GMSS 2024 4E5N EM P2 PRELIM

[Turn Over Candidate Name Class Index Number MATHEMATICS Paper 2 4052/02 4 Express 5 Normal (Academic) Candidates answer on the Question Paper. 2 hours 15 minutes Setter: Ms Nainee Ismail and Mr Kenneth Tan Thursday, 15 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 This document consists of 20 printed pages including the cover page. Geylang Methodist School (Secondary) Preliminary Examination 2024 90

GMS(S)/Math/P2/Prelim2024/4E5N 2 Mathematical Formulae Compound Interest n rP      += 1001amount Total Mensuration Curved surface area of a cone rl= 24 sphere a of area Surface r= hr 2 3 1 cone a of Volume = 3 3 4 sphere a of Volume r= CabABC sin2 1 triangleof Area = radiansin is where,length Arc r= radiansin is where,2 1 areaSector 2 r= Trigonometry C c B b A a sinsinsin == Abccba cos2222 −+= Statistics f fx  =Mean 22 deviation Standard        − = f fx f fx

GMS(S)/Math/P2/Prelim2024/4E5N 3 [Turn Over 1 A model of a thermos flask consists of a solid hemisphere attached to a solid cylinder as shown. The height of the cylinder is 14 cm. The curved surface area of the cylinder is twice the curved surface area of the hemisphere. (a) Calculate the radius of the base of the cylinder. Answer ……………………………………… cm [2] (b) Calculate the height of the model of the thermos flask. Answer ……………………………………… cm [1] (c) Find the ratio volume of hemisphere : volume of cylinder. Answer …………………… : …………………… [2] 14 cm

GMS(S)/Math/P2/Prelim2024/4E5N 4 2 (a) (i) Solve the inequalities 9 5 4 6 x−  −  . Answer …………………………………………… [2] (ii) Write down all the integers that satisfy 9 5 4 6 x−  −  . Answer …………………………………………… [1] (b) 2 2 5 xy xw −= + (i) Evaluate the value of y when 2.41x= and 1.908w= . Write your answer correct to three decimal places. Answer y= ……………………………………… [2] (ii) Rearrange the formula to make x the subject. Answer x= ……………………………………… [4]

GMS(S)/Math/P2/Prelim2024/4E5N 5 [Turn Over 2 (c) Solve 64 113 x xx −−=−− . Answer x= ……………… or ……………… [4]

GMS(S)/Math/P2/Prelim2024/4E5N 6 3 A, B, C and D are four points on horizontal ground. AB = 70 m, BC = 55 m, CA = 60 m, AD = 50 m, angle ACD = 50 and angle ADC is acute. (a) Calculate angle BAC. 2 2 255 60 70 2(60)(70)cos BAC= + −  ( )( ) 2 2 260 70 55cos 2 60 70BAC +−= [M2] Angle 49.324 49.3BAC= =  [A1] Answer Angle BAC = ……………………………… [3] (b) Calculate angle DAC. sin sin 50 60 50 ADC  = [M1] sin 50sin 60 50 0.91925 ADC  =  = Angle 66.817ADC= [M1] Angle 180 66.817 50 63.183 DAC = − −  = Angle 63.2DAC= [A1] Answer Angle DAC = ……………………………… [3] N A B D C 50 m 70 m 60 m 55 m 50o

GMS(S)/Math/P2/Prelim2024/4E5N 7 [Turn Over 3 (c) Calculate the bearing of A from D. N1 : North is drawn at D 1 63.2 49.3 112.5 ADN = +  = [M1] (Alternate Angles, AN//DN1) Bearing of A from D = 360o – 112.5o = 247.5 o [A1] (Sum of angles at a point) Answer …………………………………………… [2] (d) A man walks northwards from point A. A vertical pole is placed at point D. The greatest angle of elevation of the pole when viewed along this path is 7 . Calculate the height of the pole. Angle 180 63.2 49.30 67.5 XAD= − −  = sin 67.5 50 XD= 50sin 67.5 46.193XD= = m [M1] tan 7 46.193 h= 46.193 tan 7 5.67h=  = m [A1] Answer ………………………………………… m [2] A D X 50 m 67.5o X D h 7o 46.193 m

GMS(S)/Math/P2/Prelim2024/4E5N 8 4 Mr Tan planned to travel to Kuala Lumpur for a family trip. Before the trip, he exchanged Singapore Dollars (S$) for Malaysian Ringgit (RM). The exchange rate is S$ 1 = RM x. (a) Write down an expression, in terms of x, for the Singapore Dollars he exchanged if he received RM 2000. RM S$1 1RM 1 S$ x x = =   2000RM 2000 S$ x =  [A1] Answer S$ ………………………………………… [1] (b) After the trip, Mr Tan found out that he still had RM 300. He exchanged the Malaysian Ringgit back to Singapore Dollars. The exchange rate is now S$ 1 = RM (x + 0.05). Write down an expression, in terms of x, for the Singapore Dollars he exchanged for RM 300. ( )RM 0.05 S$1 1RM 1 S$ 0.05 xx x += =  +  300RM 300 S$ 0.05x =  + [A1] Answer S$ ………………………………………… [1] (c) Mr Tan used up S$ 485 for the trip. Form an equation in x and show that it reduces to 21940 6703 400 0xx− − = . Answer RM2000 RM300 S$485−= 2,000 300 4850.05xx −= + [M1] 2,000( 0.05) 300 485( 0.05) xx xx +− =+ [M1] 2000 100 300 485 ( 0.05)x x x x+ − = + 20 485 24.25 2000 100 300x x x x= + − − + 2 67030 485 100 4xx= − − 21940 6703 400 0xx− − = (Shown) [A1] [3]

GMS(S)/Math/P2/Prelim2024/4E5N 9 [Turn Over 4 (d) Solve the equation 21940 6703 400 0xx− − = , giving your solutions correct to two decimal places. ( ) 2 ( 6703) 6703 4(1940)( 400) 2(1940)x − −  − − −= [M1] 6703 48034209 3880 = 3.51383289 0.05867825064=− or [M1] 3.51 0.06x=− or (2 dp) [A1] Answer x= ……………… or ……………… [3] (e) From your solution to part (d), state the exchange rate in the form is S$ 1 = RM x, to two decimal places, and hence, find the total amount of Singapore Dollars (S$) he initially converted to Malaysian Ringgit (RM). S$1 = RM 3.51 [A1] 2000RM 2000 S$ 2000S$ 3.51 S$569.8005698 x =  =  = Amount of S$ 2,000 $569.803.51== (2dp) [A1] Answer S$ 1 = RM ……………………………… S$ ………………………………………… [2]

GMS(S)/Math/P2/Prelim2024/4E5N 10 5 A shop sells two flavours of ice-cream, Raisins (R) and Chunkies (C) in three different sizes, small (S), medium (M) and large (L). The sales on two successive days are given in the table below. Saturday Sunday Size small medium large small medium large Number of cups sold (Raisins) 12 17 8 14 12 10 Number of cups sold (Chunkies) 18 15 11 13 21 16 The information for Saturday’s sales can be represented by the matrix . (a) Represent the information for Sunday’s sales in a 23 matrix N. Answer =N S M L R C    [1] (b) The cost