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Mathematics Grade 10 P2 North West November 2024 QP (Eng)

Grade 10 Mathematics Paper 2 question paper from the North West November Exam 2024 (English), 10 pages. Read online or get the free PDF download, memo linked when available.

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PROVINCIAL ASSESSMENT GRADE 10 MATHEMATICS P2 NOVEMBER 2024 MARKING GUIDELINES  100 MARKS: These marking guidelines consist of 10 pages. Copyright reserved Please turn over
Mathematics/P2 2 NW/November 2024 Grade 10 – Marking Guidelines C.A is applicable in all aspects of the marking guidelines. QUESTION 1 1.1 ✓ min value,max value ✓ ✓(Combination) Q1 = 21 Q2=32 Q3= 50 (3) 1.2 IQR = Q3- Q1 Answer only full marks ✓ formula = 50 -21 ✓ answer (2) = 29 1.3 Range = max val – min val Answer only full marks ✓ formula = 68-12 ✓ answer (2) = 56 1.4 The data is skewed to the right. ✓ answer (1) [8] QUESTION 2 SCORES IN INTRODUCTORY STATISTICS Frequency Examination Scores 2.1 n  200 ✓ answer (1) 2.2 60  m  70 ✓ answer (1) 2.3 Exam scores Frequency x. f ✓Addition (200) ✓x.f (12 500) 30  m  40 12 420 ✓ substitution 40  m  50 18 810 ✓ answer 50  m  60 55 3025 (4) 60  m  70 57 3705 70  m  80 43 3225 80  m  90 11 935 90  m  100 4 380 200 12500 Copyright reserved Please turn over
Mathematics/P2 3 NW/November 2024 Grade 10 – Marking Guidelines  xf  12500 n 200  62,5 2.4 Answer only full marks ✓ 100,5 n  1 200  1 ✓ answer  2 2  100,5 (2) Q2  60  m  70 2.5 3  n  1 3  200  1 ✓ answer  4 4  150,75 [position] 70  m  80 (1) [9] QUESTION 3 3.1 BC  ( x2  x1 )2  ( y2  y1 )2 ✓sub ✓answer  (6  3)2  (1  (5)2 (2) 3 5 3.2  x  x2 y1  y 2  ✓formula D 1 ;  ✓sub  2 2  ✓✓midpoint   11  3 2  (5)   ;   2 2   3 (4)    4;   2 Copyright reserved Please turn over
Mathematics/P2 4 NW/November 2024 Grade 10 – Marking Guidelines 3.3 y2  y1 ✓formula M AB  x2  x1 ✓sub 2  (5) M AB   11  3 ✓gradient of AB 7 1 M AB     14 2 y2  y1 M BC  x2  x1 1  (5) ✓gradient of BC M BC  63 M BC  2 ✓answer M AB  M BC (5) 1   2  1 2 AB  BC   ABC  90 3.4 BC  3 5 ✓distance of AB AB  ( x2  x1 ) 2  ( y2  y1 ) 2 AB  (3  (11)) 2  (5  2) 2 ✓ formula AB  7 5 ✓sub A 1 BC  AB  ✓answer (4) 2  1 2  3 57 5  105  units 2  52,5units 2 2 [15] Copyright reserved Please turn over
Mathematics/P2 5 NW/November 2024 Grade 10 – Marking Guidelines QUESTION 4 a ✓answer (1) 4.1.1 sin A  b a 4.1.2 cot C  ✓answer (1) c 4.1.3  ✓S/R C  40  sum of ' s of a Δ. a cot C  c ✓sub 5 cot 40  c 5 ✓ c  4.20 c (3) cot 40 c  4.20 4.2.1 12 ✓ cos  13 13 cos  12 ✓diagram 12 cos  13 ✓5  y r  x  y2 2 2 5 (13) 2  (12) 2  y 2 ✓ sin   (4) 13 5 y 5 sin   13 4.2.2 tan - cosec2 ✓ 5  5 13  12    ( )2  13  12 5  ✓ 1903 5   6,34 ✓ answer (3) 300 Copyright reserved Please turn over
Mathematics/P2 6 NW/November 2024 Grade 10 – Marking Guidelines 4.3 2 cosec60.cot30  cos45.cosec45. ✓ 3 2 1  . 3 . 2 ✓ 3 3 2 1  2 1 ✓ 2 3 ✓ 2 ✓3 (5) 4.4 3  sec x  5 ✓ sec x  2 sec x  2 ✓ 1 2 1 cosx 2 ✓ x  60 cosx (3) 1 cos x  2 1 x  cos1   2 x  60 [20] QUESTION 5 a 1 ✓a=1 5.1 b  2 ✓ b=2 (2) 5.2 360 ✓answer (1) y  [0;2] or ✓✓ 5.3 y [0;2] or 0  y  2 0 y2 (combination) (2) 5.4 2 ✓answer (1) 5.5 1 solution ✓answer (1) 0  x  180 ✓✓ 5.6 0  x  180or x  (0;180 ) or (0;180 ) (combination) (2) ✓  cos x 5.7.1 h x    cos x  3 ✓ 3 (2) ✓✓Answer 5.7.2 𝑦 ∈ [−4; −2] or −4 ≤ 𝑦 ≤ −2 (Combination) (2) [13] QUESTION 6 Copyright reserved Please turn over
Mathematics/P2 7 NW/November 2024 Grade 10 – Marking Guidelines 6.1    ✓S/R B  C  BDC  180   sum of ' s of a Δ ✓answer  45  90  BDC  180   BDC  180   45  90  (2) BDC  45 6.2 DC tanB  ✓sub BC ✓answer 200 tan45   BC BC  200 m OR ✓ S✓ R (2) BC= 200m- sides opp = angles 6.3  CD ✓ratio tan A  AC 200 tan30   AC 200 AC  tan30  ✓ AC  346.41 / AC  346.41 200 3 AC  BC  AB ✓Difference 346.41  200  AB 146.41  AB ✓answer (4) [8] QUESTION 7 5m V = L.B.H SA = 2(L.B)+ 2(L.H)+2(H.B) 3m 8m Copyright reserved Please turn over
Mathematics/P2 8 NW/November 2024 Grade 10 – Marking Guidelines 7.1  2(8  5)  (2(3  5))  2(3  8) ✓ substitution  158 m 2 ✓ answer (2) 7.2 V  L.B.H  8 3 5 ✓ substitution  120 m3 ✓ answer (2) 7.3 V = L.B.H 16 x 9 x 2,5 ✓ ✓ 16 x 9 x 2,5 360 m3 ✓ answer (3) [7] QUESTION 8 8.1.1 AB  CD  opp side of parm are  ✓S/R   ✓S/R A1  C2  alt ' s AB CD ✓S/R   ✓ Conclusion E1  F1 alt ' s BE FD  ABE  CDF , AAS (4) Copyright reserved Please turn over
Mathematics/P2 9 NW/November 2024 Grade 10 – Marking Guidelines 8.1.2   ✓S/R C1  A 2  alt ' s BC AD (A)   B  D opp ' s of a parm ✓S/R   B1  D1  proven(ABE  CDF)   B2  D 2  ( A) ✓S/R   F 2  E 2  sum of ' s of a  (3) 8.1.3   C1  A2  proven ✓S/R  B2  D 2  proven ✓S/R BC  AD  opp sides of parm are  ✓conclusion (3)  CBE  ADF , AAS AF  CE  congruent Δ [10] QUESTION 9 Copyright reserved Please turn over
Mathematics/P2 10 NW/November 2024 Grade 10 – Marking Guidelines 9.2 18  4 x  8 ✓18=4x+8 18  8  4 x ✓10=4x 10 5  x ✓ answer (3) 4 2 9.3 ✓S✓R RS is a midpoint ∴ SR‖PQ ∧ ∧ 𝑄 = 𝑅 = 39°−corr ∠'s SR‖PQ(𝐴) ✓S✓R ∧ ∧ 𝑃 = 𝑆 = 55°−corr ∠'s SR‖PQ(𝐴) ∧ ∧ 𝑇 = 𝑇 − common ∠(𝐴) ✓conclusion ∴ ΔTRS|||ΔTQP,AAA (5) [10] TOTAL : 100 Copyright reserved

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