2024 HYSS S4-5 G3 MA Prelim Paper 1 MS

[Turn Over HUA YI SECONDARY SCHOOL PRELIMINARY EXAMINATION 2024 4-G3 / 5-G2 NAME CLASS INDEX NUMBER MATHEMATICS 4052/01 PAPER 1(XX) 13 August 2024 2 hour 15 minutes Candidates answer on the Question Paper. MARKING SCHEME For Examiner’s Use 90 This document consists of 22 printed pages. © HYSS 2024 No part of this document may be reproduced in any form or transmitted in any form or by any means without the prior permission of Hua Yi Secondary School. Setter: Ms Jasmine Tan

2 2024_4-G3 / 5-G2_PRELIMINARY EXAMINATION_MATHEMATICS_PAPER 1 [Turn Over 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 = Cabsin2 1 Arc length = r , where  is in radians Sector area = 2 2 1 r , where  is in radians Trigonometry C c B b A a sinsinsin == Abccba cos2222 −+= Statistics Mean =   f fx Standard deviation = 22       −     f fx f fx

3 2024_4-G3 / 5-G2_PRELIMINARY EXAMINATION_MATHEMATICS_PAPER 1 [Turn Over 1 (a) Calculate ( ) ( ) ( ) 211.8 11 7 16 40 2 16 − − − − −   −  . 1.750958 1.75 (3 s.f.) ----- B1− =− Answer ………………………. [1] (b) There are 800 people in an auditorium, correct to the nearest hundred. State the minimum number of people that could be in the auditorium at this time. 750 people ----- B1 Answer …………….… people [1] ________________________________________________________________________________ 2 (a) Express 1400 as the product of its prime factors. 321400 2 5 7 ----- B1=   Answer ………………………. [1] (b) Write down the smallest positive integer k such that 1400k is a perfect cube. 3 2 3 2 21400 2 5 7 2 5 7 (5 7 ) 245 ----- B1 kk k =    =     = Answer k = ...…………………. [1] (c) n is a number between 300 and 400. The highest common factor of n and 1400 is 35. Find the largest possible value for n. n = ?2 5 7 ?   1400 = 322 5 7 HCF = 5 7 35= By guess-and-check, 5 7 11 385 2nB=   = −−− [Condition: largest possible value between 300 – 400] Answer n = .…………………. [2]

4 2024_4-G3 / 5-G2_PRELIMINARY EXAMINATION_MATHEMATICS_PAPER 1 [Turn Over 3 (a) Simplify 0 2 79y x x − . 2 0 2 7 7 9 9 1 1 9 19 xy x x x M x A −  =   −−− = −−− Answer ………………………. [2] (b) Simplify ( ) 5 12 481a . ( ) ( ) 55 5 1212 4 44 15 81 81 243 1 aa aB  = = −−− Answer ………………………. [1] ________________________________________________________________________________ 4 (a) Express as a single fraction in its simplest form 74 2 18 3 5 81 bb c  . 7 4 7 2 2 4 3 2 18 3 18 81 5 81 5 3 486 486 M1 for , 1 for full answer55 b b b c c b b Ac  =  = −−−  Answer ………………………. [2] (b) Use the laws of indices to show that 426 100 116 36 +  can be expressed as a single power of six. Answer [2]   4 2 4 4 4 4 7 6 100 116 36 6 100 116 6 6 100 116 1 6 216 6 1 ( ) M A shown  +  =  +  = + −−− = = −−−

5 2024_4-G3 / 5-G2_PRELIMINARY EXAMINATION_MATHEMATICS_PAPER 1 [Turn Over 5 In a greenhouse, the estimated number of flowering plants increased from 4100 in January 2024 to 4980 in June 2024. The number increased by c% every month. Find the value of c. ( ) 5 5 5 4980 1 100 4980 4100 1 0.01 1 249 1 0.01205 249 1205 10.01 3.965 3.97 (3s.f.) 1 n rP cM c cM A =+  = + −−− =+ − = −−− = = −−− Answer c = .…………………. [3] ________________________________________________________________________________ 6 Kyle runs a tennis club. 54 of the members are adults and 31 are children. His aim is that at least 60% the members should be children. Form an inequality to find the smallest number of children that Kyle would still need to recruit achieve his aim. Let c be the no. of children needed to join the club. ( ) 31 60% 1 (forms inequality)54 31 31 0.6 85 31 51 0.6 0.6 20 1 (isolate unknown) 0.4 20 50 Smallest number = 50 1 c Mc cc cc c c M c c A +  −−−++ +  + +  + −  −−−    −−− Answer ……….…… children [3]

6 2024_4-G3 / 5-G2_PRELIMINARY EXAMINATION_MATHEMATICS_PAPER 1 [Turn Over 7 A car travels at an average speed of 74.5 km/h for 2.25 hours. (a) Convert 75 km/h to m/s. 75 000 5 20 20.8 m/s 1 (Accept either 1 form)3600 6 B= = −−− Answer ………………..… m/s [1] (b) By rounding the numbers correct to 1 significant figure, find an estimate of the distance travelled by the car. Show your working clearly. Distance 74.5 2.25 70 2 1 (must show rounding to 1 s.f. each) 140 km 1 M A =  =  −−− = −−− Answer ………………..… km [2] (c) Without doing any calculation, explain why the actual distance travelled by the car is greater than the answer to (b). Answer [1] Both the average speed and time is greater than the 1 significant figure rounded values of speed and time respectively. Thus, the actual distance travelled is greater than the answer in part (b). ----- B1 ________________________________________________________________________ 8 Isha has written down five numbers. The mean of these numbers is 13.2, the median is 12 and the mode is 7. The largest number is three times the smallest number. Find the five numbers in ascending order. 7 7 12 y 21 Smallest Median largest Mean 13.2 13.25 66 , so 19. Sum Sum y = = == Answer …………………..…………………. [2] Any 3 consecutive values correct → B1 ALL 5 values correct → Full mark, B2

7 2024_4-G3 / 5-G2_PRELIMINARY EXAMINATION_MATHEMATICS_PAPER 1 [Turn Over 9 Factorise completely (a) 442 32ps− , ( ) ( ) ( )( ) ( )( )( ) 224 4 2 2 2 2 2 2 2 2 2 2 2 2 2 32 2 4 1 [factorise by HCF] 2 4 4 1 [factorise by ] 2 4 2 2 1 [factorise by again] p s p s M p s p s M a b p s p s p s A a b − = − −−− = + − −−− − = + + − −−− − Answer ………………………. [3] (b) 12 9 6 8cd cx xy dy− + − . ( ) ( ) ( ) ( ) ( )( ) 3 4 3 2 3 4 3 4 3 2 4 3 1 [Factorise out sign] 3 2 4 3 1 c d x y x d c d x y d x M ve c y d x A − + − = − − − −−− − = − − −−− Answer ………………………. [2] ________________________________________________________________________________ 10 (a) Express 298 xx−+ in the form 2()a x b++ . Find the value of a and of b. ( ) ( ) 2229 8 4 9 16 7 4x x x x− + = − + − =− + − 7 1, 4 1a B b B=− −−− =− −−− Answer a = .…………………. [1] b = .…………………. [1] (b) Explain why when 4x= , the expression 298 xx−+ has its minimum value. Answer [1] For any perfect square , the smallest value is always equal to zero or greater than zero . When x = 4, ( ) 2 40x−= thus, ( ) 2 4 7 7 0x− − =−  and the coefficient of 2x is positive, implying that the quadratic expression will have a minimum value at x = 4. ------ B1

8 2024_4-G3 / 5-G2_PRELIMINARY EXAMINATION_MATHEMATICS_PAPER 1 [Turn Over 11 Solve the equation 2052 1x x+= + . ( )( ) 2 2 2052 1 5 5 2 2 20 2 7 15 0 1 2 3 5 0 1 1.5 5 1 [BOTH ans correct] x x xxx x x M x x M x or x A += + + + + = + − = −−− − + = −−− = =− −−− Answer x = ……… or ………. [3] ________________________________________________________________________________ 12 The points ( )4,20 and ( )10, 4− satisfy the curve given by the equation 2 4y ax bx= + − . Use an algebraic method to determine the values of a and b. 220 (4) (4) 4 24 16 4 6 4 (1) ab ab a b Eqn = + − =+ = + −−− 24 (10) (10) 4 4 100 10 4 100 10 10 0 (2) 1 [any 1 eqn correct] ab ab ab a b Eqn M − = + − − = + − =− + = −−− −−− *Solve by either elimination or substitution ----- Award M1 1, 10 2a b A=− = −−− Answer a = ……… , b = …..…. [4]

9 2024_4-G3 / 5-G2_PRELIMINARY EXAMINATION_MATHEMATICS_PAPER 1 [Turn Over 13 {integers :1 16}xx =   {1, 2, 4, 8,16}V = {1, 4, 7, 9,10,14}W = Some of the information is shown on the Venn diagram. (a