CCHM 2024 P2 Worked Solution

2024 Preliminary Exam/CCHMS/Secondary 4/Mathematics/4052/02 Name: Class: Class Register Number: PRELIMINARY EXAMINATION 2024 SECONDARY 4 MATHEMATICS 4052/02 Paper 2 Tuesday 20 August 2024 Candidates answer on the Question Paper 2 hours 15 minutes MARKS SCHEME This document consists of 23 printed pages and 1 blank page.

2 2024 Preliminary Exam/CCHMS/Secondary 4/Mathematics/4052/02 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 radian Sector area = , where is in radian Trigonometry Statistics Mean = f fx   Standard deviation = 22        −  f fx f fx 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 −+=

3 2024 Preliminary Exam/CCHMS/Secondary 4/Mathematics/4052/02 [Turn over TURN OVER FOR QUESTION 1

4 2024 Preliminary Exam/CCHMS/Secondary 4/Mathematics/4052/02 1 (a) The table shows the total electricity consumption in Singapore over three years. (i) In 2021, the electricity consumed by households t ook up 15.5% of the total electricity consumption. Calculate the amount of electricity consumed by households in 2021, correct to two significant figures. Electricity consumed by household =15.5% 53483 8289.865 8300 GWh (2 s.f.)  = = Answer ………………………… GWh [1] (ii) Calculate the percentage increase in the total electricity consumption from 2020 to 2022. 54884 50779Percentage increase = 100%50779 8.08% −  = Answer ……………………………% [2] (iii) Express the 2020 electricity consumption in kilowatt hours (kWh), leaving your answer in standard form, correct to two significant figures. 9 3 10 10 50779 10 10 5.0779 10 5.1 10 kWh (2 s.f.)   =  = Answer ………………………… kWh [1] Year 2020 2021 2022 Total Electricity Consumption in Gigawatt Hours (GWh) 50 779 53 483 54 884

5 2024 Preliminary Exam/CCHMS/Secondary 4/Mathematics/4052/02 [Turn over (b) A microprocessor is in the shape of a cube where the sides are 5 mm in length. (i) Find the maximum number of microprocessor s that can be placed into a container with dimensions 10 cm by 2 cm by 8 cm. 10 200.5 2 40.5 8 160.5 20 4 16 1280 = = =    = Answer ……………………………… [2] (ii) A model of the microprocessor was made to a scale of 10 : 1. Given that the surface area of microprocessor is 150 mm 2, find the surface area of the model microprocessor in square centimetres. Model : Actual (Linear Scale) 10 :1 (Area Scale) 100 :1 The surface area of the model is 100 times of the actual object. 2 2 2 Surface Area of Model 100 150 15000 mm 15000 cm100 150 cm = = = = Answer …………………………… cm2 [2]

6 2024 Preliminary Exam/CCHMS/Secondary 4/Mathematics/4052/02 2 (a) Solve ( )2 7 3 1 3xx− = − . ( )2 7 3 1 3 2 7 3 9 11 10 10 11 xx xx x x − = − − = − = = Answer x = …………………………… [2] (b) Solve the inequalities 7 1 13x− and ( )1 222 x x+ − − . 7 1 13 7 14 2 x x x −   and ( ) ( ) ( ) 1 222 1 4 2 1 4 8 57 1.4 x x xx xx x x + − − + − − + − +   1.4 2 x   Answer ………………………………… [3] (c) Rearrange the formula 2 2 5 7 xy x += to make x the subject. ( ) 2 2 22 22 2 2 5 7 75 75 7 1 5 5 71 5 71 xy x x y x x y x xy x y x y += =+ −= −= = − = − Answer x = …………………………… [3]

7 2024 Preliminary Exam/CCHMS/Secondary 4/Mathematics/4052/02 [Turn over (d) Solve the equation 21 31 2 3 x xx−=−− . Give your solutions correct to two decimal places. ( ) ( )( ) ( ) 22 22 2 2 21 31 2 3 2 2 3 3 1 2 3 1 4 6 3 2 3 2 3 1 4 6 6 9 6 9 1 10 20 8 0 5 10 4 0 x xx x x x x x x x x x x x x x x x x x xx xx −=−− − − − − = − − − − − + = − − − + + − = − − + = − + = 2( 10) ( 10) 4(5)(4) 2(5) 1.4472... or 0.552786... 1.45 or 0.55 (2 s.f.) x x x − −  − −= = = Answer x = ……………. or …………… [5]

8 2024 Preliminary Exam/CCHMS/Secondary 4/Mathematics/4052/02 3 The diagram show s two open containers , a right cylinder and a n inverted frustum. The right cylinder has a base radius of 10 cm and a height of 20 cm. The frustum is formed by cutting a smaller cone off the bottom of a larger cone. The smaller cone that was cut off has a height of 4 cm. The larger cone has a base radius of 15 cm and a height of 12 cm. Three identical spherical marbles, each of radius 3 cm, are placed into the cylindrical container and water is poured in to a depth of h cm. (a) Find the volume of each marble, leaving your answer in terms of π . ( ) 3 3 4Volume of each marble = 33 36 cm  = Answer …………………………… cm3 [1] (b) Find the volume of the frustum, leaving your answer in terms of π . ( ) 2 3 3 11 22 3 1 3 1 3 1Volume of big cone = 15 (12)3 900 cm 4 900 12 100 cm3 25Volume of frustum = 900 6 2600 cm3 Vl Vl V V       = =  = = − = Answer …………………………… cm3 [3] h cm 20 cm 10 cm 15 cm 12 cm 4 cm

9 2024 Preliminary Exam/CCHMS/Secondary 4/Mathematics/4052/02 [Turn over (c) All the water from the cylinder , without the marbles, is then poured into the empty frustum, filling it completely without any overflow. Find the value of h, correct to two decimal places. ( ) ( ) ( ) 3 2 2 2600Volume of water and marbles = 3 363 2924 cm3 292410 3 2924 3 10 9.75 h h h      + = = = = Answer h = …………………………… [3] (d) The exterior surface area of each container is painted. Find the total area painted. ( )( ) ( )( ) 2 2 2 2 2 11 22 2 1 2 1 1 1 Using Pythagoras' Theorem, 12 15 369 cm Surface area of cone 15 369 905.22036.. cm 4 1215 369 100.58004.. cm 4 15 12 5 cm Surface Area of frustum 905.22036.. 100.58 004. l l Al Al A A l l   =+ = = = =  = = = = =−( ) ( ) ( ) ( ) 2 2 2 .5 804.6403.. 78.53981 883.18013... Total Surface area = 883.18013... 2 10 20 10 883.18013... 1256.63706.. 314.1592.. 2453.9764.. 2450 cm (3s.f)   + =+ = +    + = + + = = Answer …………………………… cm2 [6]

10 2024 Preliminary Exam/CCHMS/Secondary 4/Mathematics/4052/02 4 (a) 12 students from class A took a Mathematics test. The table below shows the test marks of the students. However, two of the students’ marks are covered with ink. (i) Given that the box-and-whisker plot below shows the distribution of the results, explain why the information may not be sufficient to find the missing marks of the two students. Answer The box-and-whisker plot is not able to provide individual data point. It can only provide the minimum, maximum mark, 25th, 50th and 75th percentile and these data are insufficient to calculate the missing marks. (ii) Given further that the modal mark is 63, find the two missing marks. Since mode = 63, one of the missing marks = 63 40 47 52 52 55 61 63 63 63 _ 84 _ 95 Since upper quartile = 72, 63 722 63 144 81 x x x + = += = Answer …………… and …………… [2] 40 95 84 47 63 52 63 55 52 61 40 50 60 70 80 90 100 Mathematics Test Marks