A = R51025,96 CA answer OR n  r  A = P 1 +   100  9  8, 25  A = 25000 1 +   100  A = R51025,96 (3) 2.1.2 A = P(1+ in) A formula A = 51025,96[1+ (0, 06)(3)] CA substitution A = R 60 210, 63 CA answer OR  nr  A = P 1 +   100   (6)(3)  A = 51025,96 1 +   100  A = R 60 210, 63 OR I = pin = 51025,96  0, 06  3 = R9184, 67 A = P + I = 51025,96 + 9184, 67 = R60210, 63 (3) −3 2.2 P = 64700000(1 + 0, 025) A substitution P = 60080381 people CA answer OR 64700000 = P[1+ (0,025)]3 P = 60080381 people Penalise 1 mark if P (2) Copyright reserved Please turn over

Downloaded from Stanmorephysics.com Mathematics 4 Common Test September 2025 CAPS – Grade 10 Marking Guideline 2.3 $1 A product £1 = R23,52  R18,16 £1=$1,30 CA answer OR £1 $1 = R18,16  R23,52 $1=£0,77 (2) [10] QUESTION 3 3.1.1 P(A) + P(B) = 1 A 1 (1) 3.1.2(a) P(Aand B) = 0 A 0 (1) 3.1.2(b) P(A) = 0, 45 A answer (1) 3.2.1 18 − x + x + 27 − x + 4 = 33 A equation 49 − x = 33 CA simplification x = 16 CA answer ( x  ) OR 18 − x + x + 27 − x + 4 = 33 49 − x = 33 x = 16 (3) 3.2.2(a) 4 A answer P(none) = 33 (1) 3.2.2(b) 2 CA 2 P(N only ) = 33 CA answer (2) Copyright reserved Please turn over

Downloaded from Stanmorephysics.com Mathematics 5 Common Test September 2025 CAPS – Grade 10 Marking Guideline 3.3.1 R F A 44 A 67 67 A 7 A 2 2 (4) P ( ( not R ) and F ) = 3.3.2 7 CA 7 120 A 120 (denom) (2) [15] QUESTION 4 4.1.1 a = 35 ( also accept a = 57, 2 ) A b = 715 A c=9 A (3) 4.1.2 3350  3550  CA substitution Estimated mean =  also accept  60  60  = 55,83 ( = 59,17 ) CA answer Answer Only: Full Marks. (2) 4.1.3 40  m  50 CA answer (1) 4.1.4 50  m  60 CA answer (1) 4.2.1 43 + 45 Median = 2 = 44 A answer (1) 4.2.2 IQR = 61 − 33 A Q3 = 28 A Q1 CA answer (3) Copyright reserved Please turn over

Downloaded from Stanmorephysics.com Mathematics 6 Common Test September 2025 CAPS – Grade 10 Marking Guideline 4.2.3 30 + 33 + 33 + x + 43 + 45 + 55 + 61 + 78 + 80 A sum divided by 10 = 49,5 10 A equating to 49,5 495 = 458 + x x = 37 CA answer (3) [14] QUESTION 5 5.1(a) amplitude = 2 A answer (1) 5.1(b) period = 360 A answer (1) 5.1(c) minimum value = −1 A answer (1) 5.2 f (180 ) = 2 A f (180 ) = 2 g (180 ) = 1  f (180 ) − g (180 ) = 1 CA answer (based on f (180 ) ) Answer Only: Full Marks. (2) 5.3 0  x  90 or 270  x  360 A 0  x  90 A 270  x  360 A both have correct notation OR OR x   0 ; 90   270 ; 360 A x   0 ; 90 A   270 ; 360 A both have correct notation (3) 5.4 h( x) = − ( sin x + 1) + 3 h( x) = − sin x −1 + 3 h( x) = − sin x + 2 A h( x) = − sin x + 2 1  y  3 CA interval OR Answer Only: Full Marks. A notation y  1 ; 3 (3) [11] Copyright reserved Please turn over

Downloaded from Stanmorephysics.com Mathematics 7 Common Test September 2025 CAPS – Grade 10 Marking Guideline QUESTION 6 6.1 WE tan 30 = 15 WE A tan 30 = 15 tan 30 = WE 15 5 3 m = WE ( accept WE = 8, 66 m ) A answer (2) 6.2 WR = 7,5 m ✓A length of WR WR cos  = WT 7,5 cos  = 9 ✓CA substitution  = 33,56 ✓CA answer (3) 6.3 TR = 9 − 7,5 2 2 2 Pythag ✓A correct sub Pythag 3 11 TR = ✓CA length of TR 2 1 Area = b.h ✓A formula 2 1  3 11  = (15 )   ✓CA subst 2  2  = 37,31 m 2 ✓CA answer (5) [10] TOTAL: 75 MARKS Copyright reserved Please turn over