AISS 4N Prelim 2023 P2 MS

1 2023 Sec 4NA Math Prelim Paper 2 Marking Scheme Qn Working Mark Awarded Sub- total Remarks 1 (a) 7.004869155 = 7.00 (to 3 s.f.) B1 1 (b)(i) 62.589 10 − B1 1 (ii) ( ) ( ) 625.97 10 7.34 10  = 38.133514986 10 = 38.13 10 (to 3 s.f.) M1 A1 2 2 Sum of interior angles in a pentagon = ( )5 2 180−   = 540o Angle FED = angle EDC = angle DCG = 540o ÷ 5 = 108o x = 180o – 108o (adj angles on a straight line) = 72o 540o – 90o – 108o – 108o – 108o = 126o (2y – 3) + (y + 15) = 126 3y + 12 = 126 3y = 114 y = 38o M1 M1 A1 M1 A1 5 M1 for equation formed. 3 (a)(i) The total number of students in each class may be different so it is not accurate to compare using the size of the sector or its angle. B1 1 (ii) Grey eyes in Class 1A → 30o 30o → 3 students 1o → 1 10 student 360o → 1 36010 = 36 students M1 A1 2 (iii) 45 100%360 = 12.5% M1 A1 2

2 Qn Working Mark Awarded Sub- total Remarks (b)(i) Median = 23 24 2 + = 23.5 B1 1 (b)(ii) Total = 08 + 09 + 12 + 15 + 15 + 16 + 18 + 18 + 21 + 23 + 24 + 24 + 26 + 27 + 27 + 27 + 28 + 29 + 30 + 30 = 427 Mean = 427 ÷ 20 = 21.35 [Alternative solution] Mean = (08 + 09 + 12 + 15 + 15 + 16 + 18 + 18 + 21 + 23 + 24 + 24 + 26 + 27 + 27 + 27 + 28 + 29 + 30 + 30) ÷ 20 = 21.35 M1 A1 [M1] [A1] 2 (b)(iii) P(scored more than 25 marks) 8 20 2 5 = = M1 A1 2 4 Let r be the radius of the water surface. 2 225r = r = 15 cm By Pythagoras’ Theorem, 2217 15− = 8 cm x = 17 – 8 = 9 M1 M1 M1 A1 4 5 (a) 60 000 cm = 60 000 / 100 000 km = 0.6 km B1 1

3 Qn Working Mark Awarded Sub- total Remarks (b) 13 0.5 = 7.8 km B1 1 (c) 1 cm to 0.6 km 1 cm2 to 0.62 km2 1 cm2 to 0.36 km2 4.8 1 130.36 3= cm2 M1 A1 2 6 (5x – 4) (x + 2) = 0 5x – 4 = 0 or x + 2 = 0 x = 4 5 or x = –2 M1 A2 3 No marks given for any other methods used. 7 (a) Volume of cone = 21 3 rh = 21 12 53 = 753.9822369 cm3 Volume of cylinder = 2rh = 212 8 = 3619.114737 cm3 Volume of composite solid = 753.9822369 + 3619.114737 = 4373.096974 = 4370 cm3 (to 3 significant figures) M1 M1 A1 3 (b) By Pythagoras’ Theorem, 225 12+ = 13 cm B1 1 (c) Curved surface area (cone) = rl = 12 13 = 490.088454 cm2 Curved surface area (cylinder) = 2 rh = 2 12 8 = 603.1857895 cm2 Surface area of circle (base) = 2r = 212 M1 M1 M1

4 Qn Working Mark Awarded Sub- total Remarks = 452.3893421 cm2 490.088454 + 603.1857895 + 452.3893421 = 1545.663586 = 1550 cm2 (to 3 significant figures) A1 4 8 (a) A(–2, 5), B(4, 2) Length of AB ( ) ( ) 22 2 4 5 2= − − + − 45= = 6.7082 = 6.71 units (to 3 s.f.) B1 1 (b) y-intercept = 4 Gradient = 1 2− Equation: 1 42yx=− + B1 1 M1 for both y- intercept and gradient (c) (1, 3.5) B1 1 (d) (i) (1, –1) B1 1 (ii) Length of AB 45= = 6.7082 units (from (a)) Length of BC ( ) ( ) 22 4 7 2 ( 4)= − + − − 45= = 6.7082 AB = BC Yes, Claire is correct. M1 A1 2 M1 for calculating the length of BC. A1 for stating that Claire is correct.

5 Qn Working Mark Awarded Sub- total Remarks (iii) Area of ABCD = Area of square – Area of 4 triangles = ( ) 19 9 4 9 3 2  −   = 81 – 54 = 27 units2 M1 A1 2 9 (a) –3 B1 1 (b) Refer to annex for graph. Plotted points: • 2 marks for all 7 correctly plotted points. • 1 mark for at least 4 correctly plotted points 1 mark for smooth curve B3 3 (c) (i) 2.25 (Acceptable range of 1.75 to 2.75, inclusive) (+/- 1 square) B1 1 No marks awarded if students have conducted any form of calculations to find the values of x or y. (ii) –1.55 (Acceptable range of –1.6 to –1.5 inclusive) (+/- 1 square) B1 1 (d) 1 mark for tangent drawn. Gradient = 0 M1 A1 2 No marks awarded if no tangent is drawn. 10 (a) Total distance = 1.5 + 40 + 10 = 51.5 km B1 1

6 Qn Working Mark Awarded Sub- total Remarks (b) 135 min = 135 60 h = 2 h 15 min Silver award B1 1 (c) 40 22 9111= h M1 A1 2 (d) Shortest possible time if trains intensely for swimming = 1.5 40 10 5 22 8++ = 813 220 h = 3 h 22 min (rounded off to nearest minute) Shortest possible time if trains intensely for running = 1.5 40 10 3 22 11.9++ = 4153 2618 h = 3 h 10 min (rounded off to nearest minute) Susan should train intensively for running as 3 h 10 min < 3 h 22 min or 3 h 10 min (running) would result in the bronze award while 3 h 22 min (swimming) would result in the consolation prize. M1 M1 A1 A1 [A1] 4 A2 for any reasonable answers. 11 (a) (i) Estimated mean = 2(145) 11(155) 18(165) 6(175) 3(185) 2 11 18 6 3 + + + + + + + + = 6570 40 = 164.25 cm B1 1 (ii) Standard deviation = 9.588404455 = 9.59 cm (to 3 s.f.) B1 1 (iii) The students in Class B are taller

7 Qn Working Mark Awarded Sub- total Remarks because 169.3 > 164.25 or because the mean height of Class B is more than that of Class A. B1 1 1 mark given when both answer and explanation are given. (iv) The students in Class B have more consistent heights because 7.8 < 9.59. or because the standard deviation of Class B is less than that of Class A. B1 1 1 mark given when both answer and explanation are given. (b) (i) Probability tree diagram: B2 2 B1 for every two correct answers. (ii) P(different colours) = P(blue, red) + P(red, blue) = 3 7 7 3 10 9 10 9     +        = 7 15 M1 A1 2 12 (a) Cosine Rule 2 2 2 2 cosa b c bc A= + − 2 2 2259 213 105 2(213)(105)cos ABC= + − M1 M1 M1 for use of Cosine Rule.

8 Qn Working Mark Awarded Sub- total Remarks 2 2 2 1 259 213 105cos 2(213)(105)ABC −  −−=  − = 103.8229497o = 103.8o (to 1 d.p.) (shown) A1 3 (b) Sine Rule sin sin AC CD ADC CAD= 259 sin 79 sin 42 CD= 259sin 42 sin 79CD =  = 176.5485206 = 177 km (to 3 s.f.) M1 A1 2 M1 for use of Sine Rule. (c) Let N1 be the North of A and N2 be the North of C. 1 135N AD = 1 135 42 93N AC = − =  2 180 93ACN = −  (interior angles, N1A // N2C) = 87o 180 42 79ACD= − −  (sum of angles in a  ) = 59o Bearing of C from D = 360o – 87o – 59o (angles at a point) = 214o M1 M1 A1 3 M1 for finding 93o. M1 for finding angle ACD.