Manjusri 4045 02 Prelim 2024 MS

2024 4NA Prelim Paper 2 Answer Key 1 (a) −1.5 (b) 2.42 (3 s.f.) 2 3.29 × 109 3 (a) x = 6, y =3 (b) p = 2×3×7 = 42 4 x = 7, y = 4 5 The price in UK is cheaper. 6 (a) (i) 76 cm (ii) 61 cm (b) (i) 64.6 cm (ii) 12.7 cm 7 Rose should choose Plan A 8 QR 2 + PR 2 =162 + 282 = 1 040 PQ2,= 342, = 1156 Since 162 + 282 ≠ 342, ∠PRQ is not a right angle. Hence, PQ is not a diameter. 9 (a) 46 639 or 0.0720 (b) 145 852 or 0.170 (c) 1961 2556 or 0.767 10 (a) 23.4° (1 d.p.) (b) 35.0 km 2 (c) 3.18 km 11 (a) 5 cm (b) 31.4 cm (c) 2. 42 radians 12 (a) $568 (b) $340.80 (c) $72.58 13 (a) (i) 1.77 million (ii) 5.64 million (iii) 12.7% (c) 219.5° 14 (a) −15.5 (b) (c) When x = 0, 2 x is undefined, so y is undefined. (d) x = −2, −0.4 or 2.45. (±0.1) (e) Correct tangent drawn Gradient = 3.57 ± 0.2 15 (a) 4 3m/s2 (b) 45 (c) 14.5 m/s (d) 86.4 km/h

16 (a) (i) 20 (ii) 73 (iii) 66 (b) (i) Max mark = 5+93 = 98 B1 (ii) Interquartile range = 82 − 21 = 61 (c) 4H because 4H’s interquartile range is smaller than 4I.

2024 4NA Prelim Paper 2 Marking Scheme 1 (a) −1.5 B1 (b) 2.42 (3 s.f.) B1 2 3 292 498 114 = 3.29 × 109 B1 3 (a) 8640 = 26×33×5 x = 6, y =3 B1 B1 (b) p = 2×3×7 = 42 B1 4 5 3 23xy−= −−−−−−−(1) 2 18xy+= −−−−−−−(2) (2) × 3: 6x + 3y = 54 −−−−−−(3) (1) + (3): 11x = 77 x = 7 Sub x = 7 into (2): 2×7 + y = 18 y = 4 M1 M1 A1 5 Price of Perfume in UK: £62.00 = S$ (62 ÷ 0.58) = S$ 106.90 (2 dp) Price of perfume in Singapore: S$102 × 1.09 = S$ 111.18 The price in UK is cheaper. M1 M1 A1 Accept S$102 × 1.09 = S$ 111.18 111.18 × £0.58 =£64.48 >£62 6 (a)(i) Mode = 76 cm B1 (ii) Median = 61 cm B1 (b)(i) Mean = (43+51+57+60+61+73+76+76+84) ÷ 9 = 64.6 cm (3 s.f.) M1 A1 (ii) Standard deviation = 12.7 cm (3 s.f.) B1 7 Plan A’s total interest after 5 years = $5000 × 3% × 5 M1

= $750 Plan B’s interest after 5 years = $5000 (1+2.8%)5 − $5000 = $740.31 (2 d.p.) Or Plan A’s total amount = $5000 × 3% × 5 + $5000 = $5750 = Plan B’s total amount after 5 years = $5000 (1+2.8%) 5 = $5740.31 < $5750 Rose should choose Plan A for higher interest (or higher total amount) after 5 years M1 M1 (M1) (M1) (M1) A1 For total amt of plan B For interest For simple interest For total amount Must provide reason 8 The sum of the square of the two shorted sides =QR 2 + PR 2 = 162 + 282 = 1 040 The square of the longest side PQ 2,= 342, = 1156 Since 16 2 + 282 ≠ 342, ∠PRQ is not a right angle. Angle in semicircle is not a right angle, so PQ is not a diameter. M1 M1 A1 -1 mark If 162 + 282 =342, is seen from the beginning Must state angle in semicircle is not 90° 9 (a) Boys Girls Total 4B 23 14 37 4C 19 16 35 Total 42 30 72 23 16 72 71× = 46 639 or 0.0720 (3 s.f.) B1 (b) 30 29 72 71× =145 852 or 0.170 (3 s.f.) B1 (c) 1− 35 34 72 71× M1

= 1961 2556 0r 0.767(3 s.f.) A1 10 (a) 2 228 22 15cos 2 8 22ABC +−∠= ×× = 323 352 ∠BAC = 23.420° = 23.4° (1 d.p.) M2 A1 M1 for correct numerator M1 for correct denominator (b) Area of ∆ ABC = 1 8 22 sin 23.4202×× × ° = 34.977 = 35.0 km2 (3 s.f.) M1 A1 (c) Let h be the shortest (perpendicular) distance from A to BC. Area of ∆ ABC = 34.977 1 222 h×× = 34.977 34.977 11h= = 3.18 km (3 s.f.) M1 A1 11 (a) r2 + 122 = 132 r = 2213 12− = 5 cm M1 A1 (b) Circumference = 2π × 5 = 31.416 cm = 31.4 cm (3 s.f.) B1 (c) r × ∠BAC = 31.416 cm ∠BAC = 31.416 ÷ 13 = 2. 42 radians (3 s.f.) M1 A1 12 (a) $2840× 20% = $568 B1 (b) $2840 − $568 = $2272 Interest = $2272 × 5% × 36 12 = $340.80 M1 A1

(c) Monthly instalment = ($2272+ $340.80) ÷ 36 = $72.5777… = $72.58 M1 A1 13 (a)(i) 1.77 million B1 (a)(ii) 5.92 100105 × million = 5.64 million (3 s.f.) M1 A1 (a)(iii) The growth in non-residents from 2022 to 2023 = 1.77 − 1.57 = 0.2 million The percentage growth, z = 0.2 100%1.57× = 12.7% (3 s.f.) No, it should be 12.7% M1 A1 A1 (b) 3.61 3605.92×° = 219.5 ° (1 d.p.) M1 A1 if 3.61 5.92 is seen 14 (a) −4.25 B1 (b) G1 G1 Correct points plotted Smooth curve drawn

(c) When x = 0, 2 x is undefined, will result in “divisionby zero error” so y is undefined. B1 (d) x = −2, -0.4 or 2.45. (±0.1) B2 for all correct answers B1 for only 1 or 2 correct answers. (e) Correct tangent drawn Gradient = 3.57 ± 0.2 B1 B1 15 (a) Acceleration = 16 4 12 3= m/s2 B1 (b) 1( 37)(24) 962 k−= 9637 12k−= 37 8k = + = 45 M1 A1 (c) Total distance =1112 16 10 16 (16 24) 15 9622××+×+× + ×+ = 652 m Average speed = 652 45 = 14.5 m/s (3 s.f.) M1 M1 A1 (d) Method 1: 24 m/s = 24 1000km ÷ 1 3600h = 0.024 × 3600 km/h = 86.4 km/h Method 2: 1sec → 24 m = 24 1000 km = 0.024 km 1hour = 3600s→0.024 km × 3600 = 86.4 km M1 A1 (M1) (A1) 16 (a)(i) Interquartile range = 64 − 44 = 20 M1 A1 (ii) 90% × 30 = 27 90th percentile = 73 B1 (iii) 20% × 30 = 6 M1

Min mark for distinction = 66 A1 (b)(i) Max mark = 5+93 = 98 B1 (ii) Interquartile range = 83 − 22 = 61 B1 (c) Class 4H because 4H’s interquartile range is smaller than 4I. B1