CCHM 4048 01 AS

[Turn over Name: Class: Class Register Number: PRELIMINARY EXAMINATION 2022 SECONDARY 4 MATHEMATICS 4048/01 Paper 1 Wednesday 14 September 2022 Candidates answer on the Question Paper. 2 hours This document consists of 17 printed pages and 1 blank page. _________________________ Parent’s Signature Solutions

2 2022 Preliminary Exam/CCHMS/Secondary 4/Mathematics/4048/01 Mathematical Formulae Compound interest Total amount = Mensuration Curved surface area of a cone = Surface area of a sphere = Volume of a cone = Volume of a sphere = Area of triangle ABC = Arc length = , where is in radians Sector area = , where is in radians Trigonometry Statistics Mean = Standard deviation = nrP      + 1001 lr  2 4 r hr 3 1 2 3 3 4 r Cba sin 2 1 r  2 2 1 r  C c B b A a sin sin sin == Acbcba cos 2 222 −+=   f xf 22      −     f xf f xf

3 2022 Preliminary Exam/CCHMS/Secondary 4/Mathematics/4048/01 [Turn over Answer all the questions. 1 Evaluate 3 9.45 36.03 26.9  − , correct to 3 significant figures. 3 9.45 36.03 18.931... 26.9 18.9 (3sf)  =− − =− Answer ................................... [1] 2 Simplify 22 16 12 32 18 ab ab − − . 2 2 2 2 16 12 4(4 3 ) 32 18 2(16 9 ) a b a b a b a b −− =−− 4(4 3 ) 2(4 3 )(4 3 ) ab a b a b −= +− 2 43ab= + Answer ................................... [2] 3 Ken drives at an average speed of 85 km/h on the expressway for the first stage of his journey. After exiting the expressway, he continues to drive at an average speed of 60 km/h on the city roads for the second stage of his journey. If the average speed of his entire journey is 75 km/h , find the ratio of the time taken for the first stage of his journey to the time taken for the second stage of his journey. Let time taken for first and second part of journey be x and y hours respectively. total distance travelled = (85 60 ) km, to tal time = ( ) hoursx y x y++ 85 60 75xy xy + =+ 85 60 75( )x y x y+ = + 85 60 75 75x y x y+ = + 10 15xy= 3 2 x y = Ratio = 3: 2 Answer ................................... [2] 4 Factorise completely 22 3 12 8x x y xy+ − − . 22 3 12 8x x y xy+ − − (2 3) 4 (3 2 )x x y x= + − + ( 4 )(2 3)x y x= − + Answer ................................... [2]

4 2022 Preliminary Exam/CCHMS/Secondary 4/Mathematics/4048/01 5 The mass of an electron is 289.11 10 − g. (a) Write in standard form, the mass of an electron in kg. 289.11 10 − g = 289.11 10 1000 kg− = 319.11 10 kg− Answer ................................ kg [1] The mass of a hydrogen atom is 271.66 10 − kg. (b) Find the number of electrons that will have the same mass as 1 hydrogen atom, giving your answer to the nearest whole number. No of electrons = 27 31 1.66 10 9.11 10 − −   = 1822.1734 = 1822 (nearest whole number) Answer ................................... [1] 6 Teacher: The length of a certain rectangle is decreased by 20%, whereas its breadth is increased by 20%. Student: The area of the rectangle will remain the same as the percentage decrease of the length is equal to the percentage increase of the breadth. Therefore, there is a 0% change in the area. Is the student’s statement correct? Support your answer with mathematical calculations. Let the original length and breadth of the rectangle be x and y respectively. New length = 0.8x, New breadth = 1.2y Original Area of rectangle = xy New Area of rectangle = (0.8x)(1.2y) = 0.96 xy ……………………………………………………………………………………………….. ……………………………………………………………………………………………..[2] 7 Find the range of values of x which satisfy the inequalities 2 2 5 4 9 15 6 x x x− + + 2 2 5 2 5 4 and 9 15 15 6 x x x x− + + + 15( 2) 9(2 5) 6(2 5) 15( 4)x x x x−  + +  + 15 30 18 45 12 30 15 60x x x x−  + +  + 3 75 3 30xx−  −  25 10xx− − 10x − Answer ................................... [3] The student’s statement is incorrect. There is a change in the area and the new area of rectangle is 0.96 of the original area.

5 2022 Preliminary Exam/CCHMS/Secondary 4/Mathematics/4048/01 [Turn over 8 Solve the equation ( 1) 6xx += . ( 1) 6xx += 2 6xx+= 2 60xx+ − = ( 3)( 2) 0xx+ − = 3 or 2xx=− = Answer x = ............ or .............. [2] 9 Given that 23 2 ayx by −= + , (a) calculate x when 5, 2 and 3a b y= =− = , 2(5) 3(3) 2 2(3)x −= −+ 1 2= Answer x = .............................. [1] (b) express y in terms of a, b and x. 23 2 ayx by −= + 2 23 2 ayx by −= + 2( 2 ) 2 3x b y a y+ = − 22 2 2 3bx x y a y+ = − 223 2 2y x y a bx+ = − 22(3 2 ) 2y x a bx+ = − 2 2 2 32 a bxy x −= + Answer y = ................................. [3]

6 2022 Preliminary Exam/CCHMS/Secondary 4/Mathematics/4048/01 10 A company manufactures two sizes of the same brand of shampoo. The prices of the two sizes are indicated in the table below. Shampoo Size (ml) Price ($) 330 7.95 620 14.35 Which shampoo size is a better value for money? Explain your answer clearly. Answer Price/ml for 330ml shampoo = $7.95 330 = $0.02409… Price/ml for 620ml shampoo = $14.35 620 = $0.02314… ……………………………………………………………………………………………….. …………………………………………………………………………………………….. [2] 11 The numbers of hours spent daily online by 10 students are recorded in ascending order. 3, 4, 4, 4, 5, 6, 7, 8, 9, a The data is presented as a box and whisker plot shown below. Find the values of a, b, c and d. a = 10, b = 3 c = median = 56 2 + = 5.5 d = Upper quartile = 8 Answer a =......., b =......., c =......., d =........ [2] b 4 c d 10 Since the price/ml for the 620ml shampoo is lower than the 330ml shampoo ($0.02314… <$0.02409…), the 620 ml shampoo is a better value for money. [Can also find amount of shampoo per dollar (41.509… vs 43.205…), etc for comparison] 10 3 5.5 8

7 2022 Preliminary Exam/CCHMS/Secondary 4/Mathematics/4048/01