2024 NCHS Prelim Math P1 Suggested Solutions
Uploaded by ploopy27 ยท 27 October 2024
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Nan Chiau High School 2024 Secondary 4 Mathematics Mathematics Prelim Examination Paper 1 Solutions Qn Solution 1a False 1b False 2 105๐ ๐ = (3ร5ร7)(22ร5ร7) 2ร32ร5 = 22ร3ร52ร72 2ร32ร5 = 2ร5ร72 3 Numerator 2 ร 5 ร 72 does not include the factor 3 to reduce 105๐ 3 to a whole number. 3 Length of shorter portion of rope = 4 11 ร44 = 16 cm Length of longer portion of rope = 28 cm Let ๐ฅ be the length rope cut. 16โ๐ฅ 28โ๐ฅ = 2 5 5(16 โ ๐ฅ) = 2(28 โ ๐ฅ) 80 โ 5๐ฅ = 56 โ 2๐ฅ 24 = 3๐ฅ ๐ฅ = 8 4 79โ๐ฅ2 โ 1 = 0 79โ๐ฅ2 = 1 7(3โ๐ฅ)(3+๐ฅ) = 70 ๐ฅ = 3 ๐๐ ๐ฅ = โ3 5 ๐ฆ3 = ๐ ๐ฅ2 ๐ 32 โ ๐ 62 = 5 ๐ 9 โ ๐ 36 = 5 ๐ 12 = 5 ๐ = 60 ๐ฆ3 = 60 ๐ฅ2 23 = 60 ๐ฅ2 ๐ฅ2 = 7.5 ๐ฅ = ยฑ2.74
6 24 g โ 12.044 ร 1023 atoms 1 g โ 0.5.01833 ร 1023 atoms 5000g โ 0.5.01833 ร 1023 ร 5000 = 2509.167 ร 1023 = 2.51 ร 1026 7a Diagram A 7b Diagram B 7c Diagram D 8 9๐ฅ + 2.5๐ฆ = 3๐ฆ โ 2๐ฅ 11๐ฅ = 0.5๐ฆ 22๐ฅ = ๐ฆ or ๐ฅ ๐ฆ = 1 22 Hence, ๐ฅ = 22 and ๐ฆ = 1 is not a solution, as 22(22) โ (1) Or LHS = 9(22) + 2.5(1) = 200.5 RHS = 3(1) โ 2(22) = โ41 Since LHS โ RHS, ๐ฅ = 22 and ๐ฆ = 1 is not a solution. 9 3๐ฆ โ ๐๐ฅ โ 9 = 0 3๐ฆ = ๐๐ฅ + 9 ๐ฆ = ๐ 3 ๐ฅ + 3, where ๐ < โ3 3๐ฆ โ ๐๐ฅ โ 9 = 0 ๐ฆ = โ๐ฅ 3 y x
10 Let the first digit be ๐ฅ and second digit be ๐ฆ A number between 20 to 100 Let the number be 10๐ฅ + ๐ฆ Add the two digits together ๐ฅ + ๐ฆ Subtract the sum of the two digits from your original number to get a new number (10๐ฅ + ๐ฆ) โ (๐ฅ + ๐ฆ) = 9๐ฅ 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(๐๐)2 โ 2๐๐ = 3๐๐ + 3 2๐2๐2 โ 5๐๐ โ 3 = 0 (๐๐ โ 3)(2๐๐ + 1) = 0 ๐๐ = 3 or ๐๐ = โ 1 2 ๐ = 3 ๐ or ๐ = โ 1 2๐ 12 4 3 + 2 9๐ รท โ81๐164 = 4 3 + 2 9๐ รท 3๐4 = 4 3 + 2 9๐ ร 1 3๐4 = 4 3 + 2 27๐5 = 36๐5+2 27๐5 13 โ9๐ฅ2+12๐ฅโ4 3 โ 1 ๐ฅ โ ๐ฅ 2 = โ(3๐ฅโ2)2 3 โ 1 2๐ฅโ๐ฅ 2 = โ(3๐ฅโ2)2 3 โ 2 ๐ฅ = โ(3๐ฅโ2)2 3๐ฅโ2 ๐ฅ = โ๐ฅ(3๐ฅ โ 2) 14 ๐ฆ = 1 5 ๐ฅ2 โ 2๐ฅ + 7 ------------ [1] ๐ฆ = ๐ฅ ----------- [2] Sub eqn[1] into eqn[2] ๐ฅ = 1 5 ๐ฅ 2 โ 2๐ฅ + 7 1 5 ๐ฅ2 โ 3๐ฅ + 7 = 0 ๐ฅ = 3ยฑโ(โ3)2โ4(1 5)(7) 2(1 5) ๐ฅ = 2.89 or
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