GMSS_2024_4E5N_EM_P1_PRELIM ANS

[Turn Over Candidate Name WORKED SOLUTIONS & SUGGESTED ANSWER SCHEME Class Index Number MATHEMATICS Paper 1 4052/01 4 Express 5 Normal (Academic) Candidates answer on the Question Paper. 2 hours 15 minutes Setter: Ms Nainee Ismail Monday, 5 August 2024 READ THESE INSTRUCTIONS FIRST Write your class, index number and name on all the work you hand in. Write in dark blue or black pen. You may use an HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. Answer all the questions. The number of marks is given in brackets [ ] at the end of each question or part question. If working is needed for any question it must be shown with the answer. Omission of essential working will result in loss of marks. The total of the marks for this paper is 90. The use of an approved scientific calculator is expected, where appropriate. If the degree of accuracy is not specified in the question and if the answer is not exact, give the answer to three significant figures. Give answers in degrees to one decimal place. For , use either your calculator value or 3.142. For Examiner’s Use 90 This document consists of 20 printed pages including the cover page. Geylang Methodist School (Secondary) Preliminary Examination 2024

GMS(S)/Math/P1/Prelim2024/4E5N/Answer Scheme 2 Mathematical Formulae Compound Interest Total amount = 1 100 n rP + Mensuration Curved surface area of a cone = rl Surface area of a sphere = 4r 2 Volume of a cone = 1 3 r 2h Volume of a sphere = 4 3 r 3 Area of triangle ABC = 1 2 ab sin C Arc length = r, where  is in radians Sector area = 1 2 r 2, where  is in radians Trigonometry sin sin sin a b c A B C== a 2 = b 2 + c 2 – 2bc cos A Statistics Mean = fx f   Standard deviation = 22fx fx ff  −  

GMS(S)/Math/P1/Prelim2024/4E5N/Answer Scheme 3 [Turn Over Answer all the questions. 1 Expand and simplify (4x – y) (3x + 4y). Answer ……………………………………… [2] 2 (a) Find the lowest common multiple (LCM) of 108 and 140. Answer ……………………………………… [1] (b) Find the highest common factor (HCF) of 108 and 140. Answer ……………………………………… [1] 3 Solve 142 3 x−= . Answer x = …………………………………… [1] (4x – y) (3x + 4y) = 12x2 – 3xy + 16xy – 4y2 = 12x2 + 13xy – 4y2  4x – y 3x 12x2 – 3xy 4y 16xy – 4y2 12x2 + 13xy – 4y2 142 3 142 3 23 x x x −= −= =  x = 6 6 By Prime Factorisation, 2 108 140 2 54 70 3 27 35 3 9 35 3 3 35 5 1 7 7 1 1 By Prime Factorisation, 2 108 140 2 54 70 27 35 LCM = 22 × 33 × 5 × 7 = 4 × 27 × 5 × 7 = 3780 HCF = 22 HCF = 4 3780 4

GMS(S)/Math/P1/Prelim2024/4E5N/Answer Scheme 4 4