2024 NCHS Prelim Math P1 Suggested Solutions

Nan Chiau High School 2024 Secondary 4 Mathematics Mathematics Prelim Examination Paper 1 Solutions Qn Solution 1a False 1b False 2 105𝑏 𝑎 = (3×5×7)(22×5×7) 2×32×5 = 22×3×52×72 2×32×5 = 2×5×72 3 Numerator 2 × 5 × 72 does not include the factor 3 to reduce 105𝑏 3 to a whole number. 3 Length of shorter portion of rope = 4 11 ×44 = 16 cm Length of longer portion of rope = 28 cm Let 𝑥 be the length rope cut. 16−𝑥 28−𝑥 = 2 5 5(16 − 𝑥) = 2(28 − 𝑥) 80 − 5𝑥 = 56 − 2𝑥 24 = 3𝑥 𝑥 = 8 4 79−𝑥2 − 1 = 0 79−𝑥2 = 1 7(3−𝑥)(3+𝑥) = 70 𝑥 = 3 𝑜𝑟 𝑥 = −3 5 𝑦3 = 𝑘 𝑥2 𝑘 32 − 𝑘 62 = 5 𝑘 9 − 𝑘 36 = 5 𝑘 12 = 5 𝑘 = 60 𝑦3 = 60 𝑥2 23 = 60 𝑥2 𝑥2 = 7.5 𝑥 = ±2.74

6 24 g → 12.044 × 1023 atoms 1 g → 0.5.01833 × 1023 atoms 5000g → 0.5.01833 × 1023 × 5000 = 2509.167 × 1023 = 2.51 × 1026 7a Diagram A 7b Diagram B 7c Diagram D 8 9𝑥 + 2.5𝑦 = 3𝑦 − 2𝑥 11𝑥 = 0.5𝑦 22𝑥 = 𝑦 or 𝑥 𝑦 = 1 22 Hence, 𝑥 = 22 and 𝑦 = 1 is not a solution, as 22(22) ≠ (1) Or LHS = 9(22) + 2.5(1) = 200.5 RHS = 3(1) − 2(22) = −41 Since LHS ≠RHS, 𝑥 = 22 and 𝑦 = 1 is not a solution. 9 3𝑦 − 𝑘𝑥 − 9 = 0 3𝑦 = 𝑘𝑥 + 9 𝑦 = 𝑘 3 𝑥 + 3, where 𝑘 < −3 3𝑦 − 𝑘𝑥 − 9 = 0 𝑦 = −𝑥 3 y x

10 Let the first digit be 𝑥 and second digit be 𝑦 A number between 20 to 100 Let the number be 10𝑥 + 𝑦 Add the two digits together 𝑥 + 𝑦 Subtract the sum of the two digits from your original number to get a new number (10𝑥 + 𝑦) − (𝑥 + 𝑦) = 9𝑥 The new number is a multiple of 3 Since the new number is a multiple of 9 and 3 is a factor of 9, hence the new number is a multiple of 3. 11 2(𝑚𝑛)2 − 2𝑚𝑛 = 3𝑚𝑛 + 3 2𝑚2𝑛2 − 5𝑚𝑛 − 3 = 0 (𝑚𝑛 − 3)(2𝑚𝑛 + 1) = 0 𝑚𝑛 = 3 or 𝑚𝑛 = − 1 2 𝑛 = 3 𝑚 or 𝑛 = − 1 2𝑚 12 4 3 + 2 9𝑚 ÷ √81𝑚164 = 4 3 + 2 9𝑚 ÷ 3𝑚4 = 4 3 + 2 9𝑚 × 1 3𝑚4 = 4 3 + 2 27𝑚5 = 36𝑚5+2 27𝑚5 13 −9𝑥2+12𝑥−4 3 − 1 𝑥 − 𝑥 2 = −(3𝑥−2)2 3 − 1 2𝑥−𝑥 2 = −(3𝑥−2)2 3 − 2 𝑥 = −(3𝑥−2)2 3𝑥−2 𝑥 = −𝑥(3𝑥 − 2) 14 𝑦 = 1 5 𝑥2 − 2𝑥 + 7 ------------ [1] 𝑦 = 𝑥 ----------- [2] Sub eqn[1] into eqn[2] 𝑥 = 1 5 𝑥 2 − 2𝑥 + 7 1 5 𝑥2 − 3𝑥 + 7 = 0 𝑥 = 3±√(−3)2−4(1 5)(7) 2(1 5) 𝑥 = 2.89 or 𝑥 = 12.1

𝑦 = 2.89 or 𝑦 = 12.1 (2.89, 2.89) and (12.1, 12.1) 15 a Let the y-intercept (vertical distance) be 𝑑 By Pythagoras’ Thm 𝑑 = √172 − 82 𝑑 = 15  𝐶(0, −15) 15 b Let the 𝑥-coordinate of point B be 𝑏 𝑏 = 1 8 (−15) 𝑏 = − 15 8 Equation of line of symmetry, 𝑥 = −15 8 +(−8) 2 𝑥 = − 79 16 Since 𝑦 = −15 when 𝑥 = 0 Coefficient of 𝑥2 = −1 Since curve is 𝑦 = −(𝑥 + 8)(𝑥 + 15 8 ) or 𝑦 = − 1 8 (𝑥 + 8)(8𝑥 + 15) 𝑦 = −(− 79 16 + 8)(− 79 16 + 15 8 ) 𝑦 = 2401 256 the coordinate of the maximum point is (− 79 16 , 2401 256 ) or (- 4.94, 9.38) 16 a ( 8 8+5+𝑥)( 7 7+5+𝑥) = 4 33 ( 8 13+𝑥)( 7 12+𝑥) = 4 33 33(8)(7) = 4(13 + 𝑥)(12 + 𝑥) 462 = 𝑥2 + 25𝑥 + 156 𝑥2 + 25𝑥 − 306 = 0 (𝑥 − 9)(𝑥 + 34) = 0 𝑥 = 9 or 𝑥 = −34 (rej) 16 b P(second marble is green when three marbles are drawn) = ( 17 22) ( 5 21) ( 16 20) = 34 231

17 a 17 bi ∅ 𝑜𝑟 { }, { 2}, { 3} , { 4}, { 2, 3}, { 2, 4} , { 3, 4}, { 2, 3, 4} 17 bii Number of subsets = 2𝑛−1 18 a Mean = 12(150)+10(200)+8(300)+𝑥(400)+5(500)+2400 12+10+8+𝑥+5+1 325 = 11100+400𝑥 36+𝑥 11700 + 325𝑥 = 11100 + +400𝑥 600 = 75𝑥 𝑥 = 8  Total number of workers = 12 + 10 + 8 + 8 + 5 + 1 = 44 18 b Median is a better gauge of workers’ salary as there is salary that is an extreme value of S2400 compared to the rest. 19 a 40% + 35% + 20% = 95% As pie charts are used to visualize parts of a whole, all the parts should always add up to 100%. [or] The title exaggerated the responses of the support for Candidate C. 19 b To include in uncounted / spoiled votes (5%) to show the full result of votes [or] To re-calculate the votes according to the three candidates according to base of 95%. [or] To recraft the title to be less bias and allow readers to analysis themselves. 𝜀 𝑋 𝑌 4 2 3 5 7 11 13 1 6 8 9 10 12 14 15 16

20 a Area of rhombus = 2 × 1 2 (7.4)(7.4) sin 60 = 47.424cm2 [Or] height of rhombus = (7.4)(sin 60) = 6.4086 cm Area of rhombus = (6.4086)(7.4) = 47.424 cm2 𝐴𝐷2 = 7.42 + 7.42 − 2(7.4)(7.4) cos 120 𝐴𝐷 = 12.817 cm Area of sector = 60 360 × 𝜋(12.817)2 = 86.014 cm2 Shaded area = 86.014 − 47.424 = 38.6 cm2 20 b Arc length = 60 360 × 2𝜋(12.817) = 13.422 Perimeter = 13.422 + 2(12.817 − 7.4) + 2(7.4) = 39.1 cm 21 a 102 + 42 = (10 − 4)2 + 2(10)(4) 21 b (2𝑛)2 + (𝑛 − 1)2 = (2𝑛 − 𝑛 + 1)2 + 2(2𝑛)(𝑛 − 1) 𝑜𝑟 = 4𝑛2 + 𝑛2 − 2𝑛 + 1 or = 5𝑛2 − 2𝑛 + 1 21 c 𝑝2 + 𝑞2 = 1249 5𝑛2 − 2𝑛 + 1 = 1249 5𝑛2 − 2𝑛 − 1248 = 0 (𝑛 − 16)(5𝑛 + 78) = 0 𝑛 = 16 or 𝑛 = − 78 5 (rej) Hence, 𝑝 = 32 and 𝑞 = 15 22 a 𝐴𝐵2 + 𝐵𝐶2 = 6.42 + 4.82 = 64 𝐴𝐶2 = 82 = 64 Since 𝐴𝐵2 + 𝐵𝐶2 = 𝐴𝐶2, hence by converse of Pythagoras’ Theorem , 𝐴𝐵𝐶 = 90𝑜 sin ACB = 6.4 8 = 4 5

22 b cosDAC= − 4 5 23 a 24 m/s = 24 × 1 1000𝑘𝑚 1 × 1 3600ℎ = 86.4km/h 23 b 504 = ( 1 2 × 5 × 𝑣) + ( 1 2 (𝑣 + 24)(7)) + ( 1 2 (8 + 12)(24)) 504 = 5 2 𝑣 + 7 2 𝑣 + 84 + 240 180 = 6𝑣 𝑣 = 30 km/h 23 c Distance from 0s to 5s = 1 2 × 5 × 30 = 75 𝑚 Distance from 0s to 12s = 75 + 1 2 (30 + 24)(7) = 264𝑚 Distance from 0s to 20s= 264 + (24)(8) = 456 𝑚

24 a 24 b 24 c 5.9 x 15 = 88.5 km 25 a −20 25 b (0.8, 5.5) 25 c 𝑘 = 5.2