Mathematics Grade 10 IEB Paper 2 November 2012 Memo

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Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE GRADE 10 MATHEMATICS P2 EXEMPLAR 2012 MEMORANDUM MARKS: 100 This memorandum consists of 10 pages. Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P2 2 DBE/2012 NSC – Grade 10 Exemplar – Memorandum NOTE: • If a candidate answers a question TWICE, only mark the FIRST attempt. • If a candidate has crossed out an attempt of a question and not redone the question, mark the crossed out version. • Consistent accuracy applies in ALL aspects of the marking memorandum. • Assuming answers/values in order to solve a problem is NOT acceptable. QUESTION 1 1.1 n 929 ∑x i 929  19 Mean = i =1 = = 48,89 answer n 19 (2) 1.2 31 ; 31 ; 34 ; 36 ; 37 ; 39 ; 40 ; 43 ; 46 ; 46 ; 48 ; 52 ; 56 ; 60 ; 62 ; arranging in 63 ; 65 ; 66 ; 74. ascending order Median = 46 median (2) 1.3 Lower quartile = 37  lower quartile Upper quartile = 62  upper quartile (2) 1.4  box with 74 median 31 37 46 62 25 35 45 55 65 75  whisker (2) [8] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P2 3 DBE/2012 NSC – Grade 10 Exemplar – Memorandum QUESTION 2 2.1 The modal class is 2500 ≤ x < 4500  2500 ≤ x < 4500 (1) 2.2 Gross Vehicle Frequency Mass (GVM) Frequency Midpoint × (in kg) midpoint 2500 ≤ x < 4500 103 3500 360 500 4500 ≤ x < 6500 19 5500 104 500  midpoints 6500 ≤ x < 8500 70 7500 525 000  frequencies × midpoint 8500 ≤ x < 10500 77 9500 731 500 10500 ≤ x < 12500 85 11500 977 500 12500 ≤ x < 14500 99 13500 1 336 500 Sum 453 4 035 500 4 035 500 4035500 Estimated mean (X ) = = 8908,39 kg.  answer 453 (5) 2.3 The estimated mean.  estimated mean It is more at the centre of the data set. The modal class is found at with reason the extreme left-hand side of the data set. (1) [7] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P2 4 DBE/2012 NSC – Grade 10 Exemplar – Memorandum QUESTION 3 yy D(–3 ; 3) xx E(3 ; –5) F(–1 ; k) 3.1.1 DE = (−3 − 3) 2 + (3 − (−5)) 2  substitution into distance formula = 100 = 10 answer (2) 3.1.2 −5−3  substitution into m DE = gradient formula 3 − (−3) 4 =− answer 3 (2) 3.1.3 3 3 m EF = EF ⊥ DE  m EF = 4 4 −5−k 3 −5−k 3 =  = 3 − (−1) 4 3 − (−1) 4 −5−k 3 = 4 4  simplification − 20 − 4k = 12 − 4k = 32 k = −8  k = −8 (4) 3.1.4  (−3) + (−1) 3 + (−8)   substitution into M ;  midpoint formula  2 2   5 = − 2 ; −   2 answer (2) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P2 5 DBE/2012 NSC – Grade 10 Exemplar – Memorandum 3.1.5 If DEFG is a rectangle then M is also the midpoint of EG. Let the coordinates of G be (x ; y)  x + 3 y + (−5)   5  ;  = − 2 ; −   2 2   2 x+3  = −2 x+3 y −5 5 2 = −2 =−  x = −7 2 2 2 x + 3 = −4 and y − 5 = −5 y −5 5 x = −7 y=0  =− 2 2 y=0 ∴G( –7 ; 0) (4) OR The translation that sends E(3 ; –5) to F(–1; –8) also sends method D(–3 ; 3) to G. x–4 (–1 ; –8) = (3 – 4 ; –5 – 3) y–3 ∴ G = (– 3 – 4 ; 3 – 3) = (– 7 ; 0)  answer (4) OR The translation that sends E(3 ; –5) to D(–3 ; 3) also sends method F(–1; –8) to G. x–6 (– 3 ; 3) = (3 – 6 ; – 5 + 8) y+8 ∴ G = (–1 – 6 ; – 8 + 8) = (– 7 ; 0)  answer (4) 3.2 equation using ( x − 1) + ( 5 − (−2) ) = 53 2 2 distance formula ( x − 1) 2 + 49 = 53 x − 2 x + 1 + 49 − 53 =0 2 x2 − 2 x − 3 = 0 ( x + 1)( x − 3) = 0  standard form x −1 = = or x 3 but D is in the second quadrant  factorisation ∴ only x −1 = is valid  answer must exclude 3 (4) [18] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P2 6 DBE/2012 NSC – Grade 10 Exemplar – Memorandum QUESTION 4 4.1.1 AB  AC sin C = (1) AC 4.1.2 AB  cot A cot A = (1) BC 4.2 sin 60°.tan 30° sec 45°  3  1   substitution    2   3  = 2 1 = 2 2  simplification 1 1 = × 2 2 1 2 = × 2 2 2 2 answer = 4 (4) 4.3.1 r = (− 5) + (12 )  r = (− 5) + (12 ) 2 2 2 2 2 2 r 2 = 169  r = 13 r = 13 5 answer cos θ = − 13 (3) 4.3.2 cosec θ + 1 2 13  13  2 = =   +1 12  12  169 144 simplification = + 144 144 313 answer = (3) 144 [12] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P2 7 DBE/2012 NSC – Grade 10 Exemplar – Memorandum QUESTION 5 5.1.1 5 cos x = 3 3 cos x = 3  cos x = 5 5 3 x = cos −1   5  answer x = 53,1° (2) 5.1.2 tan 2 x = 1,19 2 x = tan −1 (1,19)  2x = 49,958....° 2 x = 49,95845.....° x = 25°  answer (3) 5.1.3 4 sec x − 3 = 5 4 sec x = 8 sec x = 2  sec x = 2 1 1 = sec x 2  inverting both 1 sides cos x = 2  cos x 1−1 x = cos   2 x = 60°  answer (4) 5.2.1 ˆ D = 8° JK alternate angles  answer (1) 5.2.2 5 5 tan8° =  tan 8° = DK DK 5 5 DK =  DK = tan 8° tan 8° DK = 35,57684..... km  answer DK = 35 577 m (3) 5.2.3 DS = 35,58 – 8 = 27,58 km  answer (1) 5.2.4 5 5 tan DŜJ =  tan DSˆ J = 27,58 27,58  5  DŜJ = tan −1    27,58  DŜJ = 10,3°  answer (2) [16] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P2 8 DBE/2012 NSC – Grade 10 Exemplar – Memorandum QUESTION 6 6.1.1 y 6 correct x-intercepts 5 correct y-intercept 4  asymptotes 3 2  shape (must pass 1 through (45° ; 2)) x 0 90° 180° 270° 360° -1 (4) -2 -3 -4 -5 -6 6.1.2 y = −2 tan x  answer (1) 6.2.1 g ( x) = a sin x 4 = a sin 90° 4 = a (1)  a=4 a=4 (1) 6.2.2 Range is − 2 ≤ y ≤ 6 .  –2 6 (2) [8] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P2 9 DBE/2012 NSC – Grade 10 Exemplar – Memorandum QUESTION 7 7.1.1 AH2 = 0,82 + 1,52  AH2 = 0,82 + 1,52 AH2 = 2,89  AH = 1,7 AH = 1,7 (2) 7.1.2 1 1 Surface area of roof = 4 × (3 × 1,7)  4 × (3 × 1,7) 2 2 = 10,2 m2 answer (2) 7.1.3 Surface area of walls = 4 × 3 × 2,1 = 25,2 m2  25,2 m2 Total surface area = 10,2 m2 + 25,2 m2 = 35,4 m2  answer (2) 7.2.1 π (8)3  π (8) 4 4 3 Volume = 3 3 = 2144,66 mm3  answer (2) 3 3 7.2.2 New volume : original volume = 2 : 1 2 =8:1  answer (2) 7.2.3 π (9)3 = 3 053,63 mm3. π (9)3 4 4 Volume including silver =  3 3 Volume of silver = 3 053,63 – 2144,66  answer = 908,97 mm3 (2) [12] QUESTION 8 8.1 OQ = 2 cm …. (the long diagonal of a kite bisects  2 cm the shorter diagonal)  correct reason (2) 8.2 PÔQ = 90° …. (the diagonals of a kite intersect at  90° right angles) correct reason (2) 8.3 QP̂O = 20° …. (the longer diagonal bisects the  QP̂O = 20° with angles of a kite) correct reason ∴ QP̂S = 20° + 20° = 40°  QP̂S = 40° (2) [6] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P2 10 DBE/2012 NSC – Grade 10 Exemplar – Memorandum QUESTION 9 A B E F O C D 9.1 O is the midpoint of BD. …. (Diagonals of parm BCDE bisect  O is the midpoint each other) of BD  reason – diagonals of parm F is the midpoint of OE. …. (Diagonals of parm AODE bisect  F is the midpoint each other) of OE ∴ OF || AB …. (The line joining the midpoints of  reason – two sides in a ∆ is || to third side) midpoint theorem (4) 9.2 AE || OD …. (Opp sides of parm AODE are  AE || OB ∴ AE || OB parallel)  reason OF || AB …. (proven above)  OE || AB ∴OE || AB ∴ ABOE is a parallelogram …. (both pairs of opposite sides of  reason – opp quad are parallel) sides parallel (4) 9.3 In ∆ABO and ∆EOD 1. AB = EO …(Opp sides of parm ABOE are equal)  AB = EO 2. AO = ED …(Opp sides of parm AODE are equal)  AO = ED 3. BO = DO …(Diagonals of parm BCDE bisect each  reason – opp other) sides are equal  BO = DO ∴∆ABO ≡ ∆EOD (S, S, S)  reason – diagonals of parm (5) [13] TOTAL: 100 Copyright reserved Powered by TCPDF (www.tcpdf.org)