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Trial SPM 2013 / Kertas Percubaan SPM 2013 SMK St George Taiping

### Transcript of Trial spm smk_st_george_taiping_2013_maths_paper1_2_[a]

• 1. PEPERIKSAAN PERCUBAAN SPM MATHEMATICS FORM 5 SGI 2013 Kertas 1 Total = Paper 1 + Paper 2 = 40 + 100 = 140 1 C 11 D 21 B 31 B 2 B 12 C 22 A 32 A 3 C 13 D 23 C 33 D 4 D 14 B 24 B 34 A 5 C 15 C 25 B 35 D 6 D 16 D 26 D 36 A 7 A 17 B 27 D 37 B 8 A 18 C 28 B 38 C 9 C 19 D 29 C 39 D 10 B 20 B 30 B 40 A Kertas 2 Q Solution Submarks Full marks 1 Dotted line y = 3 drawn and Correct shaded region Note : 1. Solid line y = 2 drawn ( 1 mark) 2. Give 1 mark if shaded region between 2 inequalities only 3M 3 2 (a) 640 = 240 (b) 05 040 or equivalent = 8 ms 1 (units not important) (c) 1 1 ( 40 5) (240) 40 ( 11) 2 2 t t t )11(40 2 1 )240()540 2 1 ( = 28 or equivalent t = 15 s (units not important) 1M 1M 1M 1M 1M 1M 1M 7 3 3 1 4 22 21 2 3 7 2 or 5.105.105.10 7 22 3 4 2 1 or 2 22 14 35 7 2 or 3577 7 22 Volume = 1 4 22 10.5 10.5 10.5 2 3 7 + 22 7 7 35 7 = 7815.5 or 2 1 7815 or 15631 2 cm3 (units not important) 1M 1M 1M 3 x + y = 8 y = 8 2x y = 2 8 2 x 0 84 y

2. 4 (a)(i) False/ Palsu (ii) True/ Benar (b) k + 1 is not an odd integer/ k + 1 bukan satu integer ganjil (c) 4n 3, n = 2, 3, 4, ... or equivalent Note : 4n 3 only (Give 1 mark) 1M 1M 2M 2M 6 5 LNM tan LNM = 8 6 or equivalent 36.87o or 36o 52 1M 1M 1M 3 6 3x2 5x 2 = 0 (3x + 1) (x 2) = 0 x = 3 1 , 2 1M 1M 1M,1M 4 7 4m n = 26 4m + 3n = 2 4n = 24 or 1 2 ( 6) 13 2 m or 4m (-6) = 26 or or 4m + 3(-6) = 2 m = 5, n = 6 Note : 5 6 m n only award 1 mark 1M 1M 1M,1M 4 8 (a) M PR = 2 and c = 4 y = 2x + 4 (b) Coordinate T : (2, 0) or x = 2 y = 2(2) + 4 or equivalent R(2, 8) 1M 1M 1M 1M 1M 5 9 (a) Sampel space/ Ruang sampel = {(S,C), (S,O), (S,R), (S,E), (C,S), (C,O), (C,R), (C,E), (O,S), (O,C), (O,R), (O,E), (R,S), (R,C), (R,O), (R,E), (E,S), (E,C), (E,O), (E,R)} (b)(i) {(C,S), (C,O), (C,R), (C,E)} 4 20 or 1 5 (ii){(S,C), (S,R), (C,S), (C,R), (O,E), (R,S), (R,C), (E,O)} 8 20 or 2 5 Note : Accept correct answer only if list out 1M 1M 1M 1M 1M 5 10 (a) 60 22 2 14 360 7 or equivalent Perimeter = 60 22 2 14 360 7 +14 + 14 + 7 + 7 or equivalent = 56.67 or 3 2 56 or 170 3 cm (units not important) 1M 1M 1M 1 132 2 2 4 3 1 1331 2 1 22(3) 4( ) 4 2 2 5 6 m n m n 3. (b) 260 22 14 360 7 or 2 7 7 22 360 60 or 290 22 7 360 7 or equivalent Area of the shaded region/ Luas kawasan berlorek = 2 2 260 22 60 22 90 22 14 7 14 7 7 360 7 360 7 360 7 or equivalent = 136.5 or 1 136 2 or 273 2 cm2 (units not important) 1M 1M 1M 6 11 (a) Q = )4(3)5(2 1 53 42 = 53 42 2 1 or 1 2 3 5 2 2 (b) 8 13 23 45 y x 8 13 53 42 2 1 y x 2 1 ,3 yx Note : 3 1 2 x y only award 1 mark 1M 1M 1M 1M 1M,1M 6 12 (a) x = 1, y = 12 x = 3, y = 4 (b) Graph All axis drawn with correct direction and uniform scale 7 points and 2 points* have been plot correctly (7 or 8 points have been plot correctly get 1 mark) Smooth curve and continue without straight line and passing through the correct 8 points for 3 x 4. (c) (i) 10 9y (ii) 0.5 0.4x (d) Identify equation y = 5x 5 Straight line y = 5x 5 correctly drawn x = 0.4 x 0.5 , x = 3.75 x 3.85 1M 1M 1M 2M 1M 1M 1M 1M 1M 1M, 1M 12 13 (a) (i) (4, 9) (ii) (4, 9) (4, 3) (b)(i)(a) U = Rotation, 900 clockwise, about centre (2, 4) Note : 1. Rotation, 900 clockwise, get 2 marks 2. Rotation, about centre (2, 4), get 2 marks 3. Rotation, get 1 mark (b) W = Enlargement, with centre (7, 4) , of scale factor 2 Note : 1. Enlargement, with centre (7, 4), get 2 marks 2. Enlargement, of scale factor 2 , get 2 marks 3. Enlargement, get 1 mark (ii) 2 2 A 2 2 120A A or equivalent A = 40 m2 (units not important) 1M 1M, 1M 3M 3M 1M 1M 1M 12 4. 14 (a) (i) 12 (ii) I II III Column I (All Correct) Column II (All Correct) Column III (All Correct) (iii) (3 112 12 117 26 122 24 127 12 132 8 137 6 142 3 147) 94 = 126.95 (b) All axis in correct direction, uniform scale, x-axis label as upper boundry All 8 points* correctly plot. (6 or 7 points* correctly plot get 1 mark) Ogive pass through point (109.5, 0) Smooth curve without any straight line and pass all the 8 points correctly (c) Median = 125.5 0.5 Height (cm) Tinggi (cm) Mid point Titik Tengah Frequency Kekerapan Cumulative Frequency Kekerapan Longgokan Upper Boundaries Sempadan atas 105 109 107 0 0 109.5 110 114 112 3 3 114.5 115 119 117 12 15 119.5 120 124 122 26 41 124.5 125 129 127 24 65 129.5 130 134 132 x = 12 77 134.5 135 139 137 8 85 139.5 140 144 142 6 91 144.5 145 - 149 147 3 94 149.5 1M 1M 1M 1M 1M 1M 1M 2M 1M 1M 1M 12 15 (a) Plan or The shape must be correct in quadrilateral form. All lines must be drawn in full. JG = NK > GN = KJ > JI = QK The measurement is accurate to 0.2 cm. (one way ) and the angles at all verticals of the rectangle are 90 1 (b) (i) Viewed from A The shape must be correct. All lines must be drawn in full. JV > VF > IH > GH > FG, UV > VZ = YU = FG Curve must be smooth The measurement is accurate to 0.2 cm. (one way ) and the angles at all verticals of the rectangle are 90 1 1M 1M 1M 1M 1M 1M 1M IHJE GF 2 cm 4 cm 5 cm KL NMQP 5 cm 2 cm 4 cm IH JE GF KL NM QP TCW JK Y ZSR UX VEL IQ FM HP GN 2 cm 5 cm 4 cm 3 cm 4 cm 6 cm 5. (i) Viewed from B The shape must be correct in quadrilateral form. All lines must be drawn in full. ICGF and QWNM are two parallel straight lines Note: Ignore straight line CW C and W are joined dotted lines to shape a rectangular CFMW VM = MQ > QI = TV > IC > CT > CG = WN The measurement is accurate to 0.2 cm. (one way ) and the angles at all verticals of the rectangle are 90 1 1M 1M 1M 1M 1M 12 16 (a) R = 70o N (b) Q(70o S, 50o W) (c) 60)9070( = 9600 n.m. (d)(i) 600 2 =1200 n.m. (ii) 70cos60 2600 = 58.48o or 58o 29 58.48o + 50o = 108.48o W or 108o 29 W 2M 2M 2M 1M 1M 2M 1M 1M 12 3 cm 2 cm 3 cm 3 cm 5 cm T IJ C ZY VU FE QK MLX NPR W GHS 6. Graph For Question 12 -1-2-3 -4 y x1 2 3 5 10 15 20 25 -5 -10 0 (-2, 16) (-1, 12) (0, 2) (1, -8) (2, -12) (3.5, 4.4) (4, 22) (3, -4) (-3, 8) 7. Graph For Question 14 Graph for Question 14 (-3, -29) 50 10 60 40 20 30 70 80 90 100 0 Upper Boundary 124.5 129.5 139.5 144.5134.5119.5114.5109.5 149.5 Cumulative Frequency