Xinmin 2024 4NA Math Prelim (P2) (MS)

This document consists of 20 printed pages and 2 blank pages. [Turn over XINMIN SECONDARY SCHOOL SEKOLAH MENENGAH XINMIN Preliminary Examination 2024 CANDIDATE NAME MARK SCHEME CLASS INDEX NUMBER MATHEMATICS (SYLLABUS A) Paper 2 Secondary 4 Normal (Academic) 4045/02 1 August 2024 2 hours Candidates answer on the Question Paper READ THESE INSTRUCTIONS FIRST Write your name, index number and class in the spaces at the top of this page. 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. Section A Answer all questions. Section B Answer one question. 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 70. 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. Errors Qn No. Errors Qn No. Accuracy Simplification Brackets Units Geometry Marks Awarded Presentation Marks Penalised Total Marks for PRWC For Examiner’s Use 70 Parent’s/Guardian’s Signature:

2 Mathematical Formulae Compound Interest Total amount = 1 100 n rP+ Mensuration Curved surface area of a cone = rl Surface area of a sphere = 24 r Volume of a cone = 21 3 rh Volume of a sphere = 34 3 r Area of triangle ABC = 1 sin2 ab C Arc length = r , where  is in radians Sector area = 21 2 r  , where  is in radians Trigonometry sin sin sin a b c ABC== 2 2 2 2 cosa b c bc A= + − Statistics Mean = fx f   Standard deviation = 22fx fx ff −  

3 [Turn over Section A (62 marks) Answer all the questions in this section. 1 The stem-and-leaf diagram shows the temperature, in °C, over 15 days in June in a city. 1 7 9 9 2 0 3 3 3 4 4 5 7 8 8 3 0 2 Key: 1 | 7 represents 17°C (a) Find (i) the range of the temperatures, (range) 32 – 17 = 15°C [B1] Answer ……………………… °C [1] (ii) the mean temperature, correct to 1 decimal place, (mean) 17 19 ... 32 15 + + + [M1 or 362 15 ] = 24.133°C = 24.1°C (1dp) [A1] *Give [A2] If students just write 24.1°C Answer ……………………… °C [2] (iii) the standard deviation of the temperatures. (sd) 4.1611 = 4.16°C (3sf) [B1] Answer ……………………… °C [1] (b) Another day in June has a temperature of 22°C. Given that this temperature value is to be included in the above diagram, would the mean temperature increase or decrease? Without the use of calculations, explain your answer. Answer …………………………………………………………………………. …………………………………………………………………………………... …………………………………………………………………………………... [1] The new mean temperature would decrease because 22°C is lower than the mean temperature of the initial set of 15 days. Therefore adding this will cause the new mean temperature to be lower. [B1]

4 2 (a) Written as the product of its prime factors, 238316 2 3 7 11=    . (i) Express 840 as the product of its prime factors. 840 = 32 3 5 7   [B1] Answer ………………………… [1] (ii) Find the highest common factor of 8316 and 840. 3840 2 3 5 7=    238316 2 3 7 11=    HCF (8316, 840) = 22 3 7 = 84 [B1] Answer ………………………… [1] (iii) Find the smallest integer value of n such that 840n is a multiple of 8316. n = 32 × 11 = 99 [B1] Answer ………………………… [1] (b) Rearrange 7 2 pq p= − to make p the subject. 7 2 pq p= − q(p – 2) = 7p [M1] pq – 2q = 7p pq – 7p = 2q p(q – 7) = 2q [M1] 2 7 qp q= − [A1] Answer p = ……………………… [3]

5 [Turn over 3 The diagram shows a rectangular garden measuring 13 m by 10 m. The garden is surrounded by a path of uniform width of x m, shown shaded in the diagram. The total area of the path is 84 m2. (a) Write down an equation in x and show that it simplifies to 2x2 + 23x – 42 = 0. Answer (10 + 2x)(13 + 2x) – (10 × 13) = 84 [M1] 130 + 20x + 26x + 4x2 – 130 = 84 4x2 + 46x – 84 = 0 2x2 + 23x – 42 = 0 (shown) Alt Mtd 2x(13 + 2x) + 2x(10) = 84 or x(10 + 2x) + x(13) = 42 [M1] 26x + 4x2 + 20x = 84 4x2 + 46x – 84 = 0 2x2 + 23x – 42 = 0 (shown) [2] (b) Solve the equation 2x2 + 23x – 42 = 0 and find the width of the path. 2x2 + 23x – 42 = 0 x = ( ) ( )( ) ( ) 2 23 23 4 2 42 22 −  − − [M1] x = 1.6027 or x = –13.102 = 1.60 (3sf) = –13.1 ∴ The width of the path is 1.60 m [A1*- with both 1.60 and –13.1 written] *Do not penalise if students rejected –13.1 due to context. Answer ……………………… m [2] Garden 13 m 10 m x m x m [A1] [A1]

6 4 Min cycled from her home to a park. The distance-time graph shows her journey. (a) Find the distance, in km, between Min’s home and the park. 30 km [B1] Answer …………………… km [1] (b) Describe her motion between 0800 and 0915. (speed) 30 114  = 24 km/h Answer …………………………………………………………………………. …………………………………………………………………………………... [1] (c) Min took a rest in the park for 15 minutes before she cycled back home at a constant speed of 20 km/h. On the same grid above, complete the graph to show Min’s journey. (time back home) 30 20 = 112 h [2] Time | | | | | | | | 0800 0830 0900 0930 1000 1030 1100 1130 1200 Distance (km) 30 – 20 – 10 – 0 Park Home Min cycled to the park at a constant speed of 24 km/h. [B1] [B1- horizontal line from (0915, 30) to (0930, 30) ] [B1- line from (0930, 30) to (1100, 0) ]

7 [Turn over 5 (a) It is given that 6x = 7y. The quantities y and z are in the ratio 9 : 2. Write the ratio x : y : z in its simplest form. 6x = 7y 7 6 x y = x : y 7 : 6 [M1- accept equivalent ratio] x : y : z 7 : 6 9 : 2 21 : 18 : 4 [A1] Answer ……… : ……… : ……… [2] (b) Simplify 3 6 2 a −   , leaving your answer in positive index form. 3 6 2 a −   = 36 2 a  [M1] = 18 8 a [A1] Answer ………………………… [2] (c) ( ) ( )1.4 10 2.5 10 10x y n k   =  , where 1 10k . (i) Find the value of k. ( ) ( )1.4 10 2.5 10xy   = 1.4 10 2.5 10 x y   = 0.56 10 xy− = 1 5.6 10 10 xy−− = 15.6 10 xy−− k = 5.6 [B1] Answer k = ……………………… [1] (ii) Write an expression for n in terms of x and y. n = x – y – 1 [B1] Answer n = ……………………… [1] Alt Mtd 3 6 2 a −   = 3 6 1 2 a  [M1] = 18 1 8 a  = 18 8 a [A1] Alt Mtd 3 6 2 a −   = ( ) 3 18 2 a − − [M1] = 18 1 8 1 a   = 18 8 a [A1]

8 6 (a) Nadia invested some money for five years at 1.2% simple interest per year. At the end of five years, it was worth $22260. How much did she invest? Let the amount invested be $P (simple interest) 1.2 5 100 P [M1] = 0.06P (total) P + 0.06P = 22260 [M1- or 22260 – 0.06P = P ] 1.06